Chapter 6, Problem 6G.6E

(a)

Interpretation Introduction

Interpretation:

The concentration of hydronium ion in a solution that is $0.12MHBrO\left(aq\right)$ and $0.160MNaBrO\left(aq\right)$ has to be calculated.

Explanation of Solution

$HBrO$ is a weak acid and its conjugate base is $Br{O}^{-}.$  The required equilibrium is given below.

  $\begin{array}{l}HBrO\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+Br{O}^{-}\left(aq\right)\\ \\ K_a=\frac{\left[H_3{O}^{+}\right]\left[Br{O}^{-}\right]}{\left[HBrO\right]}=2.0\times {10}^{-9}\end{array}$

Given that, the buffer contains $0.12MHBrO\left(aq\right)$ and $0.160MNaBrO\left(aq\right).$

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

  $\begin{array}{l}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[Br{O}^{-}\right]}{\left[HBrO\right]}=2.0\times {10}^{-9}\\ \\ \left[H_3{O}^{+}\right]=K_a\times \frac{\left[HBrO\right]}{\left[Br{O}^{-}\right]}\\ \\ \left[H_3{O}^{+}\right]\approx \left(2.0\times {10}^{-9}\right)\times \frac{0.12}{0.160}\\ \\ \left[H_3{O}^{+}\right]=1.5\times 10^{-9}M\end{array}$

Therefore, the concentration of hydronium ion is $1.5\times 10^{-9}M.$

(b)

Interpretation Introduction

Interpretation:

The concentration of hydronium ion in a solution that is $0.250MC{H}_{3}N{H}_2\left(aq\right)$ and $0.150MC{H}_{3}N{H}_{3}Cl\left(aq\right)$ has to be calculated.

Explanation of Solution

$C{H}_{3}N{H}_2$ is a weak base and its conjugate acid is $C{H}_{3}N{H}_3{}^{+}.$  The required equilibrium is given below.

  $\begin{array}{l}C{H}_{3}N{H}_2\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons O{H}^{-}\left(aq\right)+C{H}_{3}N{H}_3{}^{+}\left(aq\right)\\ \\ K_b=\frac{\left[O{H}^{-}\right]\left[C{H}_{3}N{H}_3{}^{+}\right]}{\left[C{H}_{3}N{H}_2\right]}=3.6\times {10}^{-4}\end{array}$

Given that, the buffer contains $0.250MC{H}_{3}N{H}_2\left(aq\right)$ and $0.150MC{H}_{3}N{H}_{3}Cl\left(aq\right).$

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

  $\begin{array}{l}{K}_b=\frac{\left[O{H}^{-}\right]\left[C{H}_{3}N{H}_3{}^{+}\right]}{\left[C{H}_{3}N{H}_2\right]}=3.6\times {10}^{-4}\\ \\ \left[O{H}^{-}\right]=K_b\times \frac{\left[C{H}_{3}N{H}_2\right]}{\left[C{H}_{3}N{H}_3{}^{+}\right]}\\ \\ \left[O{H}^{-}\right]\approx \left(3.6\times {10}^{-4}\right)\times \frac{0.250}{0.150}\\ \\ \left[O{H}^{-}\right]=6.0\times {10}^{-4}\text{M}\\ \\ \left[H_3{O}^{+}\right]=\frac{1.0\times {10}^{-14}}{6.0\times {10}^{-4}}=1.67\times 10^{-11}M\end{array}$

Therefore, the concentration of hydronium ion is $1.67\times 10^{-11}M.$

(c)

Interpretation Introduction

Interpretation:

The concentration of hydronium ion in a solution that is $0.160MHBrO\left(aq\right)$ and $0.320MNaBrO\left(aq\right)$ has to be calculated.

Explanation of Solution

$HBrO$ is a weak acid and its conjugate base is $Br{O}^{-}.$  The required equilibrium is given below.

  $\begin{array}{l}HBrO\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+Br{O}^{-}\left(aq\right)\\ \\ K_a=\frac{\left[H_3{O}^{+}\right]\left[Br{O}^{-}\right]}{\left[HBrO\right]}=2.0\times {10}^{-9}\end{array}$

Given that, the buffer contains $0.160MHBrO\left(aq\right)$ and $0.320MNaBrO\left(aq\right).$

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

  $\begin{array}{l}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[Br{O}^{-}\right]}{\left[HBrO\right]}=2.0\times {10}^{-9}\\ \\ \left[H_3{O}^{+}\right]=K_a\times \frac{\left[HBrO\right]}{\left[Br{O}^{-}\right]}\\ \\ \left[H_3{O}^{+}\right]\approx \left(2.0\times {10}^{-9}\right)\times \frac{0.160}{0.320}\\ \\ \left[H_3{O}^{+}\right]=1.0\times 10^{-9}M\end{array}$

Therefore, the concentration of hydronium ion is $1.0\times 10^{-9}M.$

(d)

Interpretation Introduction

Interpretation:

The concentration of hydronium ion in a solution that is $0.250MC{H}_{3}N{H}_2\left(aq\right)$ and $0.250MNaBr\left(aq\right)$ has to be calculated.

Explanation of Solution

$C{H}_{3}N{H}_2$ is a weak base and its conjugate acid is $C{H}_{3}N{H}_3{}^{+}.$  The required equilibrium is given below.

  $\begin{array}{l}C{H}_{3}N{H}_2\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons O{H}^{-}\left(aq\right)+C{H}_{3}N{H}_3{}^{+}\left(aq\right)\\ \\ K_b=\frac{\left[O{H}^{-}\right]\left[C{H}_{3}N{H}_3{}^{+}\right]}{\left[C{H}_{3}N{H}_2\right]}=3.6\times {10}^{-4}\end{array}$

Given that, the buffer contains $0.250MC{H}_{3}N{H}_2\left(aq\right)$ and $0.250MC{H}_{3}N{H}_{3}Cl\left(aq\right).$

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

An equilibrium table can be set up as given below.

  $\begin{array}{cccc}Composition(M)& C{H}_{3}N{H}_2& O{H}^{-}& C{H}_{3}N{H}_3{}^{+}\\ Initialconcentration& 0.250& 0& 0\\ Changeinconcentration& -x& +x& +x\\ Equilibriumconcentration& 0.250-x& x& x\end{array}$

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

  $\begin{array}{l}{K}_b=\frac{\left[O{H}^{-}\right]\left[C{H}_{3}N{H}_3{}^{+}\right]}{\left[C{H}_{3}N{H}_2\right]}=3.6\times {10}^{-4}\\ \\ 3.6\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{\left(0.250-x\right)}\\ \\ x^2=\left(3.6\times {10}^{-4}\right)\left(0.250\right)\left[\text{Assuming}\text{x}\ll 0.250\right]\\ \\ x=9.0\times {10}^{-5}\\ \\ \left[O{H}^{-}\right]=9.0\times {10}^{-5}M\\ \\ \left[H_3{O}^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{1.0\times {10}^{-14}}{9.0\times {10}^{-5}}=1.12\times 10^{-10}M\end{array}$

Therefore, the concentration of hydronium ion is $1.12\times 10^{-10}M.$