1. When the amount of solid is so small that a saturated solution is not produced.

3. Cadmium ions associate with ammonia molecules in solution to form the complex ion [latex]{\left[\text{Cd}{\left({\text{NH}}_{3}\right)}_{4}\right]}^{\text{2+}}[/latex], which is defined by the following equilibrium:

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

The formation of the complex ion requires 4 mol of NH₃ for each mol of Cd²⁺. First, calculate the initial amounts of Cd²⁺ and of NH₃ available for association:

[latex]\left[{\text{Cd}}^{\text{2+}}\right]=\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]\left[{\text{NH}}_{3}\right]=\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 Cd²⁺ would require 4(4.00 [latex]\times[/latex] 10^–3 mol/L) of NH₃ or a 1.6 [latex]\times[/latex] 10^–2–M solution. Due to the large value of K_f and the substantial excess of NH₃, 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 [Cd²⁺]:

	[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

[latex]{K}_{\text{f}}=4.0\times {10}^{6}=\frac{\left[\text{Cd}{\left({\text{NH}}_{3}\right)}_{4}{}^{\text{2+}}\right]}{\left[{\text{Cd}}^{\text{2+}}\right]{\left[{\text{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. For the formation reaction:

[latex]\begin{array}{l}{\text{Al}}^{\text{3+}}\left(aq\right)+6{\text{F}}^{-}\left(aq\right)\rightleftharpoons {\text{AlF}}_{6}{}^{\text{3-}}\left(aq\right)\\{K}_{\text{f}}=\frac{\left[{\text{AlF}}_{6}{}^{\text{3-}}\right]}{\left[{\text{Al}}^{\text{3+}}\right]{\left[{\text{F}}^{-}\right]}^{6}}=\frac{1}{{K}_{\text{d}}}=\frac{1}{2\times {10}^{-24}}=5\times {10}^{23}\end{array}[/latex]

7.

	[Cd(CN)42−]	[CN−]	[Cd2+]
Initial concentration (M)	0.250	0	0
Equilibrium (M)	0.250 − x	4x	x

[latex]{K}_{\text{d}}=\frac{\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{Cd}{\left(\text{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 = [Cd²⁺] = 9.5 [latex]\times[/latex] 10^–5M

4x = [CN^–] = 3.8 [latex]\times[/latex] 10^–4M

9.

	[Co(NH3)63+]	[Co3+]	[NH3]
Initial concentration (M)	0.500	0	0
Equilibrium (M)	0.500 − x	x	6x

[latex]{K}_{\text{d}}=\frac{\left[{\text{Co}}^{\text{2+}}\right]{\left[{\text{NH}}_{3}\right]}^{6}}{\left[\text{Co}{\left({\text{NH}}_{3}\right)}_{6}{}^{\text{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 = [Co³⁺] = 3.0 [latex]\times[/latex] 10^–6M

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

11. Because K_sp is small and K_f is large, most of the Ag⁺ is used to form [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex]; that is:

[latex]\begin{array}{rll}\left[\text{Ag}^{+}\right]&\lt&\left[\text{Ag}\left(\text{CN}\right)_{2}^{-}\right]\\\left[\text{Ag}\left(\text{CN}\right)_{2}^{-}\right]&\approx&2.0\times{10}^{-1}M\end{array}[/latex]

The CN^– from the dissolution and the added CN^– exist as CN^– and [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex]. Let x be the change in concentration upon addition of CN^–. Its initial concentration is approximately 0.

[CN^–] + 2 [latex]\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right][/latex] = 2 [latex]\times[/latex] 10^–1 + x

Because K_sp is small and K_f is large, most of the CN^– is used to form [latex]\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right][/latex]; that is:

[latex]\begin{array}{rll}\left[{\text{CN}}^{-}\right]&<&2\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right]\\2\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right]&\approx&2.0\times {10}^{-1}+x\end{array}[/latex]

2(2.0 [latex]\times[/latex] 10^–1) – 2.0 [latex]\times[/latex] 10^–1 = x

2.0 [latex]\times[/latex] 10^–1M [latex]\times[/latex] L = mol CN^– added

The solution has a volume of 100 mL.

2 [latex]\times[/latex] 10^–1 mol/L [latex]\times[/latex] 0.100 L = 2 [latex]\times[/latex] 10^–2 mol

mass KCN = 2.0 [latex]\times[/latex] 10^–2 mol KCN [latex]\times[/latex] 65.120 g/mol = 1.3 g

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

[latex]\begin{array}{l}\text{AgBr}\left(s\right)\rightleftharpoons\text{Ag}^{+}\left(aq\right)+\text{Br}^{-}\left(aq\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{sp}}=3.3\times{10}^{-13}\\\text{Ag}^{+}\left(aq\right)+2{\text{S}}_2\text{O}_{3}^{2-}\left(aq\right)\rightleftharpoons\text{Ag}\left(\text{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[\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}^{\text{3-}}\right]\left[{\text{Br}}^{-}\right]}{{\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right]}^{2}}=15.51[/latex]

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

[latex]\text{moles present}=\frac{0.27\text{g}\text{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]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}:[/latex]

	[Ag+]	[S2O32−]
Initial concentration (M)	0	0
Equilibrium (M)	[latex]\frac{1}{2}x[/latex]	x

[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=\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right][/latex]

The formula mass of Na₂S₂O₃•5H₂O is 248.13 g/mol. The total [latex]\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right][/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

15.

(a)

(b)

(c)

(d)

(e)

17.

(a)

(b) [latex]{\text{H}}_{3}{\text{O}}^{+}+{\text{CH}}_{3}{}^{-}\longrightarrow {\text{CH}}_{4}+{\text{H}}_{2}\text{O}[/latex]

(c) [latex]\text{CaO}+{\text{SO}}_{3}\longrightarrow \text{CaSO}4[/latex]

(d) [latex]{\text{NH}}_{4}{}^{+}+{\text{C}}_{2}{\text{H}}_{5}{\text{O}}^{-}\longrightarrow {\text{C}}_{2}{\text{H}}_{5}\text{OH}+{\text{NH}}_{3}[/latex]

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

The number of moles of AgNO₃ added is:

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

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

[latex]{\text{AgCN}}_{2}{}^{-}\left(aq\right)+{\text{Ag}}^{\text{+}}\left(aq\right)\rightleftharpoons 2\text{AgCN}\left(s\right)[/latex]

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

mol CN^– < [latex]\left[\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}\right][/latex]

mol NaCN = 2 mol [ [latex]\text{Ag}{\left(\text{CN}\right)}_{2}{}^{-}[/latex] ] = 5.74 [latex]\times[/latex] 10^–4 mol

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

21. [latex]{\text{HNO}}_{3}\left(l\right)+\text{HF}\left(l\right)\longrightarrow {\text{H}}_{2}{\text{NO}}_{3}{}^{+}+{\text{F}}^{-}[/latex]; [latex]\text{HF}\left(l\right)+{\text{BF}}_{3}\left(g\right)\longrightarrow {\text{H}}^{+}+{\text{BF}}_{4}[/latex]

23.  (a) [latex]{\text{H}}_{3}{\text{BO}}_{3}+{\text{H}}_{2}\text{O}\longrightarrow {\text{H}}_{4}{\text{BO}}_{4}{}^{-}+{\text{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.