Answer
a. $pH = 5.82$
b. $pH = 10.95$
Work Step by Step
a.
1. Analyze each ion that forms this salt:
$CH_3NH_3^+$: Conjugate pair of a weak base. So, it will affect the pH of the solution.
$Cl^-$: Conjugate pair of a strong acid (HCl). So, it will not affect the pH of the solution.
2. Calculate the $pH$ for a $CH_3NH_3^+$ 0.10M solution:
3. Since $CH_3NH_3^+$ is the conjugate acid of $CH_3NH_2$ , we can calculate its $K_a$ by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 4.4\times 10^{- 4} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 4.4\times 10^{- 4}}$
$K_a = 2.3\times 10^{- 11}$
4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$CH_3NH_3^+(aq) + H_2O(l) \lt -- \gt CH_3NH_2(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $CH_3NH_2$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [CH_3NH_2] = 0 + x = x$
-$[CH_3NH_3^+] = [CH_3NH_3^+]_{initial} - x$
For approximation, we are going to consider $[CH_3NH_3^+]_{initial} = [CH_3NH_3^+]$
5. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CH_3NH_2]}{ [CH_3NH_3^+]}$
$Ka = 2.3 \times 10^{- 11}= \frac{x * x}{ 0.1}$
$Ka = 2.3 \times 10^{- 11}= \frac{x^2}{ 0.1}$
$x^2 = 0.1 \times 2.3 \times 10^{-11} $
$x = \sqrt { 0.1 \times 2.3 \times 10^{-11}} = 1.5 \times 10^{-6} $
Percent dissociation: $\frac{ 1.5 \times 10^{- 6}}{ 0.1} \times 100\% = 0.0015\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [CH_3NH_2] = x = 1.5 \times 10^{- 6}M $
And, since 'x' has a very small value (compared to the initial concentration): $[CH_3NH_3^+] \approx 0.1M$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.5 \times 10^{- 6})$
$pH = 5.82$
-------------
b.
1. Analyze each ion that forms this salt:
$Na^+$: Conjugate pair of a strong base (NaOH). So, it will not affect the pH of the solution.
$CN^-$: Conjugate pair of a weak acid. So, it will affect the pH of the solution, because it acts as a weak base.
2. Calculate the $pH$ for a $0.050M$ $CN^-$ solution:
3. Since $CN^-$ is the conjugate base of $HCN$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 6.2\times 10^{- 10} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 6.2\times 10^{- 10}}$
$K_b = 1.6\times 10^{- 5}$
3. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$CN^-(aq) + H_2O(l) \lt -- \gt HCN(aq) + OH^-(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $HCN$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M
-$[OH^-] = [HCN] = 0 + x = x$
-$[CN^-] = [CN^-]_{initial} - x $
For approximation, we are going to consider $[CN^-]_{initial} = [CN^-]$
4. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HCN]}{ [CN^-]}$
$Kb = 1.6 \times 10^{- 5}= \frac{x * x}{ 0.05}$
$Kb = 1.6 \times 10^{- 5}= \frac{x^2}{ 0.05}$
$x^2 = 1.6 \times 10^{-5} \times 0.05$
$x = \sqrt { 1.6 \times 10^{-5} \times 0.05} = 8.9 \times 10^{-4}$
Percent ionization: $\frac{ 8.9 \times 10^{- 4}}{ 0.05} \times 100\% = 1.8\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HCN] = x = 8.9 \times 10^{- 4}M $
$[CN^-] \approx 0.050M$
5. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 8.9 \times 10^{- 4})$
$pOH = 3.05$
6. Find the pH:
$pH + pOH = 14$
$pH + 3.05 = 14$
$pH = 10.95$
