Chapter 18, Problem 84A

Interpretation Introduction

Interpretation:

The $k_a$ value for second ionization 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:

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

$pH+pOH=14$

Answer to Problem 84A

$k_a=3.2\times {10}^{-7}$

Explanation of Solution

Ionization equation is given by:

$HCr{O}_4{}^{-}+H_{2}O\rightleftharpoons Cr{O}_4{}^{2-}+H_3{O}^{+}$

Given data:

$Beforereaction,\left[HCr{O}_4{}^{-}\right]=0.040M$

$pH=3.946$

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

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

Therefore,

$\begin{array}{c}pH=3.946\\ -\log \left[H^{+}\right]=3.946\\ \left[H^{+}\right]=Anti\log (-3.946)\\ =1.13\times {10}^{-4}M\end{array}$

$\begin{array}{c}Atequilibrium,\left[HCr{O}_4{}^{-}\right]=0.040-1.13\times {10}^{-4}\\ =0.0399M\end{array}$

Therefore,

$\begin{array}{c}{k}_a=\frac{1.13\times {10}^{-4}\times 1.13\times {10}^{-4}}{0.0399}\\ =3.2\times {10}^{-7}\end{array}$

Thus using the table, acid is $HF$

Conclusion

The value of k_a of $HCr{O}_4{}^{-}$ is $k_a=3.2\times {10}^{-7}$