Chapter 16 Practice

3. There is no change. A solid has an activity of 1 whether there is a little or a lot.
5. 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.
7. [latex]\ce{CaF2}[/latex], [latex]\ce{MnCO3}[/latex], and [latex]\ce{ZnS}[/latex]; 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
9. The answers for each compound are as follows:
   1. [latex]\text{LaF}_{3}\left(s\right)\rightleftharpoons\text{La}^{3+}\left(aq\right)+{3F}^{-}\left(aq\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{sp}}=\left[\text{La}^{3+}\right]\left[\text{F}^{-}\right]^{3}[/latex]
   2. [latex]\text{CaCO}_{3}\left(s\right)\rightleftharpoons\text{Ca}^{2+}\left(aq\right)+\text{CO}_{3}^{2-}\left(aq\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{sp}}=\left[\text{Ca}^{2+}\right]\left[\text{CO}_{3}^{2-}\right][/latex]
   3. [latex]\text{Ag}_{2}\text{SO}_{4}\left(s\right)\rightleftharpoons2\text{Ag}^{+}\left(aq\right)+\text{SO}_{4}^{2-}\left(aq\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{sp}}=\left[\text{Ag}^{+}\right]^{2}\left[\text{SO}_{4}^{2-}\right][/latex]
   4. [latex]\text{Pb}\left(\text{OH}\right)_{2}\left(s\right)\rightleftharpoons\text{Pb}^{2+}\left(aq\right)+2\text{OH}^{-}\left(aq\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{sp}}=\left[\text{Pb}^{2+}\right]\left[\text{OH}^{-}\right]^{2}[/latex]
11. 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. [latex]\ce{BaSeO4}[/latex]
       1. [latex]\frac{0.{\text{118 g L}}^{-1}}{280.{\text{28 g mol}}^{-1}}=\text{4.21}\times {10}^{-4}\text{}M[/latex]
       2. K = [Ba²⁺] [latex]\left[{\text{SeO}}_{4}^{2-}\right][/latex] = (4.21 × 10^–4)(4.21 × 10^–4) = 1.77 × 10^–7
    2. [latex]\ce{Ba(BrO3)2·H2O}[/latex]:
       1. [latex]\frac{3.0{\text{g L}}^{-1}}{{\text{411.147 g mol}}^{-1}}=\text{7.3}\times {10}^{-3}\text{}M[/latex]
       2. K = [Ba²⁺] [latex]{\left[{\text{BrO}}_{3}{}^{-}\right]}^{2}[/latex] = (7.3 × 10^–3)(2 × 7.3 × 10^–3)² = 1.6 × 10^–6
    3. [latex]\ce{NH4MgAsO4·6H2O}[/latex]:
       1. [latex]\frac{0.{\text{38 g L}}^{-1}}{{\text{289.3544 g mol}}^{-1}}=\text{1.3}\times {10}^{-3}\text{}M[/latex]
       2. K = [latex]\left[{\text{NH}}_{4}{}^{\text{+}}\right][/latex] [Mg²⁺] [latex]\left[{\text{AsO}}_{4}{}^{3-}\right][/latex] = (1.3 × 10^–3)³ = 2.2 × 10^–9
    4. La₂(MoO₄)₃:
       1. [latex]\frac{0.0{\text{179 g L}}^{-1}}{{\text{757.62 g mol}}^{-1}}=\text{2.36}\times {10}^{-5}\text{}M[/latex]
       2. K = [La³⁺]² [latex]{\left[{\text{MoO}}_{4}{}^{2-}\right]}^{3}[/latex] = (2 × 2.36 × 10^–5)²(3 × 2.36 × 10^–5)³ = 2.228 × 10^–9 × 3.549 × 10^–13 = 7.91 × 10^–22
13. Let x be the molar solubility.
    1. K_sp = [latex]\left[{\text{K}}^{\text{+}}\right]\left[{\text{HC}}_{4}{\text{H}}_{4}{\text{O}}_{6}^{-}\right][/latex] = 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 = [latex]{\left[{\text{Ag}}^{\text{+}}\right]}^{4}\left[{\text{Fe(CN)}}_{6}{}^{\text{4-}}\right][/latex] = 1.55 × 10^–41 = (4x)⁴x = 256x⁵, x = 2.27 × 10^–9M;
    4. K_sp = [latex]\left[{\text{Hg}}_{2}{}^{\text{2+}}\right]{\left[{\text{I}}^{-}\right]}^{2}[/latex] = 4.5 × 10^–29 = [x][2x]² = 4x³, x = 2.2 × 10^–10M
15. The correct concentrations are as follows:
    1. K_sp = 1.8 × 10^–10 = [latex]\ce{[Ag+][Cl-]}[/latex] = x(x + 0.025), where x = [Ag⁺]. Assume that x is small when compared with 0.025 and therefore ignore it:
       • [latex]x=\frac{1.8\times {10}^{-10}}{0.025}=7.2\times {10}^{-9}\text{}M=\left[{\text{Ag}}^{\text{+}}\right],\text{}\left[{\text{Cl}}^{-}\right]=0.02\text{5}M[/latex]
       • Check: [latex]\frac{7.2\times {10}^{-9}\text{}M}{0.025\text{}M}\times \text{100%}=\text{2.9}\times {10}^{-5}%[/latex], an insignificant change;
    2. K_sp = 3.9 × 10^–11 = [latex]\ce{[Ca^2+][F-]2}[/latex] = x(2x + 0.00133 M)², where x = [latex]\ce{[Ca^2+]}[/latex]. Assume that x is small when compared with 0.0013 M and disregard it:
       • [latex]x=\frac{\text{3.9}\times {10}^{-11}}{{\left(0.00133\right)}^{2}}=2.2\times {10}^{-5}\text{}M=\left[{\text{Ca}}^{\text{2+}}\right],\text{}\left[{\text{F}}^{-}\right]=0.0013\text{}M[/latex]
       • Check: [latex]\frac{\text{2.25}\times {10}^{-5}\text{}M}{0.00133\text{}M}\times \text{100%}=\text{1.69%}[/latex]. This value is less than 5% and can be ignored.
    3. Find the concentration of [latex]\ce{K2SO4}[/latex]:
       • [latex]\frac{\text{19.50 g}}{174.2{\text{60 g mol}}^{-1}}=\text{0.1119 mol}[/latex]
         [latex]\frac{0.1119\text{mol}}{0.5\text{L}}=\text{0.2238}M={\text{[SO}}_{4}{}^{2-}][/latex]
         K_sp = 1.18 × 10^–18 = [Ag⁺]² [latex]{\text{[SO}}_{4}{}^{2-}][/latex] = 4x²(x + 0.2238)
         [latex]{x}^{2}=\frac{1.18\times {10}^{-18}}{4\left(0.2238\right)}=\text{1.32}\times {10}^{-18}[/latex]
         x = 1.15 × 10^–9[Ag⁺] = 2x = 2.30 × 10^–9M
       • Check: [latex]\frac{1.15\times {10}^{-9}}{0.2238}\times 100%=\text{5.14}\times {10}^{-7}[/latex]; the condition is satisfied.
    4. Find the concentration of [latex]\ce{OH-}[/latex] 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:
         [latex]x=\frac{4.5\times {10}^{-17}}{{\left(2.8\times {10}^{-3}\right)}^{2}}=\text{5.7}\times 10 - 12\text{}M={\text{[Zn}}^{\text{2+}}\text{]}[/latex]
       • Check: [latex]\frac{5.7\times {10}^{-12}}{2.8\times {10}^{-3}}\times \text{100%}=\text{2.0}\times {10}^{-7}%[/latex]; 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%.
17. The answers are as follows:
    1. K_sp = 1.9 × 10^–4 = [latex]\ce{[Ti+][Cl-]}[/latex]; Let x = [latex]\ce{[Cl-]}[/latex]:1.9 × 10^–4 = (x = 0.025)x
       1. Assume that x is small when compared with 0.025:
          1. [latex]x=\frac{1.9\times {10}^{-4}}{0.025}=7.6\times {10}^{-3}\text{}M[/latex]
       2. Check: [latex]\frac{7.6\times {10}^{-3}}{0.025}\times \text{100%}=\text{30%}[/latex]
       3. This value is too large to drop x. Therefore solve by using the quadratic equation:
          1. [latex]\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]
          2. x² + 0.025x – 1.9 × 10^–4 = 0
          3. [latex]\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\text{61}M\end{array}[/latex]
       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:
       2. Let x = [Ba²⁺]
       3. 1.7 × 10^–6 = x(x + 0.0313)² = 1.7 × 10^–3M
       4. Check: [latex]\frac{1.7\times {10}^{-3}}{0.0313}\times \text{100%}=\text{5.5%}[/latex]
       5. 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:
       6. x′(5.4 × 10^–5 + 0.0313)² = 1.7 × 10^–3 or
       7. [latex]x\prime =\frac{1.7\times {10}^{-6}}{1.089\times {10}^{-3}}=\text{1.6}\times {10}^{-3}[/latex]
       8. A third approximation using this last calculation is as follows:
       9. x′′(1.7 × 10^–3 + 0.0313)² = 1.7 × 10^–6 or
       10. [latex]x\prime \prime =\frac{1.7\times {10}^{-6}}{1.089\times {10}^{-3}}=\text{1.6}\times {10}^{-3}[/latex]
       11. This value is well within 5% and is acceptable.
       12. [Ba²⁺] = 1.6 × 10^–3M
       13. [F^–] = (1.6 × 10^–3 + 0.0313) = 0.0329 M;
    3. Find the molar concentration of the [latex]\ce{Mg(NO3)2}[/latex]. The molar mass of [latex]\ce{Mg(NO3)2}[/latex] is 148.3149 g/mol. The number of moles is [latex]\frac{\text{8.156 g}}{148.314{\text{9 g mol}}^{-1}}=\text{0.05499 mol}[/latex]
       1. [latex]M=\frac{\text{0.05499 mol}}{\text{2.250 L}}=\text{0.02444 M}[/latex]
       2. Let x = [latex]\left[{\text{C}}_{2}{\text{O}}_{4}^{2-}\right][/latex] and assume that x is small when compared with 0.02444 M.
       3. K_sp = 8.6 × 10^–5 = [Mg²⁺] [latex]\left[{\text{C}}_{2}{\text{O}}_{4}^{2-}\right][/latex] = (x)(x + 0.02444)
       4. 0.02444x = 8.6 × 10^–5
       5. x = [latex]\left[{\text{C}}_{2}{\text{O}}_{4}^{2-}\right][/latex] = 3.5 × 10^–3
       6. Check: [latex]\frac{3.5\times {10}^{-3}}{0.02444}\times \text{100%}=\text{14%}[/latex]
       7. This value is greater than 5%, so the quadratic equation must be used to solve for x:
       8. [latex]\begin{array}{l}{}\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\\{x}^{2}+0.02444x - 8.6\times {10}^{-5}=0\\x=\frac{-0.02444\pm \sqrt{5.973\times {10}^{-4}+3.44\times {10}^{-4}}}{2}=\text{}\frac{-0.02444\pm 0.03068}{2}\end{array}[/latex]
       9. (Use only the positive answer for physical sense.)
       10. x′ = [latex]\left[{\text{C}}_{2}{\text{O}}_{4}{}^{2-}\right][/latex] = 3.5 × 10^–3M
       11. [Mg²⁺] = 3.1 × 10^–3 + 0.02444 = 0.0275 M
       12. (d) pH = 12.700; pOH = 1.300
       13. [latex]\ce{[OH-]}[/latex] = 0.0501 M; Let x = [latex]\ce{[Ca^2+]}[/latex]
       14. K_sp = 7.9 × 10^–6 = [latex]\ce{[Ca^2+][OH-]^2}[/latex] = (x)(x + 0.050)²
       15. Assume that x is small when compared with 0.050 M:
       16. x = [latex]\ce{[Ca^2+]}[/latex] = 3.15 × 10^–3 (one additional significant figure is carried)
       17. Check: [latex]\frac{3.15\times {10}^{-3}}{0.050}\times \text{100%}=\text{6.28%}[/latex]
       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. [latex]x\prime =\frac{7.9\times {10}^{-6}}{2.836\times {10}^{-3}}=\text{2.8}\times {10}^{-3}[/latex]
       21. [latex]\ce{[Ca^2+]}[/latex] = 2.8 × 10^–3M
       22. [latex]\ce{[OH-]}[/latex] = (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%.
19. The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change.
21. [latex]\ce{Ca(OH)2: [Ca^2+][OH-]^2}[/latex] = 7.9 × 10^–6
    1. Let x be [latex]\ce{[Ca^2+]}[/latex] = molar solubility; then [latex]\ce{[OH-]}[/latex] = 2x
    2. K_sp = x(2x)² = 4x³ = 7.9 × 10^–6
    3. x³ = 1.975 × 10^–6
    4. x = 0.013 M
    5. [latex]\ce{CaCO3: [Ca^2+][CO3^2-]}[/latex] = 4.8 × 10^–9
    6. Let x be [latex]\ce{[Ca^2+]}[/latex]; then [latex]{\text{[CO}}_{3}^{2-}][/latex] = [Ca²⁺] = molar solubility
    7. K_sp = x² = 4.8 × 10^–9
    8. x = 6.9 × 10^–5M
    9. [latex]\ce{CaSO4·2H2O: [Ca^2+][SO4^2-]}\text{\cdot }\ce{[H2O]^2}[/latex] = 2.4 10^–5
    10. Let x be [latex]\ce{[Ca^2+]}[/latex] = molar solubility = [latex]{\text{[SO}}_{4}^{2-}][/latex]; then [latex]\ce{[H2O]}[/latex] = 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. [latex]\ce{CaC2O4·H2O: [Ca^2+][C2O4^2-][H2O]}[/latex] = 2.27 × 10^–9
    16. Let x be [latex]\ce{[Ca^2+]}[/latex] = molar solubility = [latex]\left[{\text{C}}_{2}{\text{O}}_{4}^{2-}\right][/latex] = [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 [latex]{\left[{\text{PO}}_{4}^{3-}\right]}^{2}[/latex] = 1 × 10^–25
    21. Upon solution there are three Ca²⁺ and two [latex]{\text{PO}}_{4}^{3-}[/latex] ions. Let the concentration of Ca²⁺ formed upon solution be x. Then [latex]\frac{2}{3}x[/latex] is the concentration of [latex]{\text{PO}}_{4}{}^{\text{3-}}:[/latex]
    22. [latex]{x}^{3}{\left(\frac{2}{3}x\right)}^{2}{x}^{3}=1\times {10}^{-25}=\text{0.4444}{x}^{5}[/latex]
    23. x = 1 × 10^–5M = [Ca²⁺]
    24. The solubility is then one-third the concentration of Ca²⁺, or 4 × 10^–6.
23. First, find the concentration in a saturated solution of CaSO₄. Before placing the CaSO₄ in water, the concentrations of Ca²⁺ and [latex]{\text{SO}}_{4}^{2-}[/latex] are 0. Let x be the change in concentration of Ca²⁺, which is equal to the concentration of [latex]{\text{SO}}_{4}^{2-}:[/latex]
    1. K_sp = [Ca²⁺] [latex]\left[{\text{SO}}_{4}^{2-}\right][/latex] = 2.4 × 10^–5
    2. x × x = x² = 2.4 × 10^–5
    3. [latex]x=\sqrt{2.4\times {10}^{-5}}[/latex]
    4. x = 4.9 × 10^–3M = [latex]\left[{\text{SO}}_{4}^{2-}\right][/latex] = [Ca²⁺]
    5. Since this concentration is higher than 2.60 × 10^–3M, “gyp” water does not meet the standards.
25. The amount of CaSO₄·2H₂O that dissolves is limited by the presence of a substantial amount of [latex]{\text{SO}}_{4}^{2-}[/latex] already in solution from the 0.010 M [latex]{\text{SO}}_{4}{}^{-}[/latex]. This is a common-ion problem. Let x be the change in concentration of Ca²⁺ and of [latex]{\text{SO}}_{4}^{2-}[/latex] that dissociates from CaSO₄:
    1. [latex]{\text{CaSO}}_{4}\left(s\right)\longrightarrow {\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)[/latex]
    2. K_sp = [Ca²⁺] [latex]{\text{[SO}}_{4}^{2-}][/latex] = 2.4 × 10^–5
    3. Addition of 0.010 M [latex]{\text{SO}}_{4}^{2-}[/latex] 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. [latex]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}[/latex]
    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, [latex]{\text{SO}}_{4}^{2-}[/latex], has caused a decrease in the concentration of Ca²⁺ that otherwise would be in solution:
    13. [latex]\sqrt{2.4\times {10}^{-5}}=\text{4.9}\times {10}^{-5}\text{}M[/latex]
27. 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 = [latex]{{\text{[Ag}}^{\text{+}}]}^{2}\left[{\text{SO}}_{4}^{2-}\right][/latex] = 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 [latex]{\text{SO}}_{4}^{2-}[/latex] ion, [Ag⁺] = 2.86 × 10^–2M, [latex]{\text{[SO}}_{4}^{2-}][/latex] = 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.
29. 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
31. The answers are as follows:
    1. [latex]{\text{CaCO}}_{3}:{\text{CaCO}}_{3}\left(s\right)\longrightarrow {\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{CO}}_{3}{}^{2-}\left(aq\right)[/latex]
       1. K_sp = [Ca²⁺] [latex]{\text{[CO}}_{3}^{2-}][/latex] = 4.8 × 10^–9
       2. test K_sp against Q = [Ca²⁺] [latex]{\text{[CO}}_{3}^{2-}][/latex]
       3. Q = [Ca²⁺] [latex]{\text{[CO}}_{3}^{2-}][/latex] = (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.
    2. [latex]{\text{Co(OH)}}_{2}:{\text{Co(OH)}}_{2}\left(s\right)\longrightarrow {\text{Co}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)[/latex]
       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
    3. The ion product does not exceed K_sp, so the compound does not precipitate.
    3. CaHPO₄: (K_sp = 5 × 10^–6):
       1. Q = [Ca²⁺] [latex]{\text{[HPO}}_{4}^{2-}][/latex] = (0.01)(2 × 10^–6) = 2 × 10^–8 < K_sp
    4. The ion product does not exceed K_sp, so compound does not precipitate.
    4. Pb₃(PO₄)₂: (K_sp = 3 × 10^–44):
       1. Q = [Pb²⁺]³ [latex]{{\text{[PO}}_{4}^{3-}]}^{2}[/latex] = (0.01)³(1 × 10^–13)² = 1 × 10^–32 > K_sp
    5. The ion product exceeds K_sp, so the compound precipitates.
33. Precipitation of [latex]{\text{SO}}_{4}^{2-}[/latex] will begin when the ion product of the concentration of the [latex]{\text{SO}}_{4}^{2-}[/latex] and Ba²⁺ ions exceeds the K_sp of BaSO₄.
    1. K_sp = 1.08 × 10^–10 = [Ba²⁺] [latex]\left[{\text{SO}}_{4}^{2-}\right][/latex] = (0.0758) [latex]{\text{[SO}}_{4}^{2-}][/latex]
    2. [latex]\left[\text{SO}_{4}^{2-}\right]=\frac{1.08\times{10}^{-10}}{0.0758}=\text{1.42}\times{10}^{-9}M[/latex]
35. Precipitation of Ag₃PO₄ will begin when the ion product of the concentrations of the Ag⁺ and [latex]{\text{PO}}_{4}{}^{\text{3-}}[/latex] ions exceeds K_sp:
    1. [latex]{\text{Ag}}_{3}{\text{PO}}_{4}\left(s\right)\longrightarrow {\text{3Ag}}^{\text{+}}\left(aq\right)+{\text{PO}}_{4}{}^{\text{3-}}\left(aq\right)[/latex]
    2. K_sp = 1.8 × 10^–18 = [Ag⁺]³ [latex]{\text{[PO}}_{4}{}^{\text{3-}}][/latex] = (0.0125)³ [latex]{\text{[PO}}_{4}{}^{3-}][/latex]
    3. [latex]\left[\text{PO}_{4}^{3-}\right]=\frac{1.08\times{10}^{-18}}{\left(0.0125\right)^{3}}=\text{9.2}\times {10}^{-13}M[/latex]
37. [latex]{\text{Ag}}_{2}{\text{CO}}_{3}\left(s\right)\longrightarrow {\text{2Ag}}^{\text{+}}\left(aq\right)+{\text{CO}}_{3}{}^{\text{2-}}\left(aq\right)[/latex]
    1. [Ag⁺]² [latex]\left[{\text{CO}}_{3}{}^{\text{2-}}\right][/latex] = K_sp = 8.2 × 10^–12
    2. [Ag⁺]²(2.5 × 10^–6) = 8.2 × 10^–12
    3. [latex]\left[\text{Ag}^{+}\right]^{2}=\frac{8.2\times{10}^{-12}}{2.50\times{10}^{-6}}=\text{3.28}\times{10}^{-6}[/latex]
    4. [Ag⁺] = 1.8 × 10^–3M
39. 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. [latex]{\left[{\text{F}}^{-}\right]}^{2}=\frac{3.9\times {10}^{-11}}{1.0\times {10}^{-4}}=\text{3.9}\times {10}^{-7}[/latex]
    3. [F^–] = 6.2 × 10^–4
41. (a) 2.28 L; (b) 7.3 × 10^–7 g
43. When 99.9% of Cu²⁺ has precipitated as CuS, then 0.1% remains in solution.
    1. [latex]\frac{0.1}{100}[/latex] × 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.
45. 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. [latex]{\text{Hg}}_{2}{}^{\text{2+}}[/latex] and Cu²⁺: Add [latex]{\text{SO}}_{4}{}^{\text{2-}}[/latex]. CuSO₄ is soluble (see Solubility Products), but K_sp for Hg₂SO₄ is 6.2 × 10^–7. When only 0.1% [latex]{\text{Hg}}_{2}{}^{\text{2+}}[/latex] remains in solution:
       1. [latex]\left[{\text{Hg}}_{2}{}^{\text{2+}}\right]=\frac{0.1%}{100%}\times \text{0.10}=1\times {10}^{-4}\text{}M[/latex]
    2. and
       1. [latex]\left[{\text{SO}}_{4}^{2-}\right]=\frac{{K}_{\text{sp}}}{\left[{\text{Hg}}_{2}{}^{\text{2+}}\right]}=\frac{6.2\times {10}^{-7}}{1\times {10}^{-4}}=\text{6.2}\times {10}^{-3}\text{}M[/latex];
    2. [latex]{\text{SO}}_{4}^{2-}[/latex] 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% [latex]{\text{SO}}_{4}^{2-}[/latex] remains in solution,
       1. [latex]{\text{[SO}}_{4}^{2-}][/latex] = 1 × 10^–4M
    3. and
       1. [latex]\left[{\text{Ba}}^{\text{2+}}\right]=\frac{1.08\times {10}^{-10}}{1\times {10}^{-4}}=1\times {10}^{-6}\text{}M[/latex];
    3. 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
    4. At that concentration:
       1. [Hg²⁺](6.7 × 10^–28) = 2 × 10^–59
       2. [Hg²⁺] = 3 × 10^–32M
    5. That is, it is virtually 100% precipitated. For a saturated (0.10 M) H₂S solution, the corresponding [latex]\left[\text{H}_{3}\text{O}^{+}\right][/latex] is:
       1. [latex]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\frac{\left[{\text{H}}_{2}\text{S}\right]}{\left[{\text{S}}^{2-}\right]}{K}_{\text{a}}=\frac{\left(0.10\right)}{\left(6.7\times {10}^{-28}\right)}\times \text{1.0}\times {10}^{-26}[/latex]
       2. [latex]\left[\text{H}_{3}\text{O}^{+}\right]=1.5M[/latex]
    6. A solution more basic than this would supply enough S^2– for CoS to precipitate.
    4. 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. [latex]{\left[{\text{OH}}^{-}\right]}^{2}=\frac{{K}_{\text{sp}}}{\left[{\text{Zn}}^{\text{2+}}\right]}\text{=}\frac{4.5\times {10}^{-11}}{1\times {10}^{-4}}=\text{4.5}\times {10}^{-7}[/latex]
       2. [OH^–] = 7 × 10^–4M
    5. When Sr(OH)₂·8H₂O just begins to precipitate:
       1. [latex]{\left[{\text{OH}}^{-}\right]}^{2}=\frac{{K}_{\text{sp}}}{\left[{\text{Sr}}^{\text{2+}}\right]}=\frac{3.2\times {10}^{-4}}{0.10}=\text{3.2}\times {10}^{-3}[/latex]
       2. [OH^–] = 0.057 M
    6. If [OH^–] is maintained less than 0.056 M, then Zn²⁺ will precipitate and Sr²⁺ will not.
    5. Ba²⁺ and Mg²⁺: Add [latex]{\text{SO}}_{4}^{2-}[/latex]. MgSO₄ is soluble and BaSO₄ is not (K_sp = 2 × 10^–10).
    6. [latex]{\text{CO}}_{3}^{2-}[/latex] 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 [latex]{\text{CO}}_{3}^{2-}[/latex] has been precipitated [latex]{\text{[CO}}_{3}^{2-}][/latex] = 1 × 10^–4M and
       1. [latex]\left[\text{Ba}^{2+}\right]=\frac{K_{\text{sp}}}{\left[\text{CO}_{3}^{2-}\right]}=\frac{8.1\times{10}^{-9}}{1\times{10}^{-4}}=\text{8.1}\times{10}^{-5}M[/latex]
    7. Ba(OH)₂·8H₂O begins to precipitate when:
       1. [latex]{\text{[Ba}}^{\text{2+}}\text{]}=\frac{{K}_{\text{sp}}}{{\left[{\text{OH}}^{-}\right]}^{2}}=\frac{5.0\times {10}^{-3}}{{\left(0.10\right)}^{2}}=\text{0.50}M[/latex]
    8. As long as [Ba²⁺] is maintained at less than 0.50 M, BaCO₃ precipitates and Ba(OH)₂·8H₂O does not.
47. 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. [latex]\left[{\text{Ag}}^{\text{+}}\right]=\frac{1.8\times {10}^{-10}}{\left[0.10\right]}\text{=1.8}\times {10}^{-9}\text{}M[/latex]
    2. For AgI: K_sp = 1.5 × 10^–16 = [Ag⁺][I^–]
       1. [latex]\left[{\text{Ag}}^{\text{+}}\right]=\frac{1.5\times {10}^{-16}}{1.0\times {10}^{-2}}=\text{1.5}\times {10}^{-9}\text{}M[/latex]
    3. As the value of [Ag⁺] is smaller for AgI, AgI will precipitate first.
49. The dissolution of Ca₃(PO₄)₂ yields:
    1. [latex]{\text{Ca}}_{3}{{\text{(PO}}_{4})}_{2}\left(s\right)\rightleftharpoons {\text{3Ca}}^{\text{2+}}\left(aq\right)+{\text{2PO}}_{4}^{3-}\left(aq\right)[/latex]
50. Given the concentration of Ca²⁺ in solution, the maximum [latex]\left[\text{PO}_{4}^{3-}\right][/latex] can be calculated by using the K_sp expression for Ca₃(PO₄)₂:
    1. K_sp = 1 × 10^–25 = [Ca²⁺]³ [latex]{{\text{[PO}}_{4}{}^{3-]}}^{2}[/latex]
    2. [latex]{\left[{\text{Ca}}^{\text{2+}}\right]}_{\text{urine}}=\frac{0.10\text{}g\left(\frac{1\text{}mol}{40.08\text{}g}\right)}{1.4\text{L}}=\text{1.8}\times {10}^{-3}\text{}M[/latex]
    3. [latex]{{\text{[PO}}_{4}{}^{3-}]}^{2}=\frac{1\times {10}^{-25}}{{\left(1.8\times {10}^{-3}\right)}^{3}}=\text{1.7}\times {10}^{-17}[/latex]
    4. [latex]{{\text{[PO}}_{4}{}^{3-}]}^{2}=4\times {10}^{-9}\text{}M[/latex]
51. 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
52. 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. [latex]\frac{1.304{\text{g L}}^{-1}}{24.305{\text{g mol}}^{-1}}=\text{0.05365}M[/latex]
53. 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
54. 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. [latex]{\text{Mg(OH)}}_{2}\left(s\right)\rightleftharpoons {\text{Mg}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{-}\left(aq\right)\left({K}_{\text{sp}}=\text{1.5}\times {10}^{-11}\right)[/latex]
55. When the initial 1.035 g/L is reduced to 0.1% of the original, the molarity is calculated as:
    1. [latex]\frac{0.001\times {\text{1.305 g L}}^{-1}}{24.305{\text{g mol}}^{-1}}=\text{5.369}\times {10}^{-5}\text{}M[/latex]
56. 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
57. Thus an additional [latex]\frac{1}{2}[/latex] × 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. [latex]{\text{mass Ca(OH)}}_{2}=2.65\times {10}^{-4}{\text{mol Ca(OH)}}_{2}{\text{L}}^{-1}\times \text{1.00}\times {10}^{3}\text{L}\times \frac{74.0946\text{g}}{{\text{mol Ca(OH)}}_{2}}=\text{20 g}[/latex]
58. The total Ca(OH)₂ required is 3.97 kg + 0.020 kg = 3.99 kg.
    1. K_sp = [La³⁺] [latex]{\left[{\text{IO}}_{3}{}^{-}\right]}^{3}[/latex] = [latex]\left(\frac{1}{3}\times \text{3.1}\times {10}^{-3}\right)[/latex] (3.1 × 10^–3)³ = (0.0010)(3.0 × 10^–8) = 3.0 × 10^–11
    2. K_sp = [Cu²⁺] [latex]{\left[{\text{IO}}_{3}{}^{-}\right]}^{2}[/latex] = 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. [latex]\left[{\text{IO}}_{3}{}^{-}\right][/latex] = 2x = 5.3 × 10^–3
55. 1.8 × 10^–5 g Pb(OH)₂
57. [latex]{\text{Mg(OH)}}_{2}\text{(s)}\rightleftharpoons {\text{Mg}}^{\text{2+}}+{\text{2OH}}^{-}{K}_{\text{sp}}=\left[{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^{2}[/latex]
58. 7.1 × 10^–12 = (x)(2x)² = 4x³
59. x = 1.21 × 10^–4M = [Mg²⁺]
60. [latex]\frac{1.21\times {10}^{-4}\text{mol}}{\text{L}}\times 0.20\text{0 L}\times \frac{{\text{1 mol Mg(OH)}}_{2}}{{\text{1 mol Mg}}^{\text{2+}}}\times \frac{{\text{58.3 g Mg(OH)}}_{2}}{{\text{1 mol Mg(OH)}}_{2}}=\text{1.14}\times {10}^{-3}{\text{Mg(OH)}}_{2}[/latex]
59. 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.

2. Cadmium ions associate with ammonia molecules in solution to form the complex ion [latex]\ce{[Cd(NH3)4]^2+}[/latex], which is defined by the following equilibrium:

[latex]\ce{Cd^2+}(aq)+\ce{4NH3}(aq)\longrightarrow \ce{[Cd(NH3)4]^2+}(aq){K}_{\text{f}}=4.0\times {10}^{6}[/latex]

The formation of the complex ion requires 4 mol of [latex]\ce{NH3}[/latex] for each mol of [latex]\ce{Cd^2+}[/latex]. First, calculate the initial amounts of Cd²⁺ and of NH₃ available for association:

[latex]\ce{[Cd^2+]}=\frac{\left(0.100\text{L}\right)\left(0.0100\text{mol}{\text{L}}^{-1}\right)}{0.250\text{L}}=4.00\times {10}^{-3}M[/latex]

[latex]\ce{[NH3]}=\frac{\left(0.150\text{L}\right)\left(0.100\text{mol}{\text{L}}^{-1}\right)}{0.250\text{L}}=6.00\times {10}^{-2}M[/latex]

For the reaction, 4.00 [latex]\times[/latex] 10^–3 mol/L of [latex]\ce{Cd^2+}[/latex] would require 4(4.00 [latex]\times[/latex] 10^–3 mol/L) of [latex]\ce{NH3}[/latex] or a 1.6 [latex]\times[/latex] 10^–2–M solution. Due to the large value of K_f and the substantial excess of [latex]\ce{NH3}[/latex], 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 [latex]\times[/latex] 10^–2 mol/L – 1.6 [latex]\times[/latex] 10^–2 mol L^–1 = 4.4 [latex]\times[/latex] 10^–2M

Let x be the change in concentration of [latex]\ce{[Cd^2+]}[/latex]:

[latex]{K}_{\text{f}}=4.0\times {10}^{6}=\frac{\left[\ce{Cd}{\left({\ce{NH}}_{3}\right)}_{4}{}^{\ce{2+}}\right]}{\left[{\ce{Cd}}^{\ce{2+}}\right]{\left[{\ce{NH}}_{3}\right]}^{4}}[/latex]

[latex]4.00\times {10}^{6}=\frac{\left(4.00\times {10}^{-3}-x\right)}{\left(x\right){\left(4.4\times {10}^{-2}+4x\right)}^{4}}[/latex]

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 [latex]\times[/latex] 10⁶x (4.4 [latex]\times[/latex] 10^–2 + 4x)⁴ = 4.00 [latex]\times[/latex] 10^–3 – x

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

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

Substitution of different values x will give a number to be compared with 4.00 [latex]\times[/latex] 10^–3. Using 2.50 [latex]\times[/latex] 10^–4 for x gives 4.35 [latex]\times[/latex] 10^–3 compared with 4.00 [latex]\times[/latex] 10^–3. Using 2.40 [latex]\times[/latex] 10^–4 gives 4.16 [latex]\times[/latex] 10^–3 compared with 4.00 [latex]\times[/latex] 10^–3. Using 2.30 [latex]\times[/latex] 10^–4 gives 3.98 [latex]\times[/latex] 10^–3 compared with 4.00 [latex]\times[/latex] 10^–3. Thus 2.30 [latex]\times[/latex] 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 [latex]\times[/latex] 10^–2, the value of x obtained is 2.35 [latex]\times[/latex] 10^–4M.

5.

[latex]{K}_{\text{d}}=\frac{\left[{\ce{Cd}}^{\ce{2+}}\right]\left[{\ce{CN}}^{-}\right]}{\left[\ce{Cd}{\left(\ce{CN}\right)}_{4}{}^{2-}\right]}=7.8\times {10}^{-18}=\frac{x{\left(4x\right)}^{4}}{0.250-x}[/latex]

Assume that x is small when compared with 0.250 M.

256x⁵ = 0.250 [latex]\times[/latex] 7.8 [latex]\times[/latex] 10^–18

x⁵ = 7.617 [latex]\times[/latex] 10^–21

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

4x =[latex]\ce{[CN-]}[/latex] = 3.8 [latex]\times[/latex] 10^–4M

7.

[latex]{K}_{\text{d}}=\frac{\left[{\ce{Co}}^{\ce{2+}}\right]{\left[{\ce{NH}}_{3}\right]}^{6}}{\left[\ce{Co}{\left({\ce{NH}}_{3}\right)}_{6}{}^{\ce{3+}}\right]}=\frac{x{\left(6x\right)}^{6}}{0.500-x}=2.2\times {10}^{-34}[/latex]

Assume that x is small when compared with 0.500 M.

4.67 [latex]\times[/latex] 104x⁷ = 0.500 [latex]\times[/latex] 2.2 [latex]\times[/latex] 10^–34

x⁷ = 2.358 [latex]\times[/latex] 10^–39

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

6x = [latex]\ce{[NH3]}[/latex] = 1.8 [latex]\times[/latex] 10^–5M

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

[latex]\begin{array}{l}\ce{AgBr}\left(s\right)\rightleftharpoons\ce{Ag}^{+}\left(aq\right)+\ce{Br}^{-}\left(aq\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{sp}}=3.3\times{10}^{-13}\\\ce{Ag}^{+}\left(aq\right)+2{\ce{S}}_2\ce{O}_{3}^{2-}\left(aq\right)\rightleftharpoons\ce{Ag}\left(\ce{S}_{2}\text{O}_{3}\right)_{2}^{3-}\left(aq\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{f}}=4.7\times{10}^{13}\end{array}[/latex]

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

[latex]\frac{\left[\ce{Ag}{\left({\ce{S}}_{2}{\ce{O}}_{3}\right)}^{\ce{3-}}\right]\left[{\ce{Br}}^{-}\right]}{{\left[{\ce{S}}_{2}{\ce{O}}_{3}{}^{2-}\right]}^{2}}=15.51[/latex]

If all [latex]\ce{Ag}[/latex] is to be dissolved, the concentration of the complex is the molar concentration of [latex]\ce{AgBr}[/latex].

formula mass ([latex]\ce{AgBr}[/latex]) = 187.772 g/mol

[latex]\text{moles present}=\frac{0.27\text{g}\ce{AgBr}}{187.772\text{g}{\text{mol}}^{-1}}=1.438\times {10}^{-3}\text{mol}[/latex]

Let x be the change in concentration of [latex]\ce{S2O3^2-}:[/latex]

[latex]\frac{\left(1.438\times {10}^{-3}\right)\left(1.438\times {10}^{-3}\right)}{{x}^{2}}=15.51[/latex]

x² = 1.333 [latex]\times[/latex] 10^–7

[latex]x=3.65\times {10}^{-4}M=\ce{[S2O3^2-]}[/latex]

The formula mass of [latex]\ce{Na2S2O3•5H2O}[/latex] is 248.13 g/mol. The total [latex]\ce{[S2O3^2-]}[/latex] needed is:

2(1.438 [latex]\times[/latex] 10^–3) + 3.65 [latex]\times[/latex] 10^–4 = 3.241 [latex]\times[/latex] 10^–3 mol

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

11.

(a)

(b) [latex]\ce{H3O+}+\ce{CH3-}\longrightarrow \ce{CH4}+\ce{H2O}[/latex]

(c) [latex]\ce{CaO}+\ce{SO3}\longrightarrow \ce{CaSO4}[/latex]

(d) [latex]\ce{NH4+}+\ce{C2H5O-}\longrightarrow \ce{C2H5OH}+\ce{NH3}[/latex]

19. The equilibrium is:[latex]\ce{Ag+}(aq)+\ce{2CN-}(aq)\rightleftharpoons \ce{Ag(CN)2-}(aq){K}_{\text{f}}=1\times {10}^{20}[/latex]

The number of moles of [latex]\ce{AgNO3}[/latex] added is:

0.02872 L [latex]\times[/latex] 0.0100 mol/L = 2.87 [latex]\times[/latex] 10^–4 mol

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

[latex]\ce{AgCN2-}(aq)+\ce{Ag+}(aq)\rightleftharpoons \ce{2AgCN}(s)[/latex]

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

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

[latex]\text{mass}(\ce{NaCN})=5.74\times {10}^{-4}\text{mol}\times \frac{49.007\text{g}}{1\text{mol}}=0.0281\text{g}[/latex]

21. [latex]\ce{HNO3}(l)+\ce{HF}(l)\longrightarrow \ce{H2NO3+}+\ce{F-}[/latex]; [latex]\ce{HF}(l)+\ce{BF3}(g)\longrightarrow \ce{H+}+\ce{BF4}[/latex]

23. (a) [latex]\ce{H3BO3}+\ce{H2O}\longrightarrow \ce{H4BO4-}+\ce{H+}[/latex];

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

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.

7. Find the amount of [latex]{\text{SO}}_{4}{}^{2-}[/latex] present from K_a for the equilibrium:

[latex]{\text{HSO}}_{4}{}^{-}+{\text{H}}_{2}\text{O}\longrightarrow {\text{H}}_{3}{\text{O}}^{+}+{\text{SO}}_{4}{}^{2-}[/latex]

Let x be the change in [latex]\left[{\text{SO}}_{4}{}^{2-}\right][/latex] :

[latex]{K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{SO}}_{4}{}^{2-}\right]}{\left[{\text{HSO}}_{4}{}^{-}\right]}=\frac{{x}^{2}}{0.250-x}=1.2\times {10}^{-2}[/latex]

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

x² + 1.2 [latex]\times[/latex] 10^–2x – 0.250(1.2 [latex]\times[/latex] 10^–2) = 0

[latex]x=\frac{-1.2\times {10}^{-2}\pm \sqrt{{\left(1.2\times {10}^{-2}\right)}^{2}+4\left(3.0\times {10}^{\text{-3}}\right)}}{2}=\frac{-1.2\times {10}^{-2}\pm 0.11}{2}=0.049M[/latex]

K_sp = [Ba²⁺] [latex]\left[{\text{SO}}_{4}{}^{2-}\right][/latex] = [Ba²⁺](0.049) = 1.08 [latex]\times[/latex] 10^–10

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

9. [latex]{\text{CH}}_{3}{\text{NH}}_{2}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{CH}}_{3}{\text{NH}}_{3}{}^{+}+{\text{OH}}^{-}[/latex]

[latex]{K}_{\text{b}}=\frac{\left[{\text{CH}}_{3}{\text{NH}}_{3}{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{3}{\text{NH}}_{2}\right]}=\frac{\left(x\right)\left(x\right)}{0.138-x}=4.4\times {10}^{-4}[/latex]

Solve the quadratic equation using the quadratic formula:

x² + 4.4 [latex]\times[/latex] 10^–4x – 0.138(4.4 [latex]\times[/latex] 10^–4) = 0

[latex]x=\frac{-4.4\times {10}^{-4}\pm \sqrt{{\left(4.4\times {10}^{-4}\right)}^{2}+4\left(6.07\times {10}^{\text{-5}}\right)}}{2}=\frac{-4.4\times {10}^{\text{-4}}\pm \sqrt{2.43\times {10}^{-4}}}{2}=\frac{0.0152}{2}=7.6\times {10}^{-3}M[/latex]

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

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

11. Two equilibria are in competition for the ions and must be considered simultaneously. Precipitation of [latex]\ce{MnS}[/latex] will occur when the concentration of [latex]\ce{S^2-}[/latex] in conjunction with 0.125 M Mn²⁺ exceeds the K_sp of MnS. The [latex]\ce{[S^2-}]}[/latex] must come from the ionization of [latex]\ce{H2S}[/latex] as defined by the equilibrium:

[latex]{\text{H}}_{2}\text{S}\left(aq\right)+2{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow 2{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)[/latex]

[latex]\frac{{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}^{2}\left[{\text{S}}^{2-}\right]}{\left[{\text{H}}_{2}\text{S}\right]}={K}_{1}{K}_{2}\left({\text{H}}_{2}\text{S}\right)=1.0\times {10}^{\text{-26}}[/latex]

As a saturated solution of [latex]\ce{H2S}[/latex] is 0.10 M, this later expression becomes:

[latex]{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}^{2}\left[{\text{S}}^{2-}\right]=1.0\times {10}^{-27}[/latex]

From the equilibrium of [latex]\ce{MnS}[/latex], the minimum concentration of [latex]\ce{S^2-}[/latex] required to cause precipitation is calculated as:

[latex]\text{MnS}\left(s\right)\longrightarrow {\text{Mn}}^{\text{2+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)[/latex]

K_sp = [latex]\ce{[Mn^2+][S2-]}[/latex] = 4.3 [latex]\times[/latex] 10^–22

[latex]\left[{\text{S}}^{2-}\right]=\frac{4.3\times {10}^{-22}}{0.125}=3.44\times {10}^{-21}[/latex]

This amount of [latex]\ce{S^2-}[/latex] will exist in solution at a pH defined by the [latex]\ce{H2S}[/latex] equilibrium:

• [latex]{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}^{2}\left(3.44\times {10}^{-21}\right)=1.0\times {10}^{-27}[/latex]
• [latex]{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}^{2}=2.91\times {10}^{-7}[/latex]
• [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=5.39\times {10}^{-4}M[/latex]
• [latex]\text{pH}=-\text{log}\left[{\text{H}}_{3}{\text{O}}^{+}\right]=3.27[/latex]

13. Three equilibria are involved:

• [latex]{\text{H}}_{2}{\text{CO}}_{3}+{\text{H}}_{2}\text{O}\longrightarrow {\text{H}}_{3}{\text{O}}^{+}+{\text{HCO}}_{3}{}^{-}{K}_{{\text{a}}_{1}}=4.3\times {10}^{-7}[/latex]
• [latex]{\text{HCO}}_{3}{}^{-}+{\text{H}}_{2}\text{O}\longrightarrow {\text{H}}_{3}{\text{O}}^{+}+{\text{CO}}_{3}{}^{2-}{K}_{{\text{a}}_{2}}=7\times {10}^{-11}[/latex]
• [latex]{\text{CdCO}}_{3}\longrightarrow {\text{Cd}}^{\text{2+}}+{\text{CO}}_{3}{}^{2-}{K}_{\text{sp}}=2.5\times {10}^{-14}[/latex]

First, find the pH of the buffer from the Henderson-Hasselbach equation. Then find [latex]\ce{[H3O+]}[/latex]:

[latex]\text{pH}=\text{p}{K}_{\text{a}}+\mathrm{log}\frac{\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[{\text{HCO}}_{3}{}^{-}\right]}[/latex]

10.155 + log [latex]\frac{0.115}{0.120}[/latex] = 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:

[latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=7.3\times {10}^{-11}[/latex]

Now, find the concentration of [latex]{\text{CO}}_{3}{}^{2-}[/latex] present in the buffer solution. Next, using [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] and K_sp, calculate the concentration of [latex]\ce{Cd^2+}[/latex]. This latter value represents the molar solubility. From [latex]{K}_{{a}_{1}}[/latex] determine [latex]\ce{[H2CO3]}[/latex]:

[latex]{K}_{{\text{a}}_{1}}=4.3\times {10}^{-7}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{HCO}}_{3}{}^{-}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}=\frac{\left(7.3\times {10}^{-11}\right)\left(0.120\right)}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}[/latex]

[H₂CO₃] = 2.05 [latex]\times[/latex] 10^–5

From [latex]{K}_{{\text{a}}_{1}}{K}_{{\text{a}}_{2}}[/latex] , find the concentration of [latex]{\text{CO}}_{3}{}^{2-}[/latex] :

[latex]\begin{array}{rll}{}{K}_{{\text{a}}_{1}}{K}_{{\text{a}}_{2}}&=&\left(4.3\times 10 - 7\right)\left(7\times 10 - 11\right)=\frac{\left[{\text{HCO}}_{3}{}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}\times \frac{\left[{\text{CO}}_{3}{}^{2-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}=\frac{{\left(7.3\times {10}^{-11}\right)}^{2}\times \left[{\text{CO}}_{3}{}^{2-}\right]}{2.04\times {10}^{-5}}\\& =&3\times {10}^{-17}\end{array}[/latex]

[latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] = 0.115 M

K_sp = [latex]\ce{[Cd^2+]}[/latex] [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] = [Cd²⁺](0.115) = 3 [latex]\times[/latex] 10^–13

[latex]\left[{\text{Cd}}^{\text{2+}}\right]=\frac{3\times {10}^{-13}}{0.115}=3\times {10}^{-12}M[/latex]

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

[latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\frac{{K}_{\text{a}}\left[\text{HOAc}\right]}{\left[{\text{OAc}}^{-}\right]}=\frac{1.8\times {10}^{-5}\left(0.250\right)}{\left(0.375\right)}=1.2\times {10}^{-5}M[/latex]

From this and K_a for [latex]\ce{H2CO3}[/latex], calculate [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] present. As [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] is fixed by K_a of acetic acid:

[latex]{K}_{\text{a}}\left({\text{H}}_{2}{\text{CO}}_{3}\right)=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{HCP}}_{3}{}^{-}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}{}^{-}\right]}[/latex]

[latex]\left[{\text{H}}_{2}{\text{CO}}_{3}{}^{-}\right]=\frac{{K}_{\text{a}}\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{4.3\times {10}^{-7}\left[0.010\right]}{\left[1.2\times {10}^{-5}\right]}=3.58\times {10}^{-4}M[/latex]

From [latex]{K}_{\text{a}}\left({\text{HCO}}_{3}{}^{-}\right)[/latex] we obtain [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] :

[latex]{K}_{a}=7\times {10}^{-11}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[{\text{HCO}}_{3}{}^{-}\right]}=\frac{1.2\times {10}^{-17}\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[3.58\times {10}^{-4}\right]}\text{ }[/latex]

[latex]\left[{\text{CO}}_{3}{}^{2-}\right]=\frac{7\times {10}^{-11}\times 3.58\times {10}^{-4}}{1.2\times {10}^{-5}}=2.09\times {10}^{-9}[/latex]

From the solubility product:

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

[latex]\left[{\text{Cd}}^{\text{2+}}\right]=\frac{2.5\times {10}^{-14}}{2.09\times {10}^{-9}}=1\times {10}^{-5}M[/latex]

17. [latex]{\text{CaCO}}_{3}:0.350\cancel{\text{g}}\times \frac{1\text{mol}}{100.09\cancel{\text{g}}}=3.50\times {10}^{-3}\text{mol}[/latex]

[latex]{\text{SrCO}}_{3}:0.350\cancel{\text{g}}\times \frac{1\text{mol}}{147.63\cancel{\text{g}}}=2.37\times {10}^{-3}\text{mol}[/latex]

[latex]{\text{BaCO}}_{3}:0.350\cancel{\text{g}}\times \frac{1\text{mol}}{197.34\cancel{\text{g}}}=1.77\times {10}^{-3}\text{mol}[/latex]

• Total: 7.64 [latex]\times[/latex] 10^–3 mol
• [latex]\ce{CaCO3}[/latex]: K_sp = 4.8 [latex]\times[/latex] 10^–9
• [latex]\ce{SrCO3}[/latex]: K_sp = 9.42 [latex]\times[/latex] 10^–10
• [latex]\ce{BaCO3}[/latex]: K_sp = 8.1 [latex]\times[/latex] 10^–9

Solubilities are approximately equal.

When the three solids dissolve, 7.64 [latex]\times[/latex] 10^–3 mol of metal ions and 7.64 [latex]\times[/latex] 10^–3 mol of [latex]{\text{CO}}_{3}{}^{2-}[/latex] are initially present. The [latex]{\text{CO}}_{3}{}^{2-}[/latex] reacts with [latex]\ce{CH3CO2H}[/latex] to form [latex]{\text{HCO}}_{3}{}^{-}[/latex] :

[latex]{\text{ CO}}_{3}{}^{2-}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)\rightleftharpoons {\text{HCO}}_{3}{}^{-}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)[/latex]

[latex]K=\frac{\left[{\text{HCO}}_{3}{}^{-}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}{\left[{\text{CO}}_{3}{}^{2-}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}\times \frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{{K}_{\text{a}}\left(\text{for}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}{{K}_{\text{a}}\left(\text{for}{\text{HCO}}_{3}{}^{-}\right)}=\frac{1.8\times {10}^{-5}}{7\times {10}^{-11}}=2.6\times {10}^{5}[/latex]

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

[latex]{\text{HCO}}_{3}{}^{-}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)\longrightarrow {\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)[/latex]

[latex]K=\frac{{K}_{\text{a}}\left(\text{for}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}{{K}_{\text{a}}\left(\text{for}{\text{HCO}}_{3}{}^{-}\right)}=\frac{1.8\times {10}^{-5}}{4.3\times {10}^{-7}}=42[/latex]

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

For each mol of [latex]{\text{CO}}_{3}{}^{2-}[/latex] produced, 2 mol of [latex]\ce{CH3CO2H}[/latex] is required for conversion to [latex]\ce{H2CO3}[/latex]. Thus 15.3 [latex]\times[/latex] 10^–3 mol [latex]\ce{CH3CO2H}[/latex] is needed:

[latex]\text{Volume}=\frac{15.3\times {10}^{-3}\text{mol}}{1.50\text{mol}{\text{L}}^{-1}}=0.0102\text{L}\left(10.2\text{mL}\right)[/latex]

19. There are two equilibria involved: [latex]\begin{array}{l}\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)\rightleftharpoons {\text{Mg}}^{\text{2+}}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right){K}_{\text{sp}}=1.5\times {10}^{-11}\\ {\text{CN}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\left({K}_{\text{b}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{1.0\times {10}^{-14}}{4\times {10}^{-10}}=2.5\times {10}^{-5}\right)\end{array}[/latex] The Mg(NO₃)₂ dissolves, and [Mg²⁺] = 0.010 M.

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

(0.010)[OH^–]² = 1.5 [latex]\times[/latex] 10^–11

[OH^–] = 3.9 [latex]\times[/latex] 10^–5M

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

[latex]\begin{array}{l}\frac{\left[\text{HCN}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CN}}^{-}\right]}={K}_{\text{b}}=2\times {10}^{-5}\\ \frac{{\left(3.9\times {10}^{-5}\right)}^{2}}{\left[{\text{CN}}^{-}\right]}=2.5\times {10}^{-5}\end{array}[/latex]

[latex]\ce{[CN-]}[/latex] = 6.1 [latex]\times[/latex] 10^–5M

mol [latex]\ce{NaCN}[/latex] = mol [latex]\ce{HCN}[/latex] + mol [latex]\ce{CN-}[/latex] = 3.9 [latex]\times[/latex] 10^–5 + 6.1 [latex]\times[/latex] 10^–5 = 1.0 [latex]\times[/latex] 10^–4

[latex]\text{mass}\left(\text{NaCN}\right)=1.0\times {10}^{-4}\text{mol}\times \frac{49.007\text{g}}{1\text{mol}}=5\times {10}^{-3}\text{g}[/latex]

21. The answers are as follows:

1. K_sp = [Mg²⁺][F^–]² = (1.21 [latex]\times[/latex] 10^–3)(2 [latex]\times[/latex] 1.21 [latex]\times[/latex] 10^–3)² = 7.09 [latex]\times[/latex] 10^–9
2. K_sp = [Mg²⁺][F^–]² = [x][0.100 + 2x]² = 7.09 [latex]\times[/latex] 10^–9
   Assume that 2x is small when compared with 0.100 M.
   0.100x = 7.09 [latex]\times[/latex] 10^–9
   x = [MgF₂] = 7.09 [latex]\times[/latex] 10^–7M
   The value 7.09 [latex]\times[/latex] 10^–7M is quite small when compared with 0.100 M, so the assumption is valid.
3. Determine the concentration of [latex]\ce{Mg^2+}[/latex] and [latex]\ce{F-}[/latex] that will be present in the final volume. Compare the value of the ion product [latex]\ce{[Mg^2+][F-]^2}[/latex] with K_sp. If this value is larger than K_sp, precipitation will occur.
   0.1000 L [latex]\times[/latex] 3.00 [latex]\times[/latex] 10^–3M [latex]\ce{Mg(NO3)2}[/latex] = 0.3000 L [latex]\times[/latex] M [latex]\ce{Mg(NO3)2}[/latex]
   M [latex]\ce{Mg(NO3)2}[/latex] = 1.00 [latex]\times[/latex] 10^–3M
   0.2000 L [latex]\times[/latex] 2.00 [latex]\times[/latex] 10^–3M NaF = 0.3000 L [latex]\times[/latex] M [latex]\ce{NaF}[/latex]
   M [latex]\ce{NaF}[/latex] = 1.33 [latex]\times[/latex] 10^–3M
   ion product = (1.00 [latex]\times[/latex] 10^–3)(1.33 [latex]\times[/latex] 10^–3)² = 1.77 [latex]\times[/latex] 10^–9
   This value is smaller than K_sp, so no precipitation will occur.
4. [latex]\ce{MgF2}[/latex] 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 [latex]\ce{Mg(OH)2}[/latex], [latex]\ce{[Mg^2+]}[/latex], [latex]\ce{[OH-]}[/latex]:

1. increase, increase, decrease;
2. increase, decrease, increase;
3. decrease, increase, decrease;
4. no effect predicted;
5. increase, no effect, no effect

25. [latex]\ce{[H2O] > [Ca^2+] > [PO4^3-] > HPO4^2- = [OH-] > [HPO4-]}[/latex]