Chapter 16 Practice

A Collective Work

Chapter 16: Equilibria of Other Reaction Classes

Chapter 16 Practice

Complete the changes in concentrations for each of the following reactions:
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2. $Math input error$
3. $Math input error$
4. $Math input error$
5. $Math input error$
How do the concentrations of $Pb^{2 +}$ and $S_{2}^{-}$ change when $K_{2} S$ is added to a saturated solution of $PbS$ ?
What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised?
Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: $CoSO_{3}$ , $CuI$ , $PbCO_{3}$ , $PbCl_{2}$ , $Tl_{2} S$ , $KClO_{4}$ ?
Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: $AgCl$ , $BaSO_{4}$ , $CaF_{2}$ , $Hg_{2} I_{2}$ , $MnCO_{3}$ , $ZnS$ , $PbS$ ?
Write the ionic equation for dissolution and the solubility product (K_sp) expression for each of the following slightly soluble ionic compounds:
1. $PbCl_{2}$
2. $Ag_{2} S$
3. $Sr_{3} (PO_{4})_{2}$
4. $SrSO_{4}$
Write the ionic equation for the dissolution and the K_sp expression for each of the following slightly soluble ionic compounds:
1. $LaF_{3}$
2. $CaCO_{3}$
3. $Ag_{2} SO_{4}$
4. $Pb (OH)_{2}$
The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
1. $BaSeO_{4}$ , 0.0118 g/100 mL
2. $Ba (BrO_{3})_{2} \cdot H_{2} O$ , 0.30 g/100 mL
3. $NH_{4} MgAsO_{4} \cdot 6 H_{2} O$ , 0.038 g/100 mL
4. $La_{2} (MoO_{4})_{2}$ , 0.00179 g/100 mL
Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: $CaF_{2}$ , $Hg_{2} Cl_{2}$ , $PbI_{2}$ , or $Sn (OH)_{2}$ .
Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
1. $KHC_{4} H_{4} O_{6}$
2. $PbI_{2}$
3. $Ag_{4} [Fe (CN)_{6}]$ , a salt containing the ${Fe(CN)}_{4}^{-}$ ion.
4. $Hg_{2} I_{2}$
Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
1. $Ag_{2} SO_{4}$
2. $PbBr_{2}$
3. $AgI$
4. $CaC_{2} O_{4} \cdot H_{2} O$
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
1. $AgCl$ (s) in 0.025 M NaCl
2. $CaF_{2}$ (s) in 0.00133 M KF
3. $Ag_{2} SO_{4}$ (s) in 0.500 L of a solution containing 19.50 g of $K_{2} SO_{4}$
4. $Zn (OH)_{2}$ (s) in a solution buffered at a pH of 11.45
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.
1. $TlCl$ (s) in 1.250 M $HCl$
2. $PbI_{2}$ (s) in 0.0355 M $CaI_{2}$
3. $Ag_{2} CrO_{4}$ (s) in 0.225 L of a solution containing 0.856 g of $K_{2} CrO_{4}$
4. $Cd (OH)_{2}$ (s) in a solution buffered at a pH of 10.995
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.
1. $TlCl$ (s) in 0.025 M $TlNO_{3}$
2. $BaF_{2}$ (s) in 0.0313 M $KF$
3. $MgC_{2} O_{4}$ in 2.250 L of a solution containing 8.156 g of $Mg (NO_{3})_{2}$
4. $Ca (OH)_{2}$ (s) in an unbuffered solution initially with a pH of 12.700
Explain why the changes in concentrations of the common ions in Question 17 can be neglected.
Explain why the changes in concentrations of the common ions in Question 18 cannot be neglected.
Calculate the solubility of aluminum hydroxide, $Al (OH)_{3}$ , in a solution buffered at pH 11.00.
Refer to Solubility Products for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.
Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure 16.1.3). This use of BaSO₄ is possible because of its low solubility. Calculate the molar solubility of $BaSO_{4}$ and the mass of barium present in 1.00 L of water saturated with $BaSO_{4}$ .
Public Health Service standards for drinking water set a maximum of 250 mg/L (2.60 × 10^–3 M) of ${SO}_{4}^{2 -}$ because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO₄ (“gyp” water) as a result or passing through soil containing gypsum, $CaSO_{4} \cdot 2 H_{2} O$ , meet these standards? What is ${SO}_{4}^{2 -}$ in such water?
Perform the following calculations:
1. Calculate $[Ag^{+}]$ in a saturated aqueous solution of $AgBr$ .
2. What will $[Ag^{+}]$ be when enough $KBr$ has been added to make $[Br^{-}]$ = 0.050 M?
3. What will $[Br^{-}]$ be when enough $AgNO_{3}$ has been added to make $[Ag^{+}]$ = 0.020 M?
The solubility product of $CaSO_{4} \cdot 2 H_{2} O$ is 2.4 × 10^–5. What mass of this salt will dissolve in 1.0 L of 0.010 M ${SO}_{4}^{2 -} ?$
Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Solubility Products).
1. $TlCl$
2. $BaF_{2}$
3. $Ag_{2} CrO_{4}$
4. $CaC_{2} O_{4} \cdot H_{2} O$
5. the mineral anglesite, $PbSO_{4}$
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate K_sp for each of the slightly soluble solids indicated:
1. $AgBr : [Ag^{+}]$ = 5.7 × 10^–7M, [Br^–] = 5.7 × 10^–7M
2. $CaCO_{3} : [Ca^{2 +}]$ = 5.3 × 10^–3M, $[{CO}_{3}^{2 -}]$ = 9.0 × 10^–7M
3. $PbF_{2} : [Pb^{2 +}]$ = 2.1 × 10^–3M, [F^–] = 4.2 × 10^–3M
4. $Ag_{2} CrO_{4} : [Ag^{+}]$ = 5.3 × 10^–5M, 3.2 × 10^–3M
5. $InF_{3} : [In^{3 +}]$ = 2.3 × 10^–3M, [F^–] = 7.0 × 10^–3M
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate K_sp for each of the slightly soluble solids indicated:
1. $TlCl : [Tl^{+}]$ = 1.21 × 10^–2M, [Cl^–] = 1.2 × 10^–2M
2. $Ce (IO_{3})_{4} : [Ce^{4 +}]$ = 1.8 × 10^–4M, $[{IO}_{3}^{-}]$ = 2.6 × 10^–13M
3. $Gd_{2} (SO_{4})_{3} : [Gd^{3 +}]$ = 0.132 M, $[{SO}_{4}^{2 -}]$ = 0.198 M
4. $Ag_{2} SO_{4} : [Ag^{+}]$ = 2.40 × 10^–2M, $[{SO}_{4}^{2 -}]$ = 2.05 × 10^–2M
5. $BaSO_{4} : [Ba^{2 +}]$ = 0.500 M, $[{SO}_{4}^{2 -}]$ = 2.16 × 10^–10M
Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Solubility Products for K_sp values.)
1. $KClO_{4} : [K^{+}]$ = 0.01 M, $[{ClO}_{4}^{-}]$ = 0.01 M
2. $K_{2} PtCl_{6} : [K^{+}]$ = 0.01 M, ${[PtCl}_{6}^{2 -}]$ = 0.01 M
3. $PbI_{2} : [Pb^{2 +}]$ = 0.003 M, [I^–] = 1.3 × 10^–3M
4. $Ag_{2} S : [Ag^{+}]$ = 1 × 10^–10M, [S^2–] = 1 × 10^–13M
Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Solubility Products for K_sp values.)
1. $CaCO_{3} : [Ca^{2 +}]$ = 0.003 M, ${[CO}_{3}^{2 -}]$ = 0.003 M
2. $Co (OH)_{2} : [Co^{2 +}]$ = 0.01 M, [OH^–] = 1 × 10^–7M
3. $CaHPO_{4} : [Ca^{2 +}]$ = 0.01 M, ${[HPO}_{4}^{2 -}]$ = 2 × 10^–6M
4. $Pb_{3} (PO_{4})_{2} : [Pb^{2 +}]$ = 0.01 M, ${[PO}_{4}^{3 -}]$ 1 × 10^–13M
Calculate the concentration of $Tl^{+}$ when $TlCl$ just begins to precipitate from a solution that is 0.0250 M in $Cl^{-}$ .
Calculate the concentration of sulfate ion when $BaSO_{4}$ just begins to precipitate from a solution that is 0.0758 M in $Ba^{2 +}$ .
Calculate the concentration of $Sr^{2 +}$ when $SrF_{2}$ starts to precipitate from a solution that is 0.0025 M in $F^{-}$ .
Calculate the concentration of ${PO}_{4}^{3 -}$ when $Ag_{3} PO_{4}$ starts to precipitate from a solution that is 0.0125 M in $Ag^{+}$ .
Calculate the concentration of $F^{-}$ required to begin precipitation of $CaF_{2}$ in a solution that is 0.010 M in $Ca^{2 +}$ .
Calculate the concentration of $Ag^{+}$ required to begin precipitation of $Ag_{2} CO_{3}$ in a solution that is 2.50 × 10^–6M in ${CO}_{3}^{2 -}$ .
What $[Ag^{+}]$ is required to reduce $[CO_{3}^{2 -}]$ to 8.2 × 10^–4M by precipitation of $Ag_{2} CO_{3}$ ?
What $[F^{-}]$ is required to reduce $[Ca^{2 +}]$ to 1.0 × 10^–4M by precipitation of $CaF_{2}$ ?
A volume of 0.800 L of a 2 × 10^–4–M $Ba (NO_{3})_{2}$ solution is added to 0.200 L of 5 × 10^–4M $Li_{2} SO_{4}$ . Does $BaSO_{4}$ precipitate? Explain your answer.
Perform these calculations for nickel(II) carbonate.
1. With what volume of water must a precipitate containing $NiCO_{3}$ be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with $NiCO_{3}$ (K_sp = 1.36 × 10^–7).
2. If the $NiCO_{3}$ were a contaminant in a sample of CoCO₃ (K_sp = 1.0 × 10^–12), what mass of $CoCO_{3}$ would have been lost? Keep in mind that both $NiCO_{3}$ and $CoCO_{3}$ dissolve in the same solution.
Iron concentrations greater than 5.4 × 10^–6M in water used for laundry purposes can cause staining. What $[OH^{-}]$ is required to reduce $[Fe^{2 +}]$ to this level by precipitation of $Fe (OH)_{2}$ ?
A solution is 0.010 M in both $Cu^{2 +}$ and $Cd^{2 +}$ . What percentage of $Cd^{2 +}$ remains in the solution when 99.9% of the $Cu^{2 +}$ has been precipitated as $CuS$ by adding sulfide?
A solution is 0.15 M in both $Pb^{2 +}$ and $Ag^{+}$ . If $Cl^{-}$ is added to this solution, what is $[Ag^{+}]$ when $PbCl_{2}$ begins to precipitate?
What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 M with respect to each ion? In some cases it may be necessary to control the $pH$ . (Hint: Consider the K_sp values given in Solubility Products.)
1. ${Hg}_{2}^{2 +}$ and $Cu^{2 +}$
2. ${SO}_{4}^{2 -}$ and $Cl^{-}$
3. $Hg^{2 +}$ and $Co^{2 +}$
4. $Zn^{2 +}$ and $Sr^{2 +}$
5. $Ba^{2 +}$ and $Mg^{2 +}$
6. ${CO}_{3}^{2 -}$ and $OH^{-}$
A solution contains 1.0 × 10^–5 mol of $KBr$ and 0.10 mol of $KCl$ per liter. $AgNO_{3}$ is gradually added to this solution. Which forms first, solid $AgBr$ or solid $AgCl$ ?
A solution contains 1.0 × 10^–2 mol of $KI$ and 0.10 mol of $KCl$ per liter. $AgNO_{3}$ is gradually added to this solution. Which forms first, solid AgI or solid $AgCl$ ?
The calcium ions in human blood serum are necessary for coagulation (Figure 16.1.4). Potassium oxalate, $K_{2} C_{2} O_{4}$ , is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of $CaC_{2} O_{4} \cdot H_{2} O$ . It is necessary to remove all but 1.0% of the $Ca^{2 +}$ in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of $Ca^{2 +}$ per 100 mL of serum, what mass of $K_{2} C_{2} O_{4}$ is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the K_sp value for $CaC_{2} O_{4}$ in serum is the same as in water.)
About 50% of urinary calculi (kidney stones) consist of calcium phosphate, $Ca_{3} (PO_{4})_{2}$ . The normal mid range calcium content excreted in the urine is 0.10 g of $Ca^{2 +}$ per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form?
The pH of normal urine is 6.30, and the total phosphate concentration $([{PO}_{4}^{3 -}] + [{HPO}_{4}^{2 -}] + [H_{2} {PO}_{4}^{-}] + [H_{3} {PO}_{4}])$ is 0.020 M. What is the minimum concentration of $Ca^{2 +}$ necessary to induce kidney stone formation? (See Question 49 for additional information.)
Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions: ${Mg}^{2+} (a q) + {Ca(OH)}_{2} (a q) ⟶ {Mg(OH)}_{2} (s) + {Ca}^{2+} (a q)$ ${Mg(OH)}_{2} (s) + 2HCl (a q) ⟶ {MgCl}_{2} (s) + {2H}_{2} O (l)$
${MgCl}_{2} (l) \overset{electrolysis}{⟶} Mg (s) + {Cl}_{2} (g)$ Sea water has a density of 1.026 g/cm³ and contains 1272 parts per million of magnesium as $Mg^{2 +}$ (aq) by mass. What mass, in kilograms, of $Ca (OH)_{2}$ is required to precipitate 99.9% of the magnesium in 1.00 × 10³ L of sea water?
Hydrogen sulfide is bubbled into a solution that is 0.10 M in both $Pb^{2 +}$ and $Fe^{2 +}$ and 0.30 M in $HCl$ . After the solution has come to equilibrium it is saturated with $H_{2} S$ ( $[H_{2} S]$ = 0.10 M). What concentrations of $Pb^{2 +}$ and $Fe^{2 +}$ remain in the solution? For a saturated solution of H₂S we can use the equilibrium: $H_{2} S (a q) + {2H}_{2} O (l) ⇌ {2H}_{3} O^{+} (a q) + S^{2 -} (a q) K = 1.0 \times 1 0^{- 26}$ (Hint: The $[H_{3} O^{+}]$ changes as metal sulfides precipitate.)
Perform the following calculations involving concentrations of iodate ions:
1. The iodate ion concentration of a saturated solution of $La (IO_{3})_{3}$ was found to be 3.1 × 10^–3 mol/L. Find the K_sp.
2. Find the concentration of iodate ions in a saturated solution of $Cu (IO_{3})_{2}$ (K_sp = 7.4 × 10^–8).
Calculate the molar solubility of $AgBr$ in 0.035 M $NaBr$ (K_sp = 5 × 10^–13).
How many grams of $Pb (OH)_{2}$ will dissolve in 500 mL of a 0.050-M $PbCl_{2}$ solution (K_sp = 1.2 × 10^–15)?
Use this simulation to study the process of salts dissolving and forming saturated solutions to complete the following exercise. Using 0.01 g $CaF_{2}$ , give the K_sp values found in a 0.2-M solution of each of the salts. Discuss why the values change as you change soluble salts.
How many grams of Milk of Magnesia, $Mg (OH)_{2}$ (s) (58.3 g/mol), would be soluble in 200 mL of water. K_sp = 7.1 × 10^–12. Include the ionic reaction and the expression for K_sp in your answer. What is the pH? (K_w = 1 × 10^–14 = $[H_{3} O^{+}]$ $[OH^{-}]$ )
Two hypothetical salts, LM₂ and LQ, have the same molar solubility in $H_{2} O$ . If K_sp for LM₂ is 3.20 × 10^–5, what is the K_sp value for LQ?
Which of the following carbonates will form first? Which of the following will form last? Explain.
1. ${MgCO}_{3} K_{sp} = 3.5 \times 1 0^{- 8}$
2. ${CaCO}_{3} K_{sp} = 4.2 \times 1 0^{- 7}$
3. ${SrCO}_{3} K_{sp} = 3.9 \times 1 0^{- 9}$
4. ${BaCO}_{3} K_{sp} = 4.4 \times 1 0^{- 5}$
5. ${MnCO}_{3} K_{sp} = 5.1 \times 1 0^{- 9}$
How many grams of $Zn (CN)_{2}$ (s) (117.44 g/mol) would be soluble in 100 mL of $H_{2} O$ ? Include the balanced reaction and the expression for K_sp in your answer. The K_sp value for Zn(CN)₂(s) is 3.0 × 10^–16.

Show Selected Solutions

There is no change. A solid has an activity of 1 whether there is a little or a lot.
The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve.
$CaF_{2}$ , $MnCO_{3}$ , and $ZnS$ ; each is a salt of a weak acid and the hydronium ion from water reacts with the anion, causing more solid to dissolve to maintain the equilibrium concentration of the anion
The answers for each compound are as follows:
1. ${LaF}_{3} (s) ⇌ {La}^{3 +} (a q) + {3 F}^{-} (a q); K_{sp} = [{La}^{3 +}] {[F^{-}]}^{3}$
2. ${CaCO}_{3} (s) ⇌ {Ca}^{2 +} (a q) + {CO}_{3}^{2 -} (a q); K_{sp} = [{Ca}^{2 +}] [{CO}_{3}^{2 -}]$
3. ${Ag}_{2} {SO}_{4} (s) ⇌ 2 {Ag}^{+} (a q) + {SO}_{4}^{2 -} (a q); K_{sp} = {[{Ag}^{+}]}^{2} [{SO}_{4}^{2 -}]$
4. $Pb {(OH)}_{2} (s) ⇌ {Pb}^{2 +} (a q) + 2 {OH}^{-} (a q); K_{sp} = [{Pb}^{2 +}] {[{OH}^{-}]}^{2}$
Convert each concentration into molar units. Multiply each concentration by 10 to determine the mass in 1 L, and then divide the molar mass.
1. $BaSeO_{4}$
  1. $\frac{0. {118 g L}^{- 1}}{280. {28 g mol}^{- 1}} = 4.21 \times 10^{- 4} M$
  2. K = [Ba²⁺] $[{SeO}_{4}^{2 -}]$ = (4.21 × 10^–4)(4.21 × 10^–4) = 1.77 × 10^–7
2. $Ba (BrO_{3})_{2} \cdot H_{2} O$ :
  1. $\frac{3.0 {g L}^{- 1}}{{411.147 g mol}^{- 1}} = 7.3 \times 10^{- 3} M$
  2. K = [Ba²⁺] ${[{BrO}_{3}^{-}]}^{2}$ = (7.3 × 10^–3)(2 × 7.3 × 10^–3)² = 1.6 × 10^–6
3. $NH_{4} MgAsO_{4} \cdot 6 H_{2} O$ :
  1. $\frac{0. {38 g L}^{- 1}}{{289.3544 g mol}^{- 1}} = 1.3 \times 10^{- 3} M$
  2. K = $[{NH}_{4}^{+}]$ [Mg²⁺] $[{AsO}_{4}^{3 -}]$ = (1.3 × 10^–3)³ = 2.2 × 10^–9
4. La₂(MoO₄)₃:
  1. $\frac{0.0 {179 g L}^{- 1}}{{757.62 g mol}^{- 1}} = 2.36 \times 10^{- 5} M$
  2. K = [La³⁺]² ${[{MoO}_{4}^{2 -}]}^{3}$ = (2 × 2.36 × 10^–5)²(3 × 2.36 × 10^–5)³ = 2.228 × 10^–9 × 3.549 × 10^–13 = 7.91 × 10^–22
Let x be the molar solubility.
1. K_sp = $[K^{+}] [{HC}_{4} H_{4} O_{6}^{-}]$ = 3 × 10^–4 = x², x = 2 × 10^–2M;
2. K_sp = [Pb²⁺][I^–]² = 8.7 × 10⁹ = x(2x)³ = 4x³, x = 1.3 × 10^–3M;
3. K_sp = ${[{Ag}^{+}]}^{4} [{Fe(CN)}_{6}^{4-}]$ = 1.55 × 10^–41 = (4x)⁴x = 256x⁵, x = 2.27 × 10^–9M;
4. K_sp = $[{Hg}_{2}^{2+}] {[I^{-}]}^{2}$ = 4.5 × 10^–29 = [x][2x]² = 4x³, x = 2.2 × 10^–10M
The correct concentrations are as follows:
1. K_sp = 1.8 × 10^–10 = $[Ag^{+}] [Cl^{-}]$ = x(x + 0.025), where x = [Ag⁺]. Assume that x is small when compared with 0.025 and therefore ignore it:
  - $x = \frac{1.8 \times 10^{- 10}}{0.025} = 7.2 \times 10^{- 9} M = [{Ag}^{+}], [{Cl}^{-}] = 0.02 5 M$
  - Check: $\frac{7.2 \times 10^{- 9} M}{0.025 M} \times 100 = 2.9 \times 10^{- 5}$ , an insignificant change;
2. K_sp = 3.9 × 10^–11 = $[Ca^{2 +}] [F^{-}]_{2}$ = x(2x + 0.00133 M)², where x = $[Ca^{2 +}]$ . Assume that x is small when compared with 0.0013 M and disregard it:
  - $x = \frac{3.9 \times 10^{- 11}}{{(0.00133)}^{2}} = 2.2 \times 10^{- 5} M = [{Ca}^{2+}], [F^{-}] = 0.0013 M$
  - Check: $\frac{2.25 \times 10^{- 5} M}{0.00133 M} \times 100 = 1.69$ . This value is less than 5% and can be ignored.
3. Find the concentration of $K_{2} SO_{4}$ :
  - $\frac{19.50 g}{174.2 {60 g mol}^{- 1}} = 0.1119 mol$
    $\frac{0.1119 mol}{0.5 L} = 0.2238 M = {[SO}_{4}^{2 -}]$
    K_sp = 1.18 × 10^–18 = [Ag⁺]² ${[SO}_{4}^{2 -}]$ = 4x²(x + 0.2238)
    $x^{2} = \frac{1.18 \times 10^{- 18}}{4 (0.2238)} = 1.32 \times 10^{- 18}$
    x = 1.15 × 10^–9[Ag⁺] = 2x = 2.30 × 10^–9M
  - Check: $\frac{1.15 \times 10^{- 9}}{0.2238} \times 100$ ; the condition is satisfied.
4. Find the concentration of $OH^{-}$ from the pH:
  - pOH = 14.00 – 11.45 = 2.55
    [OH^–] = 2.8 × 10^–3M
    K_sp = 4.5 × 10^–17 = [Zn²⁺][OH^–]² = x(2x + 2.8 × 10^–3)²
  - Assume that x is small when compared with 2.8 × 10^–3:
    $x = \frac{4.5 \times 10^{- 17}}{{(2.8 \times 10^{- 3})}^{2}} = 5.7 \times 10 - 12 M = {[Zn}^{2+}]$
  - Check: $\frac{5.7 \times 10^{- 12}}{2.8 \times 10^{- 3}} \times 100 = 2.0 \times 10^{- 7}$ ; x is less than 5% of [OH^–] and is, therefore, negligible. In each case the change in initial concentration of the common ion is less than 5%.
The answers are as follows:
1. K_sp = 1.9 × 10^–4 = $[Ti^{+}] [Cl^{-}]$ ; Let x = $[Cl^{-}]$ :1.9 × 10^–4 = (x = 0.025)x
  1. Assume that x is small when compared with 0.025:
    1. $x = \frac{1.9 \times 10^{- 4}}{0.025} = 7.6 \times 10^{- 3} M$
  2. Check: $\frac{7.6 \times 10^{- 3}}{0.025} \times 100 = 30$
  3. This value is too large to drop x. Therefore solve by using the quadratic equation:
    1. $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
    2. x² + 0.025x – 1.9 × 10^–4 = 0
    3. $\begin{array}{l} x = \frac{- 0.025 \pm \sqrt{6.25 \times 10^{- 4} + 7.6 \times 10^{- 4}}}{2} = \frac{- 0.025 \pm \sqrt{1.385 \times 10^{- 3}}}{2} \\ = \frac{- 0.025 \pm 0.0372}{2} = 0.00 61 M \end{array}$
  4. (Use only the positive answer for physical sense.)
    1. [Ti⁺] = 0.025 + 0.0061 = 3.1 × 10^–2M
    2. [Cl^–] = 6.1 × 10^–3
2. K_sp = 1.7 × 10^–6 = [Ba²⁺][F^–]²:
  1. Let x = [Ba²⁺]
  2. 1.7 × 10^–6 = x(x + 0.0313)² = 1.7 × 10^–3M
  3. Check: $\frac{1.7 \times 10^{- 3}}{0.0313} \times 100 = 5.5$
  4. This value is too large to drop x, and the entire equation must be solved. One method to find the answer is to solve by successive approximations. Begin by choosing the value of x that has just been calculated:
  5. x′(5.4 × 10^–5 + 0.0313)² = 1.7 × 10^–3 or
  6. $x' = \frac{1.7 \times 10^{- 6}}{1.089 \times 10^{- 3}} = 1.6 \times 10^{- 3}$
  7. A third approximation using this last calculation is as follows:
  8. x′′(1.7 × 10^–3 + 0.0313)² = 1.7 × 10^–6 or
  9. $x'' = \frac{1.7 \times 10^{- 6}}{1.089 \times 10^{- 3}} = 1.6 \times 10^{- 3}$
  10. This value is well within 5% and is acceptable.
  11. [Ba²⁺] = 1.6 × 10^–3M
  12. [F^–] = (1.6 × 10^–3 + 0.0313) = 0.0329 M;
3. Find the molar concentration of the $Mg (NO_{3})_{2}$ . The molar mass of $Mg (NO_{3})_{2}$ is 148.3149 g/mol. The number of moles is $\frac{8.156 g}{148.314 {9 g mol}^{- 1}} = 0.05499 mol$
  1. $M = \frac{0.05499 mol}{2.250 L} = 0.02444 M$
  2. Let x = $[C_{2} O_{4}^{2 -}]$ and assume that x is small when compared with 0.02444 M.
  3. K_sp = 8.6 × 10^–5 = [Mg²⁺] $[C_{2} O_{4}^{2 -}]$ = (x)(x + 0.02444)
  4. 0.02444x = 8.6 × 10^–5
  5. x = $[C_{2} O_{4}^{2 -}]$ = 3.5 × 10^–3
  6. Check: $\frac{3.5 \times 10^{- 3}}{0.02444} \times 100 = 14$
  7. This value is greater than 5%, so the quadratic equation must be used to solve for x:
  8. $\begin{array}{l} \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x^{2} + 0.02444 x - 8.6 \times 10^{- 5} = 0 \\ x = \frac{- 0.02444 \pm \sqrt{5.973 \times 10^{- 4} + 3.44 \times 10^{- 4}}}{2} = \frac{- 0.02444 \pm 0.03068}{2} \end{array}$
  9. (Use only the positive answer for physical sense.)
  10. x′ = $[C_{2} O_{4}^{2 -}]$ = 3.5 × 10^–3M
  11. [Mg²⁺] = 3.1 × 10^–3 + 0.02444 = 0.0275 M
  12. (d) pH = 12.700; pOH = 1.300
  13. $[OH^{-}]$ = 0.0501 M; Let x = $[Ca^{2 +}]$
  14. K_sp = 7.9 × 10^–6 = $[Ca^{2 +}] [OH^{-}]^{2}$ = (x)(x + 0.050)²
  15. Assume that x is small when compared with 0.050 M:
  16. x = $[Ca^{2 +}]$ = 3.15 × 10^–3 (one additional significant figure is carried)
  17. Check: $\frac{3.15 \times 10^{- 3}}{0.050} \times 100 = 6.28$
  18. This value is greater than 5%, so a more exact method, such as successive approximations, must be used. Begin by choosing the value of x that has just been calculated:
  19. x′(3.15 × 10^–3 + 0.0501)² = 7.9 × 10^–6 or
  20. $x' = \frac{7.9 \times 10^{- 6}}{2.836 \times 10^{- 3}} = 2.8 \times 10^{- 3}$
  21. $[Ca^{2 +}]$ = 2.8 × 10^–3M
  22. $[OH^{-}]$ = (2.8 × 10^–3 + 0.0501) = 0.053 × 10^–2M
  23. In each case, the initial concentration of the common ion changes by more than 5%.
The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change.
$Ca (OH)_{2} : [Ca^{2 +}] [OH^{-}]^{2}$ = 7.9 × 10^–6
1. Let x be $[Ca^{2 +}]$ = molar solubility; then $[OH^{-}]$ = 2x
2. K_sp = x(2x)² = 4x³ = 7.9 × 10^–6
3. x³ = 1.975 × 10^–6
4. x = 0.013 M
5. $CaCO_{3} : [Ca^{2 +}] [CO_{3}^{2 -}]$ = 4.8 × 10^–9
6. Let x be $[Ca^{2 +}]$ ; then ${[CO}_{3}^{2 -}]$ = [Ca²⁺] = molar solubility
7. K_sp = x² = 4.8 × 10^–9
8. x = 6.9 × 10^–5M
9. $Math input error$ = 2.4 10^–5
10. Let x be $[Ca^{2 +}]$ = molar solubility = ${[SO}_{4}^{2 -}]$ ; then $[H_{2} O]$ = 2x
11. K_sp = (x)(x)(2x)² = 2.4 × 10^–5
12. x⁴ = 6.0 × 10^–6
13. x = 0.049 M
14. This value is more than three times the value given by the Handbook of Chemistry and Physics of (0.014 M) and reflects the complex interaction of water within the precipitate:
15. $CaC_{2} O_{4} \cdot H_{2} O : [Ca^{2 +}] [C_{2} O_{4}^{2 -}] [H_{2} O]$ = 2.27 × 10^–9
16. Let x be $[Ca^{2 +}]$ = molar solubility = $[C_{2} O_{4}^{2 -}]$ = [H₂O]
17. x³ = 2.27 × 10^–9
18. x = 1.3 × 10^–3M
19. In this case, the interaction of water is also complex and the solubility is considerably less than that calculated.
20. Ca₃(PO₄)₂: [Ca²⁺]³ ${[{PO}_{4}^{3 -}]}^{2}$ = 1 × 10^–25
21. Upon solution there are three Ca²⁺ and two ${PO}_{4}^{3 -}$ ions. Let the concentration of Ca²⁺ formed upon solution be x. Then $\frac{2}{3} x$ is the concentration of ${PO}_{4}^{3-} :$
22. $x^{3} {(\frac{2}{3} x)}^{2} x^{3} = 1 \times 10^{- 25} = 0.4444 x^{5}$
23. x = 1 × 10^–5M = [Ca²⁺]
24. The solubility is then one-third the concentration of Ca²⁺, or 4 × 10^–6.
First, find the concentration in a saturated solution of CaSO₄. Before placing the CaSO₄ in water, the concentrations of Ca²⁺ and ${SO}_{4}^{2 -}$ are 0. Let x be the change in concentration of Ca²⁺, which is equal to the concentration of ${SO}_{4}^{2 -} :$
1. K_sp = [Ca²⁺] $[{SO}_{4}^{2 -}]$ = 2.4 × 10^–5
2. x × x = x² = 2.4 × 10^–5
3. $x = \sqrt{2.4 \times 10^{- 5}}$
4. x = 4.9 × 10^–3M = $[{SO}_{4}^{2 -}]$ = [Ca²⁺]
5. Since this concentration is higher than 2.60 × 10^–3M, “gyp” water does not meet the standards.
The amount of CaSO₄·2H₂O that dissolves is limited by the presence of a substantial amount of ${SO}_{4}^{2 -}$ already in solution from the 0.010 M ${SO}_{4}^{-}$ . This is a common-ion problem. Let x be the change in concentration of Ca²⁺ and of ${SO}_{4}^{2 -}$ that dissociates from CaSO₄:
1. ${CaSO}_{4} (s) ⟶ {Ca}^{2+} (a q) + {SO}_{4}^{2-} (a q)$
2. K_sp = [Ca²⁺] ${[SO}_{4}^{2 -}]$ = 2.4 × 10^–5
3. Addition of 0.010 M ${SO}_{4}^{2 -}$ generated from the complete dissociation of 0.010 M SO₄ gives
4. [x][x + 0.010] = 2.4 × 10^–5
5. Here, x cannot be neglected in comparison with 0.010 M; the quadratic equation must be used. In standard form:
6. x² + 0.010x – 2.4 × 10^–5 = 0
7. $x = \frac{- 0.01 \pm \sqrt{1 \times 10^{- 4} + 9.6 \times 10^{- 5}}}{2} = \frac{- 0.01 \pm 1.4 \times 10^{- 2}}{2}$
8. Only the positive value will give a meaningful answer:
9. x = 2.0 × 10^–3 = [Ca²⁺]
10. This is also the concentration of CaSO₄·2H₂O that has dissolved. The mass of the salt in 1 L is
11. Mass (CaSO₄·2H₂O) = 2.0 × 10^–3 mol/L × 172.16 g/mol = 0.34 g/L
12. Note that the presence of the common ion, ${SO}_{4}^{2 -}$ , has caused a decrease in the concentration of Ca²⁺ that otherwise would be in solution:
13. $\sqrt{2.4 \times 10^{- 5}} = 4.9 \times 10^{- 5} M$
In each of the following, allow x to be the molar concentration of the ion occurring only once in the formula.
1. K_sp = [Ag⁺][Cl^–] = 1.5 – 10^–16 = [x²], [x] = 1.2 – 10^–8M, [Ag⁺] = [I^–] = 1.2 × 10^–8M
2. K_sp = ${{[Ag}^{+}]}^{2} [{SO}_{4}^{2 -}]$ = 1.18 × 10^–5 = [2x]²[x], 4x³ = 1.18 × 10^–5, x = 1.43 × 10^–2M
  As there are 2 Ag⁺ ions for each ${SO}_{4}^{2 -}$ ion, [Ag⁺] = 2.86 × 10^–2M, ${[SO}_{4}^{2 -}]$ = 1.43 × 10^–2M; (c) Ksp = [Mn²⁺]²[OH^–]² = 4.5 × 10^–14 = [x][2x]², 4x³ = 4.5 × 10^–14, x = 2.24 × 10^–5M.
3. Since there are two OH^– ions for each Mn²⁺ ion, multiplication of x by 2 gives 4.48 × 10^–5M. If the value of x is rounded to the correct number of significant figures, [Mn²⁺] = 2.2 × 10^–5M. [OH^–] = 4.5 × 10^–5M. If the value of x is rounded before determining the value of [OH^–], the resulting value of [OH^–] is 4.4 × 10^–5M. We normally maintain one additional figure in the calculator throughout all calculations before rounding.
4. K_sp = [Sr²⁺][OH^–]² = 3.2 × 10^–4 = [x][2x]², 4x³ = 3.2 × 10^–4, x = 4.3 × 10^–2M.
  Substitution gives [Sr²⁺] = 4.3 × 10^–2 M, [OH^–] = 8.6 × 10^–2M
5. K_sp = [Mg²⁺]²[OH^–]² = 1.5 × 10^–11 = [x][2x]², 4x³ = 1.5 × 10^–11, x = 1.55 × 10^–4M, 2x = 3.1 × 10^–4.
  Substitution and taking the correct number of significant figures gives [Mg²⁺] = 1.6 × 10^–4M, [OH–] = 3.1 × 10^–4M. If the number is rounded first, the first value is still [Mg²⁺] = 1.6 × 10^–4M, but the second is [OH^–] = 3.2 × 10^–4M.
In each case the value of K_sp is found by multiplication of the concentrations raised to the ion’s stoichiometric power. Molar units are not normally shown in the value of K.
1. TlCl: K_sp = (1.4 × 10^–2)(1.2 × 10^–2) = 2.0 × 10^–4
2. Ce(IO₃)₄: K_sp = (1.8 × 10^–4)(2.6 × 10^–13)⁴ = 5.1 × 10^–17
3. Gd₂(SO₄)₃: K_sp = (0.132)²(0.198)³ = 1.35 × 10^–4
4. Ag₂SO₄: K_sp = (2.40 × 10^–2)²(2.05 × 10^–2) = 1.18 × 10^–5
5. BaSO₄: K_sp = (0.500)(2.16 × 10^–10) = 1.08 × 10^–10
The answers are as follows:
1. ${CaCO}_{3} : {CaCO}_{3} (s) ⟶ {Ca}^{2+} (a q) + {CO}_{3}^{2 -} (a q)$
  1. K_sp = [Ca²⁺] ${[CO}_{3}^{2 -}]$ = 4.8 × 10^–9
  2. test K_sp against Q = [Ca²⁺] ${[CO}_{3}^{2 -}]$
  3. Q = [Ca²⁺] ${[CO}_{3}^{2 -}]$ = (0.003)(0.003) = 9 × 10^–6
  4. K_sp = 4.8 × 10^–9 < 9 × 10^–6
2. The ion product does exceed K_sp, so CaCO₃ does precipitate.
3. ${Co(OH)}_{2} : {Co(OH)}_{2} (s) ⟶ {Co}^{2+} (a q) + {2OH}^{-} (a q)$
  1. K_sp = [Co²⁺][OH^–]² = 2 × 10^–16
  2. test K_sp against Q = [Co²⁺][OH^–]²
  3. Q = [Co²⁺][OH^–]² = (0.01)(1 × 10^–7)² = 1 × 10^–16
  4. K_sp = 2 × 10^–16 > 1 × 10^–16
4. The ion product does not exceed K_sp, so the compound does not precipitate.
5. CaHPO₄: (K_sp = 5 × 10^–6):
  1. Q = [Ca²⁺] ${[HPO}_{4}^{2 -}]$ = (0.01)(2 × 10^–6) = 2 × 10^–8 < K_sp
6. The ion product does not exceed K_sp, so compound does not precipitate.
7. Pb₃(PO₄)₂: (K_sp = 3 × 10^–44):
  1. Q = [Pb²⁺]³ ${{[PO}_{4}^{3 -}]}^{2}$ = (0.01)³(1 × 10^–13)² = 1 × 10^–32 > K_sp
8. The ion product exceeds K_sp, so the compound precipitates.
Precipitation of ${SO}_{4}^{2 -}$ will begin when the ion product of the concentration of the ${SO}_{4}^{2 -}$ and Ba²⁺ ions exceeds the K_sp of BaSO₄.
1. K_sp = 1.08 × 10^–10 = [Ba²⁺] $[{SO}_{4}^{2 -}]$ = (0.0758) ${[SO}_{4}^{2 -}]$
2. $[{SO}_{4}^{2 -}] = \frac{1.08 \times 10^{- 10}}{0.0758} = 1.42 \times 10^{- 9} M$
Precipitation of Ag₃PO₄ will begin when the ion product of the concentrations of the Ag⁺ and ${PO}_{4}^{3-}$ ions exceeds K_sp:
1. ${Ag}_{3} {PO}_{4} (s) ⟶ {3Ag}^{+} (a q) + {PO}_{4}^{3-} (a q)$
2. K_sp = 1.8 × 10^–18 = [Ag⁺]³ ${[PO}_{4}^{3-}]$ = (0.0125)³ ${[PO}_{4}^{3 -}]$
3. $[{PO}_{4}^{3 -}] = \frac{1.08 \times 10^{- 18}}{{(0.0125)}^{3}} = 9.2 \times 10^{- 13} M$
${Ag}_{2} {CO}_{3} (s) ⟶ {2Ag}^{+} (a q) + {CO}_{3}^{2-} (a q)$
1. [Ag⁺]² $[{CO}_{3}^{2-}]$ = K_sp = 8.2 × 10^–12
2. [Ag⁺]²(2.5 × 10^–6) = 8.2 × 10^–12
3. ${[{Ag}^{+}]}^{2} = \frac{8.2 \times 10^{- 12}}{2.50 \times 10^{- 6}} = 3.28 \times 10^{- 6}$
4. [Ag⁺] = 1.8 × 10^–3M
In the K_sp expression, substitute the concentration of Ca²⁺ and solve for [F^–].
1. K_sp = 3.9 × 10^–11 = [Ca²⁺][F^–]² = (1.0 × 10^–4)[F^–]²
2. ${[F^{-}]}^{2} = \frac{3.9 \times 10^{- 11}}{1.0 \times 10^{- 4}} = 3.9 \times 10^{- 7}$
3. [F^–] = 6.2 × 10^–4
(a) 2.28 L; (b) 7.3 × 10^–7 g
When 99.9% of Cu²⁺ has precipitated as CuS, then 0.1% remains in solution.
1. $\frac{0.1}{100}$ × 0.010 mol/L = 1 × 10^–5M = [Cu²⁺]
2. [Cu²⁺][S^2–] = K_sp = 6.7 × 10^–42
3. (1 × 10^–5)[S^2–] = 6.7 × 10^–42
4. [S^2–] = 7 × 10^–37M
5. [Cd²⁺][S^2–]K_sp = 2.8 × 10^–35
6. [Cd²⁺](7 × 10^–37) = 2.8 × 10^–35
7. [Cd²⁺] = 4 × 10¹M
8. Thus [Cd²⁺] can increase to about 40 M before precipitation begins. [Cd²⁺] is only 0.010 M, so 100% of it is dissolved.
To compare ions of the same oxidation state, look for compounds with a common counter ion that have very different K_sp values, one of which has a relatively large K_sp—that is, a compound that is somewhat soluble.
1. ${Hg}_{2}^{2+}$ and Cu²⁺: Add ${SO}_{4}^{2-}$ . CuSO₄ is soluble (see Solubility Products), but K_sp for Hg₂SO₄ is 6.2 × 10^–7. When only 0.1% ${Hg}_{2}^{2+}$ remains in solution:
  1. $[{Hg}_{2}^{2+}] = \frac{0.1}{100} \times 0.10 = 1 \times 10^{- 4} M$
2. and
  1. $[{SO}_{4}^{2 -}] = \frac{K_{sp}}{[{Hg}_{2}^{2+}]} = \frac{6.2 \times 10^{- 7}}{1 \times 10^{- 4}} = 6.2 \times 10^{- 3} M$ ;
3. ${SO}_{4}^{2 -}$ and Cl^–: Add Ba²⁺. BaCl₂ is soluble (see the section on catalysis), but K_sp for BaSO₄ is 1.08 × 10^–10. When only 0.1% ${SO}_{4}^{2 -}$ remains in solution,
  1. ${[SO}_{4}^{2 -}]$ = 1 × 10^–4M
4. and
  1. $[{Ba}^{2+}] = \frac{1.08 \times 10^{- 10}}{1 \times 10^{- 4}} = 1 \times 10^{- 6} M$ ;
5. Hg²⁺ and Co²⁺: Add S^2–: For the least soluble form of CoS, K_sp = 6.7 × 10^–29 and for HgS, K_sp = 2 × 10^–59. CoS will not begin to precipitate until:
  1. [Co²⁺][S^2–] = K_sp = 6.7 × 10^–29
  2. (0.10)[S^2–] = 6.7 × 10^–29
  3. [S^2–] = 6.7 × 10^–28
6. At that concentration:
  1. [Hg²⁺](6.7 × 10^–28) = 2 × 10^–59
  2. [Hg²⁺] = 3 × 10^–32M
7. That is, it is virtually 100% precipitated. For a saturated (0.10 M) H₂S solution, the corresponding $[H_{3} O^{+}]$ is:
  1. $[H_{3} O^{+}] = \frac{[H_{2} S]}{[S^{2 -}]} K_{a} = \frac{(0.10)}{(6.7 \times 10^{- 28})} \times 1.0 \times 10^{- 26}$
  2. $[H_{3} O^{+}] = 1.5 M$
8. A solution more basic than this would supply enough S^2– for CoS to precipitate.
9. Zn²⁺ and Sr²⁺: Add OH^– until [OH^–] = 0.050 M. For Sr(OH)₂·8H₂O, K_sp = 3.2 × 10^–4. For Zn(OH)₂, K_sp = 4.5 × 10^–11. When Zn²⁺ is 99.9% precipitated, then [Zn²⁺] = 1 × 10^–4M and
  1. ${[{OH}^{-}]}^{2} = \frac{K_{sp}}{[{Zn}^{2+}]} = \frac{4.5 \times 10^{- 11}}{1 \times 10^{- 4}} = 4.5 \times 10^{- 7}$
  2. [OH^–] = 7 × 10^–4M
10. When Sr(OH)₂·8H₂O just begins to precipitate:
  1. ${[{OH}^{-}]}^{2} = \frac{K_{sp}}{[{Sr}^{2+}]} = \frac{3.2 \times 10^{- 4}}{0.10} = 3.2 \times 10^{- 3}$
  2. [OH^–] = 0.057 M
11. If [OH^–] is maintained less than 0.056 M, then Zn²⁺ will precipitate and Sr²⁺ will not.
12. Ba²⁺ and Mg²⁺: Add ${SO}_{4}^{2 -}$ . MgSO₄ is soluble and BaSO₄ is not (K_sp = 2 × 10^–10).
13. ${CO}_{3}^{2 -}$ and OH^–: Add Ba²⁺. For Ba(OH)₂, 8H₂O, K_sp = 5.0 × 10^–3; for BaCO₃, K_sp = 8.1 × 10^–9. When 99.9% of ${CO}_{3}^{2 -}$ has been precipitated ${[CO}_{3}^{2 -}]$ = 1 × 10^–4M and
  1. $[{Ba}^{2 +}] = \frac{K_{sp}}{[{CO}_{3}^{2 -}]} = \frac{8.1 \times 10^{- 9}}{1 \times 10^{- 4}} = 8.1 \times 10^{- 5} M$
14. Ba(OH)₂·8H₂O begins to precipitate when:
  1. ${[Ba}^{2+}] = \frac{K_{sp}}{{[{OH}^{-}]}^{2}} = \frac{5.0 \times 10^{- 3}}{{(0.10)}^{2}} = 0.50 M$
15. As long as [Ba²⁺] is maintained at less than 0.50 M, BaCO₃ precipitates and Ba(OH)₂·8H₂O does not.
Compare the concentration of Ag⁺ as determined from the two solubility product expressions. The one requiring the smaller [Ag⁺] will precipitate first.
1. For AgCl: K_sp = 1.8 × 10^–10 = [Ag⁺][Cl^–]
  1. $[{Ag}^{+}] = \frac{1.8 \times 10^{- 10}}{[0.10]} =1.8 \times 10^{- 9} M$
2. For AgI: K_sp = 1.5 × 10^–16 = [Ag⁺][I^–]
  1. $[{Ag}^{+}] = \frac{1.5 \times 10^{- 16}}{1.0 \times 10^{- 2}} = 1.5 \times 10^{- 9} M$
3. As the value of [Ag⁺] is smaller for AgI, AgI will precipitate first.
The dissolution of Ca₃(PO₄)₂ yields:
1. ${Ca}_{3} {{(PO}_{4})}_{2} (s) ⇌ {3Ca}^{2+} (a q) + {2PO}_{4}^{3 -} (a q)$
Given the concentration of Ca²⁺ in solution, the maximum $[{PO}_{4}^{3 -}]$ can be calculated by using the K_sp expression for Ca₃(PO₄)₂:
1. K_sp = 1 × 10^–25 = [Ca²⁺]³ ${[PO}_{4}^{3 -]}^{2}$
2. ${[{Ca}^{2+}]}_{urine} = \frac{0.10 g (\frac{1 m o l}{40.08 g})}{1.4 L} = 1.8 \times 10^{- 3} M$
3. ${{[PO}_{4}^{3 -}]}^{2} = \frac{1 \times 10^{- 25}}{{(1.8 \times 10^{- 3})}^{3}} = 1.7 \times 10^{- 17}$
4. ${{[PO}_{4}^{3 -}]}^{2} = 4 \times 10^{- 9} M$
Calculate the amount of Mg²⁺ present in sea water; then use K_sp to calculate the amount of Ca(OH)₂ required to precipitate the magnesium.
1. mass Mg = 1.00 × 10³ L × 1000 cm³/L × 1.026 g/cm³ × 1272 ppm × 10^–6 ppm^–1 = 1.305 × 10³ g
The concentration is 1.305 g/L. If 99.9% is to be recovered 0.999 × 1.305 g/L = 1.304 g/L will be obtained. The molar concentration is:
1. $\frac{1.304 {g L}^{- 1}}{24.305 {g mol}^{- 1}} = 0.05365 M$
As the Ca(OH)₂ reacts with Mg²⁺ on a 1:1 mol basis, the amount of Ca(OH)₂ required to precipitate 99.9% of the Mg²⁺ in 1 L is:
1. 0.05365 M × 74.09 g/mol Ca(OH)₂ = 3.97 g/L
For treatment of 1000 L, 1000 L × 3.97 g/L = 3.97 × 10³ g = 3.97 kg. However, additional [OH^–] must be added to maintain the equilibrium:
1. ${Mg(OH)}_{2} (s) ⇌ {Mg}^{2+} (a q) + {2OH}^{-} (a q) (K_{sp} = 1.5 \times 10^{- 11})$
When the initial 1.035 g/L is reduced to 0.1% of the original, the molarity is calculated as:
1. $\frac{0.001 \times {1.305 g L}^{- 1}}{24.305 {g mol}^{- 1}} = 5.369 \times 10^{- 5} M$
The added amount of OH^– required is found from the solubility product:
1. [Mg²⁺][OH^–]² = (5.369 × 10^–5)[OH^–]² = 1.5 × 10^–11 = K_sp
2. [OH^–] = 5.29 × 10^–4
Thus an additional $\frac{1}{2}$ × 5.29 × 10^–4 mol (2.65 × 10^–4 mol) of Ca(OH)₂ per liter is required to supply the OH^–. For 1000 L:
1. ${mass Ca(OH)}_{2} = 2.65 \times 10^{- 4} {mol Ca(OH)}_{2} L^{- 1} \times 1.00 \times 10^{3} L \times \frac{74.0946 g}{{mol Ca(OH)}_{2}} = 20 g$
The total Ca(OH)₂ required is 3.97 kg + 0.020 kg = 3.99 kg.
1. K_sp = [La³⁺] ${[{IO}_{3}^{-}]}^{3}$ = $(\frac{1}{3} \times 3.1 \times 10^{- 3})$ (3.1 × 10^–3)³ = (0.0010)(3.0 × 10^–8) = 3.0 × 10^–11
2. K_sp = [Cu²⁺] ${[{IO}_{3}^{-}]}^{2}$ = x(2x)² = 7.4 × 10^–8
3. 4x³ = 7.4 × 10^–8
4. x³ = 1.85 × 10^–8
5. x = 2.64 × 10^–3
6. [Cu²⁺] = 2.6 × 10^–3
7. $[{IO}_{3}^{-}]$ = 2x = 5.3 × 10^–3
1.8 × 10^–5 g Pb(OH)₂
${Mg(OH)}_{2} (s) ⇌ {Mg}^{2+} + {2OH}^{-} K_{sp} = [{Mg}^{2+}] {[{OH}^{-}]}^{2}$
7.1 × 10^–12 = (x)(2x)² = 4x³
x = 1.21 × 10^–4M = [Mg²⁺]
$\frac{1.21 \times 10^{- 4} mol}{L} \times 0.20 0 L \times \frac{{1 mol Mg(OH)}_{2}}{{1 mol Mg}^{2+}} \times \frac{{58.3 g Mg(OH)}_{2}}{{1 mol Mg(OH)}_{2}} = 1.14 \times 10^{- 3} {Mg(OH)}_{2}$
SrCO₃ will form first, since it has the smallest K_sp value it is the least soluble. BaCO₃ will be the last to precipitate, it has the largest K_sp value.

16.2 Lewis Acids and Bases

Explain why the addition of $NH_{3}$ or $HNO_{3}$ to a saturated solution of $Ag_{2} CO_{3}$ in contact with solid $Ag_{2} CO_{3}$ increases the solubility of the solid.
Calculate the cadmium ion concentration, $[Cd^{2 +}]$ , in a solution prepared by mixing 0.100 L of 0.0100 M $Cd (NO_{3})_{2}$ with 1.150 L of 0.100 NH₃(aq).
Explain why addition of $NH_{3}$ or $HNO_{3}$ to a saturated solution of $Cu (OH)_{2}$ in contact with solid $Cu (OH)_{2}$ increases the solubility of the solid.
Using the value of the formation constant for the complex ion $Co (NH_{3})_{6}^{2 +}$ , calculate the dissociation constant.
Using the dissociation constant, K_d = 3.4 $\times$ 10^–15, calculate the equilibrium concentrations of $Zn^{2 +}$ and $OH^{-}$ in a 0.0465-M solution of $Zn (OH)_{4}^{2 -}$ .
Using the dissociation constant, K_d = 2.2 $\times$ 10^–34, calculate the equilibrium concentrations of $Co^{3 +}$ and $NH_{3}$ in a 0.500-M solution of $Co (NH_{3})_{6}^{3 +}$ .
Using the dissociation constant, K_d = 1 $\times$ 10^–44, calculate the equilibrium concentrations of $Fe^{3 +}$ and $CN^{-}$ in a 0.333 M solution of $Fe (CN)_{6}^{3 -}$ .
Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 $\times$ 10^–3 mol of silver bromide.
A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of $Na_{2} S_{2} O_{3} \cdot 5 H_{2} O$ (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as $Ag (S_{2} O_{3})_{2}^{3 -}$ (K_f = 4.7 $\times$ 10¹³)?
We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, $H^{+}$ ) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions.
Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:
1. $CS_{2} + SH^{-} ⟶ HCS_{3}^{-}$
2. $BF_{3} + F^{-} ⟶ BF_{4}^{-}$
3. $I^{-} + SnI_{2} ⟶ SnI_{3}^{-}$
4. $Al (OH)_{3} + OH^{-} ⟶ Al (OH)_{4}^{-}$
5. $F^{-} + SO_{3} ⟶ SFO_{3}^{-}$
Using Lewis structures, write balanced equations for the following reactions:
1. $HCl (g) + PH_{3} (g) ⟶$
2. $H_{3} O^{+} + CH_{3}^{-} ⟶$
3. $CaO + SO_{3} ⟶$
4. $NH_{4}^{+} + C_{2} H_{5} O^{-} ⟶$
Calculate $[{HgCl}_{4}^{2 -}]$ in a solution prepared by adding 0.0200 mol of $NaCl$ to 0.250 L of a 0.100-M $HgCl_{2}$ solution.
In a titration of cyanide ion, 28.72 mL of 0.0100 M $AgNO_{3}$ is added before precipitation begins. [The reaction of $Ag^{+}$ with $CN^{-}$ goes to completion, producing the $Ag (CN)_{2}^{-}$ complex.] Precipitation of solid $AgCN$ takes place when excess $Ag^{+}$ is added to the solution, above the amount needed to complete the formation of $Ag (CN)_{2}^{-}$ . How many grams of $NaCN$ were in the original sample?
What are the concentrations of $Ag^{+}$ , $CN^{-}$ , and $Ag (CN)_{2}^{-}$ in a saturated solution of $AgCN$ ?
In dilute aqueous solution $HF$ acts as a weak acid. However, pure liquid $HF$ (boiling point = 19.5 °C) is a strong acid. In liquid $HF$ , $HNO_{3}$ acts like a base and accepts protons. The acidity of liquid $HF$ can be increased by adding one of several inorganic fluorides that are Lewis acids and accept $F^{-}$ ion (for example, $BF_{3}$ or $SbF_{5}$ ). Write balanced chemical equations for the reaction of pure $HNO_{3}$ with pure $HF$ and of pure $HF$ with $BF_{3}$ . Write the Lewis structures of the reactants and products, and identify the conjugate acid-base pairs.
The simplest amino acid is glycine, $H_{2} NCH_{2} CO_{2} H$ . The common feature of amino acids is that they contain the functional groups: an amine group, $- NH_{2}$ , and a carboxylic acid group, $- CO_{2} H$ . An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, $CH_{3} CO_{2} H$ , and the base strength of the amino group is slightly greater than that of ammonia, $NH_{3}$ .
1. Write the Lewis structures of the ions that form when glycine is dissolved in 1 M $HCl$ and in 1 M $KOH$ .
2. Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the $- NH_{2}$ and $- CO_{2} -$ groups.)
Boric acid, $H_{3} BO_{3}$ , is not a Brønsted-Lowry acid but a Lewis acid.
1. Write an equation for its reaction with water.
2. Predict the shape of the anion thus formed.
3. What is the hybridization on the boron consistent with the shape you have predicted?

Show Selected Solutions

2. Cadmium ions associate with ammonia molecules in solution to form the complex ion $[Cd (NH_{3})_{4}]^{2 +}$ , which is defined by the following equilibrium:

$Cd^{2 +} (a q) + 4 NH_{3} (a q) ⟶ [Cd (NH_{3})_{4}]^{2 +} (a q) K_{f} = 4.0 \times 10^{6}$

The formation of the complex ion requires 4 mol of $NH_{3}$ for each mol of $Cd^{2 +}$ . First, calculate the initial amounts of Cd²⁺ and of NH₃ available for association:

$[Cd^{2 +}] = \frac{(0.100 L) (0.0100 mol L^{- 1})}{0.250 L} = 4.00 \times 10^{- 3} M$

$[NH_{3}] = \frac{(0.150 L) (0.100 mol L^{- 1})}{0.250 L} = 6.00 \times 10^{- 2} M$

For the reaction, 4.00 $\times$ 10^–3 mol/L of $Cd^{2 +}$ would require 4(4.00 $\times$ 10^–3 mol/L) of $NH_{3}$ or a 1.6 $\times$ 10^–2–M solution. Due to the large value of K_f and the substantial excess of $NH_{3}$ , it can be assumed that the reaction goes to completion with only a small amount of the complex dissociating to form the ions. After reaction, concentrations of the species in the solution are

[NH₃] = 6.00 $\times$ 10^–2 mol/L – 1.6 $\times$ 10^–2 mol L^–1 = 4.4 $\times$ 10^–2M

Let x be the change in concentration of $[Cd^{2 +}]$ :

[Cd(NH₃)₄²⁺]

[Cd²⁺]

[NH₃]

Initial concentration (M)

4.00 × 10⁻³

0

4.4 × 10⁻²

Equilibrium (M)

4.00 × 10⁻³ − x

x

4.4 × 10⁻² + 4x

$K_{f} = 4.0 \times 10^{6} = \frac{[Cd {({NH}_{3})}_{4}^{2 +}]}{[{Cd}^{2 +}] {[{NH}_{3}]}^{4}}$

$4.00 \times 10^{6} = \frac{(4.00 \times 10^{- 3} - x)}{(x) {(4.4 \times 10^{- 2} + 4 x)}^{4}}$

As x is expected to be about the same size as the number from which it is subtracted, the entire expression must be expanded and solved, in this case, by successive approximations where substitution of values for x into the equation continues until the remainder is judged small enough. This is a slightly different method than used in most problems. We have:

4.0 $\times$ 10⁶x (4.4 $\times$ 10^–2 + 4x)⁴ = 4.00 $\times$ 10^–3 – x

4.0 $\times$ 10⁶x (3.75 $\times$ 10^–6 + 1.36 $\times$ 10^–3x + 0.186x² + 11.264x³ +256x⁴) = 4.00 $\times$ 10^–3

16x + 5440x² + 7.44 $\times$ 10⁵x³ + 4.51 $\times$ 10⁷x⁴ + 1.024 $\times$ 10⁹x ⁵ = 4.00 $\times$ 10^–3

Substitution of different values x will give a number to be compared with 4.00 $\times$ 10^–3. Using 2.50 $\times$ 10^–4 for x gives 4.35 $\times$ 10^–3 compared with 4.00 $\times$ 10^–3. Using 2.40 $\times$ 10^–4 gives 4.16 $\times$ 10^–3 compared with 4.00 $\times$ 10^–3. Using 2.30 $\times$ 10^–4 gives 3.98 $\times$ 10^–3 compared with 4.00 $\times$ 10^–3. Thus 2.30 $\times$ 10^–4 is close enough to the true value of x to make the difference equal to zero. If the approximation to drop 4x is compared with 4.4 $\times$ 10^–2, the value of x obtained is 2.35 $\times$ 10^–4M.

5.

[Cd(CN)₄²⁻]

[CN⁻]

[Cd²⁺]

Initial concentration (M)

0.250

0

Equilibrium (M)

0.250 − x

4x

x

$K_{d} = \frac{[{Cd}^{2 +}] [{CN}^{-}]}{[Cd {(CN)}_{4}^{2 -}]} = 7.8 \times 10^{- 18} = \frac{x {(4 x)}^{4}}{0.250 - x}$

Assume that x is small when compared with 0.250 M.

256x⁵ = 0.250 $\times$ 7.8 $\times$ 10^–18

x⁵ = 7.617 $\times$ 10^–21

x = $[Cd^{2 +}]$ = 9.5 $\times$ 10^–5M

4x = $[CN^{-}]$ = 3.8 $\times$ 10^–4M

7.

[Co (NH_{3})_{6}^{3 +}]

[Co^{3 +}]

[NH_{3}]

Initial concentration (M)

0.500

0

Equilibrium (M)

0.500 − x

x

6x

$K_{d} = \frac{[{Co}^{2 +}] {[{NH}_{3}]}^{6}}{[Co {({NH}_{3})}_{6}^{3 +}]} = \frac{x {(6 x)}^{6}}{0.500 - x} = 2.2 \times 10^{- 34}$

Assume that x is small when compared with 0.500 M.

4.67 $\times$ 104x⁷ = 0.500 $\times$ 2.2 $\times$ 10^–34

x⁷ = 2.358 $\times$ 10^–39

x = $[Co^{3 +}]$ = 3.0 $\times$ 10^–6M

6x = $[NH_{3}]$ = 1.8 $\times$ 10^–5M

13. The reaction is governed by two equilibria, both of which must be satisfied:

$\begin{array}{l} AgBr (s) ⇌ {Ag}^{+} (a q) + {Br}^{-} (a q); K_{sp} = 3.3 \times 10^{- 13} \\ {Ag}^{+} (a q) + 2 S_{2} O_{3}^{2 -} (a q) ⇌ Ag {(S_{2} O_{3})}_{2}^{3 -} (a q); K_{f} = 4.7 \times 10^{13} \end{array}$

The overall equilibrium is obtained by adding the two equations and multiplying their Ks:

$\frac{[Ag {(S_{2} O_{3})}^{3 -}] [{Br}^{-}]}{{[S_{2} O_{3}^{2 -}]}^{2}} = 15.51$

If all $Ag$ is to be dissolved, the concentration of the complex is the molar concentration of $AgBr$ .

formula mass ( $AgBr$ ) = 187.772 g/mol

$moles present = \frac{0.27 g AgBr}{187.772 g {mol}^{- 1}} = 1.438 \times 10^{- 3} mol$

Let x be the change in concentration of $S_{2} O_{3}^{2 -} :$

[Ag^{+}]

[S_{2} O_{3}^{2 -}]

Initial concentration (M)

0

Equilibrium (M)

\frac{1}{2} x

x

$\frac{(1.438 \times 10^{- 3}) (1.438 \times 10^{- 3})}{x^{2}} = 15.51$

x² = 1.333 $\times$ 10^–7

$x = 3.65 \times 10^{- 4} M = [S_{2} O_{3}^{2 -}]$

The formula mass of $Na_{2} S_{2} O_{3} \cdot 5 H_{2} O$ is 248.13 g/mol. The total $[S_{2} O_{3}^{2 -}]$ needed is:

2(1.438 $\times$ 10^–3) + 3.65 $\times$ 10^–4 = 3.241 $\times$ 10^–3 mol

g(hypo) = 3.241 $\times$ 10^–3 mol $\times$ 248.13 g/mol = 0.80 g

11.

(a)

(b) $H_{3} O^{+} + CH_{3}^{-} ⟶ CH_{4} + H_{2} O$

(c) $CaO + SO_{3} ⟶ CaSO_{4}$

(d) $NH_{4}^{+} + C_{2} H_{5} O^{-} ⟶ C_{2} H_{5} OH + NH_{3}$

19. The equilibrium is: $Ag^{+} (a q) + 2 CN^{-} (a q) ⇌ Ag (CN)_{2}^{-} (a q) K_{f} = 1 \times 10^{20}$

The number of moles of $AgNO_{3}$ added is:

0.02872 L $\times$ 0.0100 mol/L = 2.87 $\times$ 10^–4 mol

This compound reacts with $CN^{-}$ to form $Ag (CN)_{2}^{-}$ , so there are 2.87 $\times$ 10^–4 mol $Ag (CN)_{2}^{-}$ . This amount requires 2 $\times$ 2.87 $\times$ 10^–4 mol, or 5.74 $\times$ 10^–4 mol, of $CN^{-}$ . The titration is stopped just as precipitation of $AgCN$ begins:

$AgCN_{2}^{-} (a q) + Ag^{+} (a q) ⇌ 2 AgCN (s)$

so only the first equilibrium is applicable. The value of K_f is very large.

mol $CN^{-} < [Ag (CN)_{2}^{-}]$ mol $NaCN$ = 2 mol [ $Ag (CN)_{2}^{-}$ ] = 5.74 $\times$ 10^–4 mol

$mass (NaCN) = 5.74 \times 10^{- 4} mol \times \frac{49.007 g}{1 mol} = 0.0281 g$

21. $HNO_{3} (l) + HF (l) ⟶ H_{2} NO_{3}^{+} + F^{-}$ ; $HF (l) + BF_{3} (g) ⟶ H^{+} + BF_{4}$

23. (a) $H_{3} BO_{3} + H_{2} O ⟶ H_{4} BO_{4}^{-} + H^{+}$ ;

(b) First, form a symmetrical structure with the unique atom, B, as the central atom. Then include the 32e^– to form the Lewis structure:

An H atom is bonded to an O atom. The O atom has 2 dots above it and 2 dots below it. The O atom is bonded to a B atom, which has three additional O atoms bonded to it as well. Each of these additional O atoms has 4 dots arranged around it, and is bonded to an H atom. This entire molecule is contained in brackets, to the right of which is a superscripted negative sign.

Because there are four bonds and no lone pair (unshared pair) on B, the electronic and molecular shapes are the same—both tetrahedral.

(c) The tetrahedral structure is consistent with sp³ hybridization.

16.3 Coupled Equilibria

A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?
Calculate the equilibrium concentration of $Zn^{2 +}$ in a 0.30-M solution of $Zn (CN)_{4}^{2 -}$ .
Calculate the equilibrium concentration of $Zn^{2 +}$ in a solution initially with 0.150 M $Zn^{2 +}$ and 2.50 M $CN^{-}$ .
Calculate the $Co^{2 +}$ equilibrium concentration when 0.100 mole of $[Co (NH_{3})_{6}] (NO_{3})_{2}$ is added to a solution with 0.025 M $NH_{3}$ . Assume the volume is 1.00 L.
Calculate the molar solubility of $Sn (OH)_{2}$ in a buffer solution containing equal concentrations of $NH_{3}$ and $NH_{4}^{+}$ .
What is the molar solubility of $CaF_{2}$ in a 0.100-M solution of $HF$ ? K_a for $HF$ = 7.2 $\times$ 10^–4.
What is the molar solubility of $BaSO_{4}$ in a 0.250-M solution of $NaHSO_{4}$ ? K_a for $HSO_{4}^{-}$ = 1.2 $\times$ 10^–2.
What is the molar solubility of $Tl (OH)_{3}$ in a 0.10-M solution of $NH_{3}$ ?
What is the molar solubility of $Pb (OH)_{2}$ in a 0.138-M solution of $CH_{3} NH_{2}$ ?
A solution of 0.075 M $CoBr_{2}$ is saturated with $H_{2} S$ ( $[H_{2} S]$ = 0.10 M). What is the minimum pH at which $CoS$ begins to precipitate?
$CoS (s) ⇌ Co^{2 +} (a q) + S^{2 -} (a q) K_{sp} = 4.5 \times 10^{- 27}$ $H_{2} S (a q) + 2 H_{2} O (l) ⇌ 2 H_{3} O^{+} (a q) + S^{2 -} (a q) K = 1.0 \times 10^{-26}$
A 0.125-M solution of $Mn (NO_{3})_{2}$ is saturated with $H_{2} S$ ( $[H_{2} S]$ = 0.10 M). At what pH does MnS begin to precipitate? $MnS (s) ⇌ Mn^{2 +} (a q) + S^{2 -} (a q) K_{sp} = 4.3 \times 10^{- 22}$ $H_{2} S (a q) + 2 H_{2} O (l) ⇌ 2 H_{3} O^{+} (a q) + S^{2 -} (a q) K = 1.0 \times 10^{- 26}$
Calculate the molar solubility of $BaF_{2}$ in a buffer solution containing 0.20 M $HF$ and 0.20 M $NaF$ .
Calculate the molar solubility of $CdCO_{3}$ in a buffer solution containing 0.115 M $Na_{2} CO_{3}$ and 0.120 M $NaHCO_{3}$
To a 0.10-M solution of $Pb (NO_{3})_{2}$ is added enough $HF$ (g) to make $[HF]$ = 0.10 M.
1. Does $PbF_{2}$ precipitate from this solution? Show the calculations that support your conclusion.
2. What is the minimum pH at which $PbF_{2}$ precipitates?
Calculate the concentration of $Cd^{2 +}$ resulting from the dissolution of $CdCO_{3}$ in a solution that is 0.250 M in $CH_{3} CO_{2} H$ , 0.375 M in $NaCH_{3} CO_{2}$ , and 0.010 M in $H_{2} CO_{3}$ .
Both $AgCl$ and $AgI$ dissolve in $NH_{3}$ .
1. What mass of $AgI$ dissolves in 1.0 L of 1.0 M $NH_{3}$ ?
2. What mass of $AgCl$ dissolves in 1.0 L of 1.0 M $NH_{3}$ ?
Calculate the volume of 1.50 M $CH_{3} CO_{2} H$ required to dissolve a precipitate composed of 350 mg each of $CaCO_{3}$ , $SrCO_{3}$ , and $BaCO_{3}$ .
Even though $Ca (OH)_{2}$ is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of $Ca (OH)_{2}$ ?
What mass of $NaCN$ must be added to 1 L of 0.010 M $Mg (NO_{3})_{2}$ in order to produce the first trace of $Mg (OH)_{2}$ ?
Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.)
The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. Solve the following problem: $MgF_{2} (s) ⇌ Mg^{2 +} (a q) + 2 F^{-} (a q)$ In a saturated solution of $MgF_{2}$ at 18 °C, the concentration of $Mg^{2 +}$ is 1.21 $\times$ 10^–3M. The equilibrium is represented by the equation above.
1. Write the expression for the solubility-product constant, K_sp, and calculate its value at 18 °C.
2. Calculate the equilibrium concentration of $Mg^{2 +}$ in 1.000 L of saturated $MgF_{2}$ solution at 18 °C to which 0.100 mol of solid $KF$ has been added. The $KF$ dissolves completely. Assume the volume change is negligible.
3. Predict whether a precipitate of $MgF_{2}$ will form when 100.0 mL of a 3.00 $\times$ 10^–3–M solution of $Mg (NO_{3})_{2}$ is mixed with 200.0 mL of a 2.00 $\times$ 10^–3–M solution of $NaF$ at 18 °C. Show the calculations to support your prediction.
4. At 27 °C the concentration of $Mg^{2 +}$ in a saturated solution of $MgF_{2}$ is 1.17 $\times$ 10^–3M. Is the dissolving of $MgF_{2}$ in water an endothermic or an exothermic process? Give an explanation to support your conclusion.
Which of the following compounds, when dissolved in a 0.01-M solution of $HClO_{4}$ , has a solubility greater than in pure water: $AgBr$ , $BaF_{2}$ , $Ca_{3} (PO_{4})_{3}$ , $ZnS$ , $PbI_{2}$ ? Explain your answer.
What is the effect on the amount of solid $Mg (OH)_{2}$ that dissolves and the concentrations of $Mg^{2 +}$ and $OH^{-}$ when each of the following are added to a mixture of solid $Mg (OH)_{2}$ and water at equilibrium?
1. $MgCl_{2}$
2. $KOH$
3. $HClO_{4}$
4. $NaNO_{3}$
5. $Mg (OH)_{2}$
What is the effect on the amount of $CaHPO_{4}$ that dissolves and the concentrations of $Ca^{2 +}$ and $HPO_{4}^{-}$ when each of the following are added to a mixture of solid $CaHPO_{4}$ and water at equilibrium?
1. $CaCl_{2}$
2. $HCl$
3. $KClO_{4}$
4. $NaOH$
5. $CaHPO_{4}$
Identify all chemical species present in an aqueous solution of $Ca_{3} (PO_{4})_{2}$ and list these species in decreasing order of their concentrations. (Hint: Remember that the $PO_{4}^{3 -}$ ion is a weak base.)
A volume of 50 mL of 1.8 M $NH_{3}$ is mixed with an equal volume of a solution containing 0.95 g of $MgCl_{2}$ . What mass of $NH_{4} Cl$ must be added to the resulting solution to prevent the precipitation of $Mg (OH)_{2}$ ?

Show Selected Solutions

7. Find the amount of ${SO}_{4}^{2 -}$ present from K_a for the equilibrium:

${HSO}_{4}^{-} + H_{2} O ⟶ H_{3} O^{+} + {SO}_{4}^{2 -}$

Let x be the change in $[{SO}_{4}^{2 -}]$ :

$K_{a} = \frac{[H_{3} O^{+}] [{SO}_{4}^{2 -}]}{[{HSO}_{4}^{-}]} = \frac{x^{2}}{0.250 - x} = 1.2 \times 10^{- 2}$

Because K_a is too large to disregard x in the expression 0.250 – x, we must solve the quadratic equation:

x² + 1.2 $\times$ 10^–2x – 0.250(1.2 $\times$ 10^–2) = 0

$x = \frac{- 1.2 \times 10^{- 2} \pm \sqrt{{(1.2 \times 10^{- 2})}^{2} + 4 (3.0 \times 10^{-3})}}{2} = \frac{- 1.2 \times 10^{- 2} \pm 0.11}{2} = 0.049 M$

K_sp = [Ba²⁺] $[{SO}_{4}^{2 -}]$ = [Ba²⁺](0.049) = 1.08 $\times$ 10^–10

[Ba²⁺] = 2.2 $\times$ 10^–9 (molar solubility)

9. ${CH}_{3} {NH}_{2} + H_{2} O ⇌ {CH}_{3} {NH}_{3}^{+} + {OH}^{-}$

$K_{b} = \frac{[{CH}_{3} {NH}_{3}^{+}] [{OH}^{-}]}{[{CH}_{3} {NH}_{2}]} = \frac{(x) (x)}{0.138 - x} = 4.4 \times 10^{- 4}$

Solve the quadratic equation using the quadratic formula:

x² + 4.4 $\times$ 10^–4x – 0.138(4.4 $\times$ 10^–4) = 0

$x = \frac{- 4.4 \times 10^{- 4} \pm \sqrt{{(4.4 \times 10^{- 4})}^{2} + 4 (6.07 \times 10^{-5})}}{2} = \frac{- 4.4 \times 10^{-4} \pm \sqrt{2.43 \times 10^{- 4}}}{2} = \frac{0.0152}{2} = 7.6 \times 10^{- 3} M$

K_sp = [Pb²⁺][OH^–]² = [Pb²⁺](7.6 $\times$ 10^–3)² = 2.8 $\times$ 10^–16

[Pb²⁺] = 4.8 $\times$ 10^–12 (molar solubility)

11. Two equilibria are in competition for the ions and must be considered simultaneously. Precipitation of $MnS$ will occur when the concentration of $S^{2 -}$ in conjunction with 0.125 M Mn²⁺ exceeds the K_sp of MnS. The $Math input error$ must come from the ionization of $H_{2} S$ as defined by the equilibrium:

$H_{2} S (a q) + 2 H_{2} O (l) ⟶ 2 H_{3} O^{+} (a q) + S^{2 -} (a q)$

$\frac{{[H_{3} O^{+}]}^{2} [S^{2 -}]}{[H_{2} S]} = K_{1} K_{2} (H_{2} S) = 1.0 \times 10^{-26}$

As a saturated solution of $H_{2} S$ is 0.10 M, this later expression becomes:

${[H_{3} O^{+}]}^{2} [S^{2 -}] = 1.0 \times 10^{- 27}$

From the equilibrium of $MnS$ , the minimum concentration of $S^{2 -}$ required to cause precipitation is calculated as:

$MnS (s) ⟶ {Mn}^{2+} (a q) + S^{2 -} (a q)$

K_sp = $[Mn^{2 +}] [S_{2}^{-}]$ = 4.3 $\times$ 10^–22

$[S^{2 -}] = \frac{4.3 \times 10^{- 22}}{0.125} = 3.44 \times 10^{- 21}$

This amount of $S^{2 -}$ will exist in solution at a pH defined by the $H_{2} S$ equilibrium:

${[H_{3} O^{+}]}^{2} (3.44 \times 10^{- 21}) = 1.0 \times 10^{- 27}$
${[H_{3} O^{+}]}^{2} = 2.91 \times 10^{- 7}$
$[H_{3} O^{+}] = 5.39 \times 10^{- 4} M$
$pH = - log [H_{3} O^{+}] = 3.27$

13. Three equilibria are involved:

$H_{2} {CO}_{3} + H_{2} O ⟶ H_{3} O^{+} + {HCO}_{3}^{-} K_{a_{1}} = 4.3 \times 10^{- 7}$
${HCO}_{3}^{-} + H_{2} O ⟶ H_{3} O^{+} + {CO}_{3}^{2 -} K_{a_{2}} = 7 \times 10^{- 11}$
${CdCO}_{3} ⟶ {Cd}^{2+} + {CO}_{3}^{2 -} K_{sp} = 2.5 \times 10^{- 14}$

First, find the pH of the buffer from the Henderson-Hasselbach equation. Then find $[H_{3} O^{+}]$ :

$pH = p K_{a} + \log \frac{[{CO}_{3}^{2 -}]}{[{HCO}_{3}^{-}]}$

10.155 + log $\frac{0.115}{0.120}$ = 10.155 – 0.018 = 10.137

In this case, several more significant figures are carried than justified so that the value of the log ratio is meaningful:

$[H_{3} O^{+}] = 7.3 \times 10^{- 11}$

Now, find the concentration of ${CO}_{3}^{2 -}$ present in the buffer solution. Next, using $[{CO}_{3}^{2 -}]$ and K_sp, calculate the concentration of $Cd^{2 +}$ . This latter value represents the molar solubility. From $K_{a_{1}}$ determine $[H_{2} CO_{3}]$ :

$K_{a_{1}} = 4.3 \times 10^{- 7} = \frac{[H_{3} O^{+}] [{HCO}_{3}^{-}]}{[H_{2} {CO}_{3}]} = \frac{(7.3 \times 10^{- 11}) (0.120)}{[H_{2} {CO}_{3}]}$

[H₂CO₃] = 2.05 $\times$ 10^–5

From $K_{a_{1}} K_{a_{2}}$ , find the concentration of ${CO}_{3}^{2 -}$ :

$\begin{aligned} K_{a_{1}} K_{a_{2}} & = & (4.3 \times 10 - 7) (7 \times 10 - 11) = \frac{[{HCO}_{3}^{-}] [H_{3} O^{+}]}{[H_{2} {CO}_{3}]} \times \frac{[{CO}_{3}^{2 -}] [H_{3} O^{+}]}{[H_{2} {CO}_{3}]} = \frac{[H_{3} O^{+}] [{CO}_{3}^{2 -}]}{[H_{2} {CO}_{3}]} = \frac{{(7.3 \times 10^{- 11})}^{2} \times [{CO}_{3}^{2 -}]}{2.04 \times 10^{- 5}} \\ = & 3 \times 10^{- 17} \end{aligned}$

$[{CO}_{3}^{2 -}]$ = 0.115 M

K_sp = $[Cd^{2 +}]$ $[{CO}_{3}^{2 -}]$ = [Cd²⁺](0.115) = 3 $\times$ 10^–13

$[{Cd}^{2+}] = \frac{3 \times 10^{- 13}}{0.115} = 3 \times 10^{- 12} M$

15. For K_sp, (CdCO₃) = [Cd²⁺] $[{CO}_{3}^{2 -}]$ = 2.5 $\times$ 10^–14, the amount of ${CO}_{3}^{2 -}$ is governed by K_a of H₂CO₃, 4.3 $\times$ 10^–7, and K_a of ${HCO}_{3}^{-}$ , 7 $\times$ 10^–11, and $[H_{3} O^{+}]$ by K_a of acetic acid. First, calculate the $[H_{3} O^{+}]$ from K_a of acetic acid (HOAc):

$[H_{3} O^{+}] = \frac{K_{a} [HOAc]}{[{OAc}^{-}]} = \frac{1.8 \times 10^{- 5} (0.250)}{(0.375)} = 1.2 \times 10^{- 5} M$

From this and K_a for $H_{2} CO_{3}$ , calculate $[{CO}_{3}^{2 -}]$ present. As $[H_{3} O^{+}]$ is fixed by K_a of acetic acid:

$K_{a} (H_{2} {CO}_{3}) = \frac{[H_{3} O^{+}] [{HCP}_{3}^{-}]}{[H_{2} {CO}_{3}^{-}]}$

$[H_{2} {CO}_{3}^{-}] = \frac{K_{a} [H_{2} {CO}_{3}]}{[H_{3} O^{+}]} = \frac{4.3 \times 10^{- 7} [0.010]}{[1.2 \times 10^{- 5}]} = 3.58 \times 10^{- 4} M$

From $K_{a} ({HCO}_{3}^{-})$ we obtain $[{CO}_{3}^{2 -}]$ :

$K_{a} = 7 \times 10^{- 11} = \frac{[H_{3} O^{+}] [{CO}_{3}^{2 -}]}{[{HCO}_{3}^{-}]} = \frac{1.2 \times 10^{- 17} [{CO}_{3}^{2 -}]}{[3.58 \times 10^{- 4}]}$

$[{CO}_{3}^{2 -}] = \frac{7 \times 10^{- 11} \times 3.58 \times 10^{- 4}}{1.2 \times 10^{- 5}} = 2.09 \times 10^{- 9}$

From the solubility product:

K_sp = [Cd²⁺] $[{CO}_{3}^{2 -}]$ = 2.5 $\times$ 10^–14 = Cd²⁺ (2.09 $\times$ 10^–9)

$[{Cd}^{2+}] = \frac{2.5 \times 10^{- 14}}{2.09 \times 10^{- 9}} = 1 \times 10^{- 5} M$

17. ${CaCO}_{3} : 0.350 g \times \frac{1 mol}{100.09 g} = 3.50 \times 10^{- 3} mol$

${SrCO}_{3} : 0.350 g \times \frac{1 mol}{147.63 g} = 2.37 \times 10^{- 3} mol$

${BaCO}_{3} : 0.350 g \times \frac{1 mol}{197.34 g} = 1.77 \times 10^{- 3} mol$

Total: 7.64 $\times$ 10^–3 mol
$CaCO_{3}$ : K_sp = 4.8 $\times$ 10^–9
$SrCO_{3}$ : K_sp = 9.42 $\times$ 10^–10
$BaCO_{3}$ : K_sp = 8.1 $\times$ 10^–9

Solubilities are approximately equal.

When the three solids dissolve, 7.64 $\times$ 10^–3 mol of metal ions and 7.64 $\times$ 10^–3 mol of ${CO}_{3}^{2 -}$ are initially present. The ${CO}_{3}^{2 -}$ reacts with $CH_{3} CO_{2} H$ to form ${HCO}_{3}^{-}$ :

${CO}_{3}^{2 -} (a q) + {CH}_{3} {CO}_{2} H (a q) ⇌ {HCO}_{3}^{-} (a q) + {CH}_{3} {CO}_{2}^{-} (a q)$

$K = \frac{[{HCO}_{3}^{-}] [{CH}_{3} {CO}_{2}^{-}]}{[{CO}_{3}^{2 -}] [{CH}_{3} {CO}_{2} H]} \times \frac{[H_{3} O^{+}]}{[H_{3} O^{+}]} = \frac{K_{a} (for {CH}_{3} {CO}_{2} H)}{K_{a} (for {HCO}_{3}^{-})} = \frac{1.8 \times 10^{- 5}}{7 \times 10^{- 11}} = 2.6 \times 10^{5}$

This K value is large, so virtually all ${CO}_{3}^{2 -}$ undergoes this reaction and approximately 7.64 $\times$ 10^–3 mol of ${HCO}_{3}^{-}$ forms. The ${HCO}_{3}^{-}$ reacts with CH₃CO₂H:

${HCO}_{3}^{-} (a q) + {CH}_{3} {CO}_{2} H (a q) ⟶ H_{2} {CO}_{3} (a q) + {CH}_{3} {CO}_{2}^{-} (a q)$

$K = \frac{K_{a} (for {CH}_{3} {CO}_{2} H)}{K_{a} (for {HCO}_{3}^{-})} = \frac{1.8 \times 10^{- 5}}{4.3 \times 10^{- 7}} = 42$

This reaction is virtually complete as well (K is large).

For each mol of ${CO}_{3}^{2 -}$ produced, 2 mol of $CH_{3} CO_{2} H$ is required for conversion to $H_{2} CO_{3}$ . Thus 15.3 $\times$ 10^–3 mol $CH_{3} CO_{2} H$ is needed:

$Volume = \frac{15.3 \times 10^{- 3} mol}{1.50 mol L^{- 1}} = 0.0102 L (10.2 mL)$

19. There are two equilibria involved: $\begin{array}{l} Mg {(OH)}_{2} (s) ⇌ {Mg}^{2+} (a q) + 2 {OH}^{-} (a q) K_{sp} = 1.5 \times 10^{- 11} \\ {CN}^{-} (a q) + H_{2} O (l) ⇌ HCN (a q) + {OH}^{-} (a q) (K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 \times 10^{- 14}}{4 \times 10^{- 10}} = 2.5 \times 10^{- 5}) \end{array}$ The Mg(NO₃)₂ dissolves, and [Mg²⁺] = 0.010 M.

$[Mg^{2 +}] [OH^{-}]^{2}$ = K_sp = 1.5 $\times$ 10^–11

(0.010)[OH^–]² = 1.5 $\times$ 10^–11

[OH^–] = 3.9 $\times$ 10^–5M

We need to add enough $CN^{-}$ to make $[OH^{-}]$ = 3.9 $\times$ 10^–5M. Both $OH^{-}$ and $HCN$ come from $CN^{-}$ , so $[OH^{-}]$ = $[HCN]$ :

$\begin{array}{l} \frac{[HCN] [{OH}^{-}]}{[{CN}^{-}]} = K_{b} = 2 \times 10^{- 5} \\ \frac{{(3.9 \times 10^{- 5})}^{2}}{[{CN}^{-}]} = 2.5 \times 10^{- 5} \end{array}$

$[CN^{-}]$ = 6.1 $\times$ 10^–5M

mol $NaCN$ = mol $HCN$ + mol $CN^{-}$ = 3.9 $\times$ 10^–5 + 6.1 $\times$ 10^–5 = 1.0 $\times$ 10^–4

$mass (NaCN) = 1.0 \times 10^{- 4} mol \times \frac{49.007 g}{1 mol} = 5 \times 10^{- 3} g$

21. The answers are as follows:

K_sp = [Mg²⁺][F^–]² = (1.21 $\times$ 10^–3)(2 $\times$ 1.21 $\times$ 10^–3)² = 7.09 $\times$ 10^–9
K_sp = [Mg²⁺][F^–]² = [x][0.100 + 2x]² = 7.09 $\times$ 10^–9Assume that 2x is small when compared with 0.100 M.
0.100x = 7.09 $\times$ 10^–9x = [MgF₂] = 7.09 $\times$ 10^–7M
The value 7.09 $\times$ 10^–7M is quite small when compared with 0.100 M, so the assumption is valid.
Determine the concentration of $Mg^{2 +}$ and $F^{-}$ that will be present in the final volume. Compare the value of the ion product $[Mg^{2 +}] [F^{-}]^{2}$ with K_sp. If this value is larger than K_sp, precipitation will occur.
0.1000 L $\times$ 3.00 $\times$ 10^–3M $Mg (NO_{3})_{2}$ = 0.3000 L $\times$ M $Mg (NO_{3})_{2}$ M $Mg (NO_{3})_{2}$ = 1.00 $\times$ 10^–3M
0.2000 L $\times$ 2.00 $\times$ 10^–3M NaF = 0.3000 L $\times$ M $NaF$
M $NaF$ = 1.33 $\times$ 10^–3M
ion product = (1.00 $\times$ 10^–3)(1.33 $\times$ 10^–3)² = 1.77 $\times$ 10^–9This value is smaller than K_sp, so no precipitation will occur.
$MgF_{2}$ is less soluble at 27 °C than at 18 °C. Because added heat acts like an added reagent, when it appears on the product side, the Le Châtelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.

23. Effect on amount of solid $Mg (OH)_{2}$ , $[Mg^{2 +}]$ , $[OH^{-}]$ :

increase, increase, decrease;
increase, decrease, increase;
decrease, increase, decrease;
no effect predicted;
increase, no effect, no effect

25. $[H_{2} O] > [Ca^{2 +}] > [PO_{4}^{3 -}] > HPO_{4}^{2 -} = [OH^{-}] > [HPO_{4}^{-}]$

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Chemistry Fundamentals Copyright © by Dr. Julie Donnelly, Dr. Nicole Lapeyrouse, and Dr. Matthew Rex is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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