Chapter 18.3, Problem 32PP

(a)

Interpretation Introduction

Interpretation:

$k_a$ value of given acid is to be calculated.

Concept introduction:

Acid ionization constant tells about equilibrium that whether reactants are favored or products.

For a reaction,

$HA+H_{2}O\rightleftharpoons A^{-}+H_3{O}^{+}$

Acid ionization constant is given by:

$\begin{array}{l}{k}_a=\frac{\left[A^{-}\right]\times \left[H_3{O}^{+}\right]}{\left[HA\right]}\\ where,\\ \left[A^{-}\right]isconcentrationof{A}^{-}\\ \left[H_3{O}^{+}\right]isconcentrationof{H}_3{O}^{+}\\ \left[HA\right]isconcentrationofHA\end{array}$

Answer to Problem 32PP

$k_a=5.4\times {10}^{-3}$

Explanation of Solution

Ionization equation is given by:

$H_{3}As{O}_4+H_{2}O\rightleftharpoons H_{2}As{O}_4{}^{-}+H_3{O}^{+}$

Given data:

$Beforereaction,\left[H_{3}As{O}_4\right]=0.220M$

$pH=1.50$

Now $\left[H_{2}As{O}_4{}^{-}\right]=\left[H^{+}\right]$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=1.50\\ -\log \left[H^{+}\right]=1.50\\ \left[H^{+}\right]=Anti\log (-1.50)\\ =3.2\times {10}^{-2}M\end{array}$

$\begin{array}{c}Atequilibrium,\left[H_{3}As{O}_4\right]=0.220-3.2\times {10}^{-2}\\ =0.188M\end{array}$

Therefore,

$\begin{array}{c}{k}_a=\frac{3.2\times {10}^{-2}\times 3.2\times {10}^{-2}}{0.188}\\ =5.4\times {10}^{-3}\end{array}$

(b)

Interpretation Introduction

Interpretation:

$k_a$ value of given acid is to be calculated.

Concept introduction:

Acid ionization constant tells about equilibrium that whether reactants are favored or products.

For a reaction,

$HA+H_{2}O\rightleftharpoons A^{-}+H_3{O}^{+}$

Acid ionization constant is given by:

$\begin{array}{l}{k}_a=\frac{\left[A^{-}\right]\times \left[H_3{O}^{+}\right]}{\left[HA\right]}\\ where,\\ \left[A^{-}\right]isconcentrationof{A}^{-}\\ \left[H_3{O}^{+}\right]isconcentrationof{H}_3{O}^{+}\\ \left[HA\right]isconcentrationofHA\end{array}$

Answer to Problem 32PP

$k_a=1.1\times {10}^{-2}$

Explanation of Solution

Ionization equation is given by:

$HCl{O}_2+H_{2}O\rightleftharpoons Cl{O}_2{}^{-}+H_3{O}^{+}$

Given data:

$Beforereaction,\left[HCl{O}_2\right]=0.040M$

$pH=1.80$

Now $\left[Cl{O}_2{}^{-}\right]=\left[H^{+}\right]$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=1.80\\ -\log \left[H^{+}\right]=1.80\\ \left[H^{+}\right]=Anti\log (-1.80)\\ =1.6\times {10}^{-2}M\end{array}$

$\begin{array}{c}Atequilibrium,\left[HCl{O}_2\right]=0.04-1.6\times {10}^{-2}\\ =0.024M\end{array}$

Therefore,

$\begin{array}{c}{k}_a=\frac{1.6\times {10}^{-2}\times 1.6\times {10}^{-2}}{0.024}\\ =1.1\times {10}^{-2}\end{array}$