Chapter 18.3, Problem 34PP

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:

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

And $pH+pOH=14$

Answer to Problem 34PP

$k_a=6.3\times {10}^{-4}$

The acid is HF

Explanation of Solution

Ionization equation is given by:

$HX+H_{2}O\rightleftharpoons X^{-}+H_3{O}^{+}$

Given data:

$Beforereaction,\left[HX\right]=0.0091M$

$pOH=11.32$

$\begin{array}{c}pH=14-pOH\\ =14-11.32\\ =2.68\end{array}$

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

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

Therefore,

$\begin{array}{c}pH=2.68\\ -\log \left[H^{+}\right]=2.68\\ \left[H^{+}\right]=Anti\log (-2.68)\\ =2.1\times {10}^{-3}M\end{array}$

$\begin{array}{c}Atequilibrium,\left[HX\right]=0.0091-2.1\times {10}^{-3}\\ =0.0070M\end{array}$

Therefore,

$\begin{array}{c}{k}_a=\frac{2.1\times {10}^{-3}\times 2.1\times {10}^{-3}}{0.0070}\\ =6.3\times {10}^{-4}\end{array}$

Thus using the table, acid is $HF$

Conclusion

The value of k_a of HF is $k_a=6.3\times {10}^{-4}$