1. In dissolution, one unit of substance produces a quantity of discrete ions or polyatomic ions that equals the number of times that the subunit appears in the formula.

1. [latex]\begin{array}{ccc}\text{AgI}\left(s\right)\rightleftharpoons & {\text{Ag}}^{\text{+}}\left(aq\right)+& {\text{I}}^{-}\left(aq\right)\\ & x& {x}\end{array}[/latex]
   Dissolving AgI(s) must produce the same amount of I^– ion as it does Ag⁺ ion.
2. [latex]\begin{array}{ccc}{\text{CaCO}}_{3}\left(s\right)\rightleftharpoons & {\text{Ca}}^{\text{2+}}\left(aq\right)+& {\text{CO}}_{3}^{2-}\left(aq\right)\\ & {x}& x\end{array}[/latex]
   Dissolving CaCO₃(s) must produce the same amount of Ca²⁺ ion as it does [latex]{\text{CO}}_{3}^{2-}[/latex] ion.
3. [latex]\begin{array}{lll}\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)\longrightarrow \hfill & {\text{Mg}}^{\text{2+}}\left(aq\right)\hfill & +2{\text{OH}}^{-}\left(aq\right)\hfill \\ \hfill & x\hfill & {\text{2}x}\hfill \end{array}[/latex]
   When one unit of Mg(OH)₂ dissolves, two ions of OH^– are formed for each Mg²⁺ ion.
4. [latex]\begin{array}{ccc}{\text{Mg}}_{3}{\left({\text{PO}}_{4}\right)}_{2}\left(s\right)\rightleftharpoons & {\text{3Mg}}^{\text{2+}}\left(aq\right)+& {\text{2PO}}_{4}^{3-}\left(aq\right)\\ & {3x}& 2x\end{array}[/latex]
   One unit of Mg₃(PO₄)₂ provides two units of [latex]{\text{PO}}_{4}^{3-}[/latex] ion and three units of Mg²⁺ ion.
5. [latex]\begin{array}{cccc}{\text{Ca}}_{5}{\left({\text{PO}}_{4}\right)}_{3}\text{OH}\left(s\right)\rightleftharpoons & {\text{5Ca}}^{\text{2+}}\left(aq\right)+& {\text{3PO}}_{4}^{3-}\left(aq\right)+& {\text{OH}}^{-}\left(aq\right)\\ & {5x}& {3x}& x\end{array}[/latex]
   One unit of Ca₅(PO₄)₃OH dissolves into five units of Ca²⁺ ion, three units of [latex]{\text{PO}}_{4}^{3-}[/latex] ion, and one unit of OH^– ion.

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. CaF₂, MnCO₃, 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

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. BaSeO₄
   • [latex]\frac{0.{\text{118 g L}}^{-1}}{280.{\text{28 g mol}}^{-1}}=\text{4.21}\times {10}^{-4}\text{}M[/latex]
   • K = [Ba²⁺] [latex]\left[{\text{SeO}}_{4}^{2-}\right][/latex] = (4.21 × 10^–4)(4.21 × 10^–4) = 1.77 × 10^–7
2. Ba(BrO₃)₂·H₂O:
   • [latex]\frac{3.0{\text{g L}}^{-1}}{{\text{411.147 g mol}}^{-1}}=\text{7.3}\times {10}^{-3}\text{}M[/latex]
   • 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. NH₄MgAsO₄·6H₂O:
   • [latex]\frac{0.{\text{38 g L}}^{-1}}{{\text{289.3544 g mol}}^{-1}}=\text{1.3}\times {10}^{-3}\text{}M[/latex]
   • 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:
   • [latex]\frac{0.0{\text{179 g L}}^{-1}}{{\text{757.62 g mol}}^{-1}}=\text{2.36}\times {10}^{-5}\text{}M[/latex]
   • 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 = [Ag⁺][Cl^–] = 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 = [Ca²⁺][F^–]² = x(2x + 0.00133 M)², where x = [Ca²⁺]. 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 K₂SO₄:
   • [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 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:
     [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.  (a) K_sp = 1.9 × 10^–4 = [Ti⁺][Cl^–]; Let x = [Cl^–]:

1.9 × 10^–4 = (x = 0.025)x

Assume that x is small when compared with 0.025:

[latex]x=\frac{1.9\times {10}^{-4}}{0.025}=7.6\times {10}^{-3}\text{}M[/latex]

Check: [latex]\frac{7.6\times {10}^{-3}}{0.025}\times \text{100%}=\text{30%}[/latex]

This value is too large to drop x. Therefore solve by using the quadratic equation:

[latex]\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]

x² + 0.025x – 1.9 × 10^–4 = 0

[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]

(Use only the positive answer for physical sense.)

[Ti⁺] = 0.025 + 0.0061 = 3.1 × 10^–2M

[Cl^–] = 6.1 × 10^–3

(b) K_sp = 1.7 × 10^–6 = [Ba²⁺][F^–]²; Let x = [Ba²⁺]

1.7 × 10^–6 = x(x + 0.0313)² = 1.7 × 10^–3M

Check: [latex]\frac{1.7\times {10}^{-3}}{0.0313}\times \text{100%}=\text{5.5%}[/latex]

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:

x′(5.4 × 10^–5 + 0.0313)² = 1.7 × 10^–3 or

[latex]x\prime =\frac{1.7\times {10}^{-6}}{1.089\times {10}^{-3}}=\text{1.6}\times {10}^{-3}[/latex]

A third approximation using this last calculation is as follows:

x′′(1.7 × 10^–3 + 0.0313)² = 1.7 × 10^–6 or

[latex]x\prime \prime =\frac{1.7\times {10}^{-6}}{1.089\times {10}^{-3}}=\text{1.6}\times {10}^{-3}[/latex]

This value is well within 5% and is acceptable.

[Ba²⁺] = 1.6 × 10^–3M

[F^–] = (1.6 × 10^–3 + 0.0313) = 0.0329 M;

(c) Find the molar concentration of the Mg(NO₃)₂. The molar mass of Mg(NO₃)₂ 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]

[latex]M=\frac{\text{0.05499 mol}}{\text{2.250 L}}=\text{0.02444 M}[/latex]

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.

K_sp = 8.6 × 10^–5 = [Mg²⁺] [latex]\left[{\text{C}}_{2}{\text{O}}_{4}^{2-}\right][/latex] = (x)(x + 0.02444)

0.02444x = 8.6 × 10^–5

x = [latex]\left[{\text{C}}_{2}{\text{O}}_{4}^{2-}\right][/latex] = 3.5 × 10^–3

Check: [latex]\frac{3.5\times {10}^{-3}}{0.02444}\times \text{100%}=\text{14%}[/latex]

This value is greater than 5%, so the quadratic equation must be used to solve for x:

[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]

(Use only the positive answer for physical sense.)

x′ = [latex]\left[{\text{C}}_{2}{\text{O}}_{4}{}^{2-}\right][/latex] = 3.5 × 10^–3M

[Mg²⁺] = 3.1 × 10^–3 + 0.02444 = 0.0275 M

(d) pH = 12.700; pOH = 1.300

[OH^–] = 0.0501 M; Let x = [Ca²⁺]

K_sp = 7.9 × 10^–6 = [Ca²⁺][OH^–]² = (x)(x + 0.050)²

Assume that x is small when compared with 0.050 M:

x = [Ca²⁺] = 3.15 × 10^–3 (one additional significant figure is carried)

Check: [latex]\frac{3.15\times {10}^{-3}}{0.050}\times \text{100%}=\text{6.28%}[/latex]

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:

x′(3.15 × 10^–3 + 0.0501)² = 7.9 × 10^–6 or

[latex]x\prime =\frac{7.9\times {10}^{-6}}{2.836\times {10}^{-3}}=\text{2.8}\times {10}^{-3}[/latex]

[Ca²⁺] = 2.8 × 10^–3M

[OH^–] = (2.8 × 10^–3 + 0.0501) = 0.053 × 10^–2M

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. Ca(OH)₂: [Ca²⁺][OH^–]² = 7.9 × 10^–6

Let x be [Ca²⁺] = molar solubility; then [OH^–] = 2x

K_sp = x(2x)² = 4x³ = 7.9 × 10^–6

x³ = 1.975 × 10^–6

x = 0.013 M

CaCO₃: [Ca²⁺] [latex]{\text{[CO}}_{3}^{2-}][/latex] = 4.8 × 10^–9

Let x be [Ca²⁺]; then [latex]{\text{[CO}}_{3}^{2-}][/latex] = [Ca²⁺] = molar solubility

K_sp = x² = 4.8 × 10^–9

x = 6.9 × 10^–5M

CaSO₄·2H₂O: [Ca²⁺] [latex]{\text{[SO}}_{4}^{2-}]\text{\cdot }[/latex] [H₂O]² = 2.4 10^–5

Let x be [Ca²⁺] = molar solubility = [latex]{\text{[SO}}_{4}^{2-}][/latex]; then [H₂O] = 2x

K_sp = (x)(x)(2x)² = 2.4 × 10^–5

x⁴ = 6.0 × 10^–6

x = 0.049 M

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:

CaC₂O₄·H₂O: [Ca²⁺] [latex]\left[{\text{C}}_{2}{\text{O}}_{4}^{2-}\right][/latex] [H₂O] = 2.27 × 10^–9

Let x be [Ca²⁺] = molar solubility = [latex]\left[{\text{C}}_{2}{\text{O}}_{4}^{2-}\right][/latex] = [H₂O]

x³ = 2.27 × 10^–9

x = 1.3 × 10^–3M

In this case, the interaction of water is also complex and the solubility is considerably less than that calculated.

Ca₃(PO₄)₂: [Ca²⁺]³ [latex]{\left[{\text{PO}}_{4}^{3-}\right]}^{2}[/latex] = 1 × 10^–25

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]

[latex]{x}^{3}{\left(\frac{2}{3}x\right)}^{2}{x}^{3}=1\times {10}^{-25}=\text{0.4444}{x}^{5}[/latex]

x = 1 × 10^–5M = [Ca²⁺]

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]

K_sp = [Ca²⁺] [latex]\left[{\text{SO}}_{4}^{2-}\right][/latex] = 2.4 × 10^–5

x × x = x² = 2.4 × 10^–5

[latex]x=\sqrt{2.4\times {10}^{-5}}[/latex]

x = 4.9 × 10^–3M = [latex]\left[{\text{SO}}_{4}^{2-}\right][/latex] = [Ca²⁺]

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:

[latex]{\text{CaSO}}_{4}\left(s\right)\longrightarrow {\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)[/latex]

K_sp = [Ca²⁺] [latex]{\text{[SO}}_{4}^{2-}][/latex] = 2.4 × 10^–5

Addition of 0.010 M [latex]{\text{SO}}_{4}^{2-}[/latex] generated from the complete dissociation of 0.010 M SO₄ gives

[x][x + 0.010] = 2.4 × 10^–5. Here, x cannot be neglected in comparison with 0.010 M; the quadratic equation must be used. In standard form:

x² + 0.010x – 2.4 × 10^–5 = 0

[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]

Only the positive value will give a meaningful answer:

x = 2.0 × 10^–3 = [Ca²⁺]

This is also the concentration of CaSO₄·2H₂O that has dissolved. The mass of the salt in 1 L is

Mass (CaSO₄·2H₂O) = 2.0 × 10^–3 mol/L × 172.16 g/mol = 0.34 g/L

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:

[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. (a) [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]

K_sp = [Ca²⁺] [latex]{\text{[CO}}_{3}^{2-}][/latex] = 4.8 × 10^–9

test K_sp against Q = [Ca²⁺] [latex]{\text{[CO}}_{3}^{2-}][/latex]

Q = [Ca²⁺] [latex]{\text{[CO}}_{3}^{2-}][/latex] = (0.003)(0.003) = 9 × 10^–6

K_sp = 4.8 × 10^–9 < 9 × 10^–6

The ion product does exceed K_sp, so CaCO₃ does precipitate.

(b) [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]

K_sp = [Co²⁺][OH^–]² = 2 × 10^–16

test K_sp against Q = [Co²⁺][OH^–]²

Q = [Co²⁺][OH^–]² = (0.01)(1 × 10^–7)² = 1 × 10^–16

K_sp = 2 × 10^–16 > 1 × 10^–16

The ion product does not exceed K_sp, so the compound does not precipitate.

(c) CaHPO₄: (K_sp = 5 × 10^–6):

Q = [Ca²⁺] [latex]{\text{[HPO}}_{4}^{2-}][/latex] = (0.01)(2 × 10^–6) = 2 × 10^–8 < K_sp

The ion product does not exceed K_sp, so compound does not precipitate.

(d) Pb₃(PO₄)₂: (K_sp = 3 × 10^–44):

Q = [Pb²⁺]³ [latex]{{\text{[PO}}_{4}^{3-}]}^{2}[/latex] = (0.01)³(1 × 10^–13)² = 1 × 10^–32 > K_sp

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.

K_sp = 1.08 × 10^–10 = [Ba²⁺] [latex]\left[{\text{SO}}_{4}^{2-}\right][/latex] = (0.0758) [latex]{\text{[SO}}_{4}^{2-}][/latex]

[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:

[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]

K_sp = 1.8 × 10^–18 = [Ag⁺]³ [latex]{\text{[PO}}_{4}{}^{\text{3-}}][/latex] = (0.0125)³ [latex]{\text{[PO}}_{4}{}^{3-}][/latex]

[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]

[Ag⁺]² [latex]\left[{\text{CO}}_{3}{}^{\text{2-}}\right][/latex] = K_sp = 8.2 × 10^–12

[Ag⁺]²(2.5 × 10^–6) = 8.2 × 10^–12

[latex]\left[\text{Ag}^{+}\right]^{2}=\frac{8.2\times{10}^{-12}}{2.50\times{10}^{-6}}=\text{3.28}\times{10}^{-6}[/latex]

[Ag⁺] = 1.8 × 10^–3M

39. In the K_sp expression, substitute the concentration of Ca²⁺ and solve for [F^–].

K_sp = 3.9 × 10^–11 = [Ca²⁺][F^–]² = (1.0 × 10^–4)[F^–]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]

[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.

[latex]\frac{0.1}{100}[/latex] × 0.010 mol/L = 1 × 10^–5M = [Cu²⁺]

[Cu²⁺][S^2–] = K_sp = 6.7 × 10^–42

(1 × 10^–5)[S^2–] = 6.7 × 10^–42

[S^2–] = 7 × 10^–37M

[Cd²⁺][S^2–]K_sp = 2.8 × 10^–35

[Cd²⁺](7 × 10^–37) = 2.8 × 10^–35

[Cd²⁺] = 4 × 10¹M

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.

(a) [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:

[latex]\left[{\text{Hg}}_{2}{}^{\text{2+}}\right]=\frac{0.1%}{100%}\times \text{0.10}=1\times {10}^{-4}\text{}M[/latex]

and

[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];

(b) [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, [latex]{\text{[SO}}_{4}^{2-}][/latex] = 1 × 10^–4M and

[latex]\left[{\text{Ba}}^{\text{2+}}\right]=\frac{1.08\times {10}^{-10}}{1\times {10}^{-4}}=1\times {10}^{-6}\text{}M[/latex];

(c) 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:

[Co²⁺][S^2–] = K_sp = 6.7 × 10^–29

(0.10)[S^2–] = 6.7 × 10^–29

[S^2–] = 6.7 × 10^–28

At that concentration:

[Hg²⁺](6.7 × 10^–28) = 2 × 10^–59

[Hg²⁺] = 3 × 10^–32M

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:

[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]

[latex]\left[\text{H}_{3}\text{O}^{+}\right]=1.5M[/latex]

A solution more basic than this would supply enough S^2– for CoS to precipitate.

(d) 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

[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]

[OH^–] = 7 × 10^–4M

When Sr(OH)₂·8H₂O just begins to precipitate:

[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]

[OH^–] = 0.057 M

If [OH^–] is maintained less than 0.056 M, then Zn²⁺ will precipitate and Sr²⁺ will not.

(e) Ba²⁺ and Mg²⁺: Add [latex]{\text{SO}}_{4}^{2-}[/latex]. MgSO₄ is soluble and BaSO₄ is not (K_sp = 2 × 10^–10).

(f) [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

[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]

Ba(OH)₂·8H₂O begins to precipitate when:

[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]

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.

For AgCl: K_sp = 1.8 × 10^–10 = [Ag⁺][Cl^–]

[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]

For AgI: K_sp = 1.5 × 10^–16 = [Ag⁺][I^–]

[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]

As the value of [Ag⁺] is smaller for AgI, AgI will precipitate first.

49. The dissolution of Ca₃(PO₄)₂ yields:

[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]

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:

K_sp = 1 × 10^–25 = [Ca²⁺]³ [latex]{{\text{[PO}}_{4}{}^{3-]}}^{2}[/latex]

[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]

[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]

[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.

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:

[latex]\frac{1.304{\text{g L}}^{-1}}{24.305{\text{g mol}}^{-1}}=\text{0.05365}M[/latex]

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:

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:

[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]

When the initial 1.035 g/L is reduced to 0.1% of the original, the molarity is calculated as:

[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]

The added amount of OH^– required is found from the solubility product:

[Mg²⁺][OH^–]² = (5.369 × 10^–5)[OH^–]² = 1.5 × 10^–11 = K_sp

[OH^–] = 5.29 × 10^–4

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:

[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]

The total Ca(OH)₂ required is 3.97 kg + 0.020 kg = 3.99 kg.

53. (a) 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;

(b) K_sp = [Cu²⁺] [latex]{\left[{\text{IO}}_{3}{}^{-}\right]}^{2}[/latex] = x(2x)² = 7.4 × 10^–8

4x³ = 7.4 × 10^–8

x³ = 1.85 × 10^–8

x = 2.64 × 10^–3

[Cu²⁺] = 2.6 × 10^–3

[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]

7.1 × 10^–12 = (x)(2x)² = 4x³

x = 1.21 × 10^–4M = [Mg²⁺]

[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.