Chapter 13, Problem 108E

Arsenic acid (H₃AsO₄) is a triprotic acid with $K_{a_1}$ = 5.5 × 10⁻³, $K_{a_2}$ = 1.7 × 10⁻⁷, and $K_{a_3}$ = 5.1 × 10⁻¹². Calculate [H⁺], [OH⁻], [H₃AsO₄], [H₂AsO₄⁻], [HAsO₄²⁻], and [AsO₄³⁻] in a 0.20-M arsenic acid solution.

Interpretation Introduction

Interpretation: The value of $\left[{\text{H}}^{+}\right],\left[{\text{OH}}^{-}\right],\left[{\text{H}}_2{\text{AsO}}_4{}^{-}\right],\left[{\text{HAsO}}_4^{2-}\right]$ and $\left[{\text{AsO}}_4^{3-}\right]$ for a $0.20\text{M}$ solution of arsenic acid is to be calculated.

Concept introduction: Arsenic acid is a weak acid; therefore, it dissociates up to a very small extent and slowly in an aqueous solution.

To determine: The value of $\left[{\text{H}}^{+}\right],\left[{\text{OH}}^{-}\right],\left[{\text{H}}_3{\text{AsO}}_4{}^{-}\right],\left[{\text{HAsO}}_4^{2-}\right]$ and $\left[{\text{AsO}}_4^{3-}\right]$ for a $0.20\text{M}$ solution of arsenic acid.

Answer to Problem 108E

Answer

• The value of $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{0.03064M}$.
• The value of $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{3.27\times 10^{-13}M}$.
• The value of $\left[{\text{H}}_2{\text{AsO}}_4^{-}\right]$ is $\underset{\_}{0.0305M}$.
• The value of $\left[{\text{HAsO}}_4^{-}\right]$ is $\underset{\_}{7.18\times 10^{-5}M}$.
• The value of $\left[{\text{AsO}}_4^{3-}\right]$ is $\underset{\_}{1.9\times 10^{-8}M}$.
• The value of $\left[{\text{H}}_3{\text{AsO}}_4\right]$ is $\underset{\_}{1.69M}$.

Explanation of Solution

Explanation

Given

The standard value of $K_{{\text{a}}_1}$ for arsenic acid is $5.5\times {10}^{-3}$.

The standard value of $K_{{\text{a}}_2}$ for arsenic acid is $1.7\times {10}^{-7}$.

The standard value of $K_{{\text{a}}_3}$ for arsenic acid is $5.1\times {10}^{-12}$.

The dissociated amount of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{H}}_2{\text{AsO}}_4^{-}\right]$ from the arsenic acid $\left({\text{H}}_3{\text{AsO}}_4\right)$ is assumed to be $x$. The concentration of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{H}}_2{\text{AsO}}_4^{-}\right]$ is calculated by using the ICE (Initial Change Equilibrium) table.

$\begin{array}{cccccc}& {\text{H}}_3{\text{AsO}}_4& \rightleftharpoons & {\text{H}}^{+}& +& {\text{H}}_2{\text{AsO}}_4^{-}\\ \text{Initial}\left(\text{M}\right):& 0.20& & 0& & 0\\ \text{Change}\left(\text{M}\right):& -x& & x& & x\\ \text{Equilibrium}\left(\text{M}\right):& 0.20-x& & x& & x\end{array}$

The value of acid dissociation equilibrium constant $K_{\text{a}}$ is calculated by the formula,

$K_{\text{a}}=\frac{\left[\text{Concentration}\text{of}\text{products}\right]}{\left[\text{Concentration}\text{of}\text{reactants}\right]}$

Thus, $K_{\text{a}}$ for the above stated reaction is,

$K_{{\text{a}}_1}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{H}}_2{\text{AsO}}_4^{2-}\right]}{\left[{\text{H}}_3{\text{AsO}}_4\right]}$

Substitute the values of $K_{{\text{a}}_1}$, concentration of reactants and products from the ICE table in the above expression.

$\begin{array}{c}5.5\times {10}^{-3}=\frac{\left[x\right]\left[x\right]}{\left[0.20-x\right]}\\ x^2+5.5\times {10}^{-3}x-1.1\times {10}^{-3}=0\\ x=\underset{\_}{0.0305M}\end{array}$

Hence, the values of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{H}}_2{\text{AsO}}_4^{-}\right]$ at equilibrium is $\underset{\_}{0.0305M}$.

The dissociated amount of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{HAsO}}_4^{2-}\right]$ from ${\text{H}}_2{\text{AsO}}_4^{1-}$ is assumed to be $y$. The concentration of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{HAsO}}_4^{2-}\right]$ is calculated by using the ICE (Initial Change Equilibrium) table.

$\begin{array}{cccccc}& {\text{H}}_2{\text{AsO}}_4^{-}& \rightleftharpoons & {\text{H}}^{+}& +& {\text{HAsO}}_4^{2-}\\ \text{Initial}\left(\text{M}\right):& 0.0305& & 0& & 0\\ \text{Change}\left(\text{M}\right):& -y& & y& & y\\ \text{Equilibrium}\left(\text{M}\right):& 0.0305-y& & y& & y\end{array}$

The value of acid dissociation equilibrium constant $K_{\text{a}}$ is calculated by the formula.

$K_{\text{a}}=\frac{\left[\text{Concentration}\text{of}\text{products}\right]}{\left[\text{Concentration}\text{of}\text{reactants}\right]}$

Thus, $K_{\text{a}}$ for the above stated reaction is,

$K_{{\text{a}}_2}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{HAsO}}_4^{2-}\right]}{\left[{\text{H}}_2{\text{AsO}}_4^{1-}\right]}$

Substitute the values of $K_{{\text{a}}_2}$, concentration of reactants and products from the ICE table in the above expression.

$\begin{array}{c}1.7\times {10}^{-7}=\frac{\left[x\right]\left[x\right]}{\left[0.0305-x\right]}\\ x^2+1.7\times {10}^{-7}x-5.18\times {10}^{-9}=0\\ x=\underset{\_}{7.18\times 10^{-5}M}\end{array}$

Hence, the values of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{HAsO}}_4^{2-}\right]$ at equilibrium is $\underset{\_}{7.18\times 10^{-5}M}$.

The dissociated amount of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{AsO}}_4^{3-}\right]$ from the ${\text{HAsO}}_4^{2-}$ is assumed to be $z$. The concentration of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{AsO}}_4^{3-}\right]$ is calculated by using the ICE (Initial Change Equilibrium) table.

$\begin{array}{cccccc}& {\text{HAsO}}_4^{2-}& \rightleftharpoons & {\text{H}}^{+}& +& {\text{AsO}}_4^{3-}\\ \text{Initial}\left(\text{M}\right):& 7.18\times {10}^{-5}& & 0& & 0\\ \text{Change}\left(\text{M}\right):& -z& & z& & z\\ \text{Equilibrium}\left(\text{M}\right):& 7.18\times {10}^{-5}-z& & z& & z\end{array}$

The value of acid dissociation equilibrium constant $K_{\text{a}}$ is calculated by the formula.

$K_{\text{a}}=\frac{\left[\text{Concentration}\text{of}\text{products}\right]}{\left[\text{Concentration}\text{of}\text{reactants}\right]}$

Thus, $K_{\text{a}}$ for the above stated reaction is,

$K_{{\text{a}}_3}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{AsO}}_4^{3-}\right]}{\left[{\text{HAsO}}_4^{2-}\right]}$

Substitute the values of $K_{{\text{a}}_3}$, concentration of reactants and products from the ICE table in the above expression.

$\begin{array}{c}5.1\times {10}^{-12}=\frac{\left[x\right]\left[x\right]}{\left[7.18\times {10}^{-5}-x\right]}\\ x^2+5.1\times {10}^{-12}x-3.66\times {10}^{-16}=0\\ x=\underset{\_}{1.9\times 10^{-8}M}\end{array}$

Hence, the values of $\left[{\text{H}}^{+}\right]$ and $\left[{\text{AsO}}_4^{3-}\right]$ at equilibrium is $\underset{\_}{1.9\times 10^{-8}M}$.

Thus the total value of $\begin{array}{c}\left[{\text{H}}^{+}\right]=0.0305\text{M}+7.18\times {10}^{-5}\text{M}+1.9\times {10}^{-8}\text{M}\\ =\underset{\_}{0.03064M}\end{array}$

The value of ionic-product of water is given by the formula.

$\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}$

Where,

• $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion.
• $\left[{\text{OH}}^{-}\right]$ is concentration of hydroxide ion.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above expression.

$\begin{array}{l}\left[0.03064\text{M}\right]\times \left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}\\ \left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{0.03064}\\ \left[{\text{OH}}^{-}\right]=\underset{\_}{3.27\times 10^{-13}M}\end{array}$

The dissociation reaction of ${\text{H}}_3{\text{AsO}}_4$ is,

${\text{H}}_3{\text{AsO}}_4\rightleftharpoons {\text{H}}^{+}+{\text{H}}_2{\text{AsO}}_4^{2-}$

The acid dissociation constant $K_{{\text{a}}_1}$ for the above stated reaction is,

$K_{{\text{a}}_1}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{H}}_2{\text{AsO}}_4^{2-}\right]}{\left[{\text{H}}_3{\text{AsO}}_4\right]}$

Substitute the values of concentration of reactants, products and $K_{{\text{a}}_1}$ in the above expression.

$\begin{array}{c}5.5\times {10}^{-3}=\frac{\left[0.0305\right]\left[0.0305\right]}{\left[{\text{H}}_3{\text{PO}}_4\right]}\\ \left[{\text{H}}_3{\text{AsO}}_4\right]=\frac{0.00093025}{5.5\times {10}^{-3}}\\ \left[{\text{H}}_3{\text{AsO}}_4\right]=\underset{\_}{1.69M}\end{array}$

Conclusion

Conclusion

• The value of $\left[{\text{H}}^{+}\right]$ is $\underset{\_}{0.03064M}$.
• The value of $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{3.27\times 10^{-13}M}$.
• The value of $\left[{\text{H}}_2{\text{AsO}}_4^{-}\right]$ is $\underset{\_}{0.0305M}$.
• The value of $\left[{\text{HAsO}}_4^{-}\right]$ is $\underset{\_}{7.18\times 10^{-5}M}$.
• The value of $\left[{\text{AsO}}_4^{3-}\right]$ is $\underset{\_}{1.9\times 10^{-8}M}$.
• The value of $\left[{\text{H}}_3{\text{AsO}}_4\right]$ is $\underset{\_}{1.69M}$.