Chapter 13, Problem 62QAP

Selenious acid, H₂SeO₃, is primarily used to chemically darken copper, brass, and bronze. It is a diprotic acid with the following K_a values: $K_{a1}=2.7\times {10}^{-3}$ and $K_{a2}=5.0\times {10}^{-8}$. What is the pH of a 2.89 M solution of selenious acid? Estimate [HSeO₃^-] and [SeO₃^2-].

Interpretation Introduction

Interpretation:

The pH of solution, the concentration of ${\text{HSeO}}_3{}^{-}\left(aq\right)$ and that of ${\text{SeO}}_3{}^{2-}\left(aq\right)$ needs to be calculated.

Concept introduction:

The dissociation reaction of a weak acid is represented as follows:

$\text{HB}\rightleftarrows {\text{H}}^{+}+{\text{B}}^{-}$

The expression for the acid dissociation constant will be as follows:

$K_a=\frac{\left[{\text{H}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{HB}\right]}$

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion, $\left[{\text{B}}^{-}\right]$ is concentration of conjugate base and $\left[\text{HB}\right]$ is the concentration of weak acid.

The pH of the solution is calculated as follows:

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

Here, $\left[{\text{H}}^{+}\right]$ is concentration of hydrogen ion.

Answer to Problem 62QAP

The pH of solution is 1.054, the concentration of ${\text{HSeO}}_3{}^{-}\left(aq\right)$ is $0.088\text{ M}$ and that of ${\text{SeO}}_3{}^{2-}\left(aq\right)$ is $5.0\times {10}^{-8}$.

Explanation of Solution

The molecular formula of selenium acid is ${\text{H}}_{\text{2}}{\text{SeO}}_3$. The first and second ionization constant is $2.7\times {10}^{-3}$ and $5.0\times {10}^{-8}$ respectively.

The molarity of solution is 2.89 M.

The concentration of each species can be calculated using the ICE table as follows:

$\begin{array}{l}{\text{ H}}_{\text{2}}{\text{SeO}}_3\left(aq\right)\rightleftharpoons {\text{HSeO}}_3{}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\\ I2.89\text{ - -}\\ C-xxx\\ E2.89-xxx\end{array}$

Now, the second ionization reaction is represented as follows:

$\begin{array}{l}{\text{ HSeO}}_3{}^{-}\left(aq\right)\rightleftharpoons {\text{SeO}}_3{}^{2-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\\ Ix\text{ - }x\\ C-yyy\\ Ex-yyx+y\end{array}$

According to above two ICE tables, the equilibrium concentration of ${\text{H}}_{\text{2}}{\text{SeO}}_3\left(aq\right)$, ${\text{HSeO}}_3{}^{-}\left(aq\right)$, ${\text{SeO}}_3{}^{2-}\left(aq\right)$ and ${\text{H}}^{+}\left(aq\right)$ is $2.89-x$, $x-y$, y and $x+y$ respectively.

The expression for dissociation constant is as follows:

${\text{K}}_{a1}=\frac{\left[{\text{HSeO}}_3{}^{-}\left(aq\right)\right]\left[{\text{H}}^{+}\left(aq\right)\right]}{\left[{\text{H}}_{\text{2}}{\text{SeO}}_3\left(aq\right)\right]}$

Putting the values,

$2.7\times {10}^{-3}=\frac{\left(x-y\right)\left(x+y\right)}{\left(2.89-x\right)}$.. ..... (1)

Similarly, the expression for the second ionization constant is as follows:

${\text{K}}_{a2}=\frac{\left[{\text{SeO}}_3{}^{2-}\left(aq\right)\right]\left[{\text{H}}^{+}\left(aq\right)\right]}{\left[{\text{HSeO}}_3{}^{-}\left(aq\right)\right]}$

Putting the values,

$5.0\times {10}^{-8}=\frac{y\left(x+y\right)}{\left(x-y\right)}$.. ..... (2)

Since, the second dissociation constant is very low thus; the value of y can be neglected from the denominator in the equation (2)

$5.0\times {10}^{-8}=\frac{y\left(x\right)}{\left(x\right)}=y$

Thus, the value of y is $5.0\times {10}^{-8}$.

From the ICE table, it is concentration of ${\text{SeO}}_3{}^{2-}\left(aq\right)$, thus,

$\left[{\text{SeO}}_3{}^{2-}\left(aq\right)\right]=5.0\times {10}^{-8}\text{ M}$

Now, in the equation (1),

$2.7\times {10}^{-3}=\frac{\left(x-y\right)\left(x+y\right)}{\left(2.89-x\right)}$

Calculate the ratio of acid dissociation constant and concentration of acid; if it is less than 0.05 then the value of x from the denominator can be neglected.

$\frac{2.7\times {10}^{-3}}{2.89}=9.34\times {10}^{-4}$

The calculated value is less than 0.05 thus,

$2.7\times {10}^{-3}=\frac{\left(x-y\right)\left(x+y\right)}{2.89}$

Or,

$2.7\times {10}^{-3}=\frac{x^2-y^2}{2.89}$

Putting the value of y in the above equation,

$2.7\times {10}^{-3}=\frac{x^2-{\left(5.0\times {10}^{-8}\right)}^2}{2.89}=\frac{x^2-\left(2.5\times {10}^{-15}\right)}{\left(2.89\right)}$

Or,

$\begin{array}{l}{x}^2=\left(2.7\times {10}^{-3}\right)\left(2.89\right)+\left(2.5\times {10}^{-15}\right)=7.803\times {10}^{-3}\\ x=\sqrt{3.48\times {10}^{-3}}=0.088\end{array}$

Thus, the concentration of ${\text{HSeO}}_3{}^{-}\left(aq\right)$ will be:

$\left[{\text{HSeO}}_3{}^{-}\left(aq\right)\right]=x-y$

Putting the values,

$\left[{\text{HSeO}}_3{}^{-}\left(aq\right)\right]=0.088-\left(5.0\times {10}^{-8}\right)=0.088\text{ M}$

The concentration of ${\text{H}}^{+}\left(aq\right)$ will be:

$\left[{\text{H}}^{+}\left(aq\right)\right]=x+y$

Putting the values,

$\left[{\text{H}}^{+}\left(aq\right)\right]=0.088+\left(5.0\times {10}^{-8}\right)=0.088\text{ M}$

The pH of the solution can be calculated as follows:

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

Putting the values,

$\text{pH}=-\log \left(0.088\right)=1.054$

Therefore, the pH of solution is 1.054, the concentration of ${\text{HSeO}}_3{}^{-}\left(aq\right)$ is $0.088\text{ M}$ and that of ${\text{SeO}}_3{}^{2-}\left(aq\right)$ is $5.0\times {10}^{-8}$.

Conclusion

Therefore, the pH of solution is 1.054, the concentration of ${\text{HSeO}}_3{}^{-}\left(aq\right)$ is $0.088\text{ M}$ and that of ${\text{SeO}}_3{}^{2-}\left(aq\right)$ is $5.0\times {10}^{-8}$.