Chapter 18, Problem 101A

Interpretation Introduction

Interpretation:

The $k_a$ value for monoprotic acid $HA$ 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:

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

$pH+pOH=14$

Answer to Problem 101A

$k_a=3.1\times {10}^{-6}$

Explanation of Solution

Ionization equation is given by:

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

Given data:

$Beforereaction,\left[HA\right]=0.200M$

$pH=3.10$

Now $\left[A^{-}\right]=\left[H^{+}\right]$

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

Therefore,

$\begin{array}{c}pH=3.10\\ -\log \left[H^{+}\right]=3.10\\ \left[H^{+}\right]=Anti\log (-3.10)\\ =7.9\times {10}^{-4}M\end{array}$

$\begin{array}{c}Atequilibrium,\left[HA\right]=0.200-7.9\times {10}^{-4}\\ =0.199M\end{array}$

Therefore,

$\begin{array}{c}{k}_a=\frac{7.9\times {10}^{-4}\times 7.9\times {10}^{-4}}{0.199}\\ =3.1\times {10}^{-6}\end{array}$

Conclusion

The value of $k_a=3.1\times {10}^{-6}$