Chapter 18, Problem 6STP

Interpretation Introduction

Interpretation:

Acid dissociation constant of propanoic 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:

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

Answer to Problem 6STP

A

Explanation of Solution

The equation is given by:

$C_3{H}_{7}COOH\rightleftharpoons C_3{H}_{7}CO{O}^{-}+H^{+}$

Therefore,

$k_a=\frac{\left[C_3{H}_{7}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C_3{H}_{7}COOH\right]}$

Given data:

$pH=2.43$

$\left[C_3{H}_{7}COOH\right]=1.0M$

Now,

$\begin{array}{c}pH=2.43\\ -\log \left[H^{+}\right]=2.43\\ \left[H^{+}\right]=Anti\log (-2.43)\\ =0.0037\end{array}$

$\left[C_3{H}_{7}CO{O}^{-}\right]=\left[H^{+}\right]=0.0037$

Therefore,

$\begin{array}{c}{k}_a=\frac{0.0037\times 0.0037}{1}\\ =1.4\times {10}^{-5}\end{array}$

Conclusion

Mathematical calculations show that option A is correct.