Chapter 6, Problem 6G.5E

(a)

Interpretation Introduction

Interpretation:

The concentration of hydronium ion in a solution that is $0.075MHCN\left(aq\right)$ and $0.060MNaCN\left(aq\right)$ has to be calculated.

Explanation of Solution

$HCN$ is a weak acid and its conjugate base is $C{N}^{-}.$  The required equilibrium is given below.

  $\begin{array}{l}HCN\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C{N}^{-}\left(aq\right)\\ \\ K_a=\frac{\left[H_3{O}^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}=4.9\times {10}^{-10}\end{array}$

Given that, the buffer contains $0.075MHCN\left(aq\right)$ and $0.060MNaCN\left(aq\right).$

The equilibrium concentration of $H_3{O}^{+}$ ions can be calculated as shown below.

  $\begin{array}{l}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}=4.9\times {10}^{-10}\\ \\ \left[H_3{O}^{+}\right]=K_a\times \frac{\left[HCN\right]}{\left[C{N}^{-}\right]}\\ \\ \left[H_3{O}^{+}\right]\approx \left(4.9\times {10}^{-10}\right)\times \frac{0.075}{0.060}\\ \\ \left[H_3{O}^{+}\right]=6.1\times 10^{-10}M\end{array}$

Therefore, the concentration of hydronium ion is $6.1\times 10^{-10}M.$

(b)

Interpretation Introduction

Interpretation:

The concentration of hydronium ion in a solution that is $0.20MN{H}_{2}N{H}_2\left(aq\right)$ and $0.30MNaCl\left(aq\right)$ has to be calculated.

Explanation of Solution

$N{H}_{2}N{H}_2$ is a weak base and its conjugate acid is $N{H}_{2}N{H}_3{}^{+}.$  The required equilibrium is given below.

  $\begin{array}{l}N{H}_{2}N{H}_2\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons O{H}^{-}\left(aq\right)+N{H}_{2}N{H}_3{}^{+}\left(aq\right)\\ \\ K_b=\frac{\left[O{H}^{-}\right]\left[N{H}_{2}N{H}_3{}^{+}\right]}{\left[N{H}_{2}N{H}_2\right]}=1.7\times {10}^{-6}\end{array}$

Given that, the buffer contains $0.20MN{H}_{2}N{H}_2\left(aq\right)$ and $0.30MNaCl\left(aq\right).$

$NaCl$ will have no effect on the concentration of $H_3{O}^{+}.$  Therefore, no need to consider $NaCl$ for the calculation of concentration of $H_3{O}^{+}$ ions.

An equilibrium table can be set up as given below.

  $\begin{array}{cccc}Composition(M)& N{H}_{2}N{H}_2& O{H}^{-}& N{H}_{2}N{H}_3{}^{+}\\ Initialconcentration& 0.20& 0& 0\\ Changeinconcentration& -x& +x& +x\\ Equilibriumconcentration& 0.20-x& x& x\end{array}$

Now, the above equilibrium expression can be solved for $x.$

  $\begin{array}{l}{K}_b=\frac{\left[O{H}^{-}\right]\left[N{H}_{2}N{H}_3{}^{+}\right]}{\left[N{H}_{2}N{H}_2\right]}=1.7\times {10}^{-6}\\ \\ 1.7\times {10}^{-6}=\frac{\left(x\right)\left(x\right)}{\left(0.20-x\right)}\\ \\ x^2=\left(1.7\times {10}^{-6}\right)\left(0.20\right)\left[\text{Assuming}\text{x}\ll 0.20\right]\\ \\ x=5.83\times {10}^{-4}\\ \\ \left[O{H}^{-}\right]=5.83\times {10}^{-4}M\\ \\ \left[H_3{O}^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{1.0\times {10}^{-14}}{5.83\times {10}^{-4}}=1.7\times 10^{-11}M\end{array}$

Therefore, the concentration of hydronium ion is $1.7\times 10^{-11}M.$

(c)

Interpretation Introduction

Interpretation:

The concentration of hydronium ion in a solution that is $0.015MHCN\left(aq\right)$ and $0.030MNaCN\left(aq\right)$ has to be calculated.

Explanation of Solution

$HCN$ is a weak acid and its conjugate base is $C{N}^{-}.$  The required equilibrium is given below.

  $\begin{array}{l}HCN\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C{N}^{-}\left(aq\right)\\ \\ K_a=\frac{\left[H_3{O}^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}=4.9\times {10}^{-10}\end{array}$

Given that, the buffer contains $0.015MHCN\left(aq\right)$ and $0.030MNaCN\left(aq\right).$

The equilibrium concentration of $H_3{O}^{+}$ ions can be calculated as shown below.

  $\begin{array}{l}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}=4.9\times {10}^{-10}\\ \\ \left[H_3{O}^{+}\right]=K_a\times \frac{\left[HCN\right]}{\left[C{N}^{-}\right]}\\ \\ \left[H_3{O}^{+}\right]\approx \left(4.9\times {10}^{-10}\right)\times \frac{0.015}{0.030}\\ \\ \left[H_3{O}^{+}\right]=2.5\times 10^{-10}M\end{array}$

Therefore, the concentration of hydronium ion is $2.5\times 10^{-10}M.$

(d)

Interpretation Introduction

Interpretation:

The concentration of hydronium ion in a solution that is $0.125MN{H}_{2}N{H}_2\left(aq\right)$ and $0.125MN{H}_{2}N{H}_{3}Br\left(aq\right)$ has to be calculated.

Explanation of Solution

$N{H}_{2}N{H}_2$ is a weak base and its conjugate acid is $N{H}_{2}N{H}_3{}^{+}.$  The required equilibrium is given below.

  $\begin{array}{l}N{H}_{2}N{H}_2\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons O{H}^{-}\left(aq\right)+N{H}_{2}N{H}_3{}^{+}\left(aq\right)\\ \\ K_b=\frac{\left[O{H}^{-}\right]\left[N{H}_{2}N{H}_3{}^{+}\right]}{\left[N{H}_{2}N{H}_2\right]}=1.7\times {10}^{-6}\end{array}$

Given that, the buffer contains $0.125MN{H}_{2}N{H}_2\left(aq\right)$ and $0.125MN{H}_{2}N{H}_{3}Br\left(aq\right).$

The equilibrium concentration of $H_3{O}^{+}$ ions can be calculated as shown below.

  $\begin{array}{l}{K}_b=\frac{\left[O{H}^{-}\right]\left[N{H}_{2}N{H}_3{}^{+}\right]}{\left[N{H}_{2}N{H}_2\right]}=1.7\times {10}^{-6}\\ \\ \left[O{H}^{-}\right]=K_b\times \frac{\left[N{H}_{2}N{H}_2\right]}{\left[N{H}_{2}N{H}_3{}^{+}\right]}\\ \\ \left[O{H}^{-}\right]\approx \left(1.7\times {10}^{-6}\right)\times \frac{0.125}{0.125}\\ \\ \left[O{H}^{-}\right]=1.7\times {10}^{-6}\text{M}\\ \\ \left[H_3{O}^{+}\right]=\frac{1.0\times {10}^{-14}}{1.7\times {10}^{-6}}=5.9\times 10^{-9}M\end{array}$

Therefore, the concentration of hydronium ion is $5.9\times 10^{-9}M.$