Question

Calculate K, for a dibasic acid if its concentration is ${\text{0}}{\text{.05 N}}$ and hydrogen ion concentration is ${\text{1}} \times {\text{1}}{{\text{0}}^{ - {\text{3}}}}{\text{mol/L}}$ .

Answer 1

From the given value of the normality, calculate the initial concentration of the dibasic acid. Then write an expression for the equilibrium constant ${{\text{K}}_a}$ and substitute concentration values in this expression and calculate the equilibrium constant in terms of the degree of dissociation. You can calculate the degree of dissociation from the hydrogen ion concentration.

Complete step-by-step solution: Let us represent the dibasic acid as ${{\text{H}}_2}{\text{A}}$
You can represent the initial concentration of dibasic acid as C and the degree of dissociation as $\alpha$ .
Write an equation for the dissociation of dibasic acid.
${{\text{H}}_2}{\text{A }} \rightleftharpoons {\text{ 2 }}{{\text{H}}^ + }{\text{ + }}{{\text{A}}^{2 - }}$
Prepare a table showing initial concentration and equilibrium concentration for various species.

| ${{\text{H}}_2}{\text{A}}$| ${{\text{H}}^ + }$ | ${{\text{A}}^{2 - }}$
---|---|---|---
Initial concentration(M)| C| 0| 0
Equilibrium concentration (M)| $C - C\alpha$| $2C\alpha$| $C\alpha$

But you can calculate the initial concentration from normality
The initial concentration $C = \dfrac{{0.05}}{2} = 0.025{\text{M}}$
Substitute the value of C in the table and calculate various equilibrium concentrations.

| ${{\text{H}}_2}{\text{A}}$| ${{\text{H}}^ + }$ | ${{\text{A}}^{2 - }}$
---|---|---|---
Initial concentration (M) | 0.025| 0| 0
Equilibrium concentration (M)| $0.025 - 0.025\alpha$| $0.050\alpha$| $0.025\alpha$

Write the expression for the acid dissociation constant ${{\text{K}}_a}$
${{\text{K}}_a} = \dfrac{{{{\left[ {{{\text{H}}^ + }} \right]}^2} \times \left[ {{{\text{A}}^{2 - }}} \right]}}{{\left[ {{{\text{H}}_2}{\text{A}}} \right]}}$
Substitute values in the above expression

$${{\text{K}}_a} = \dfrac{{{{\left[ {{{\text{H}}^ + }} \right]}^2} \times \left[ {{{\text{A}}^{2 - }}} \right]}}{{\left[ {{{\text{H}}_2}{\text{A}}} \right]}} \\\ {\Rightarrow {\text{K}}_a} = \dfrac{{{{\left( {0.050\alpha } \right)}^2} \times \left( {0.025\alpha } \right)}}{{0.025 - 0.025\alpha }} \\\$$

Assume that the degree of dissociation has small value so that in the denominator, you can approximate $0.025 - 0.025\alpha$ to $0.025$

$${\Rightarrow {\text{K}}_a} = \dfrac{{{{\left( {0.050\alpha } \right)}^2} \times \left( {0.025\alpha } \right)}}{{0.025}} \\\ {\Rightarrow {\text{K}}_a} = \dfrac{{{{\left( {0.050\alpha } \right)}^2} \times \left( {0.025\alpha } \right)}}{{0.025}} \\\ {\Rightarrow {\text{K}}_a} = 0.0025{\alpha ^3} \\\$$

From the hydrogen ion concentration, obtain the degree of dissociation as shown below:
But hydrogen ion concentration is ${\text{1}} \times {\text{1}}{{\text{0}}^{ - {\text{3}}}}{\text{mol/L}}$ .

$$\Rightarrow \left[ {{{\text{H}}^ + }} \right]{\text{ = 0}}{\text{.050}}\alpha \\\ \Rightarrow \left[ {{{\text{H}}^ + }} \right]{\text{ = 1}} \times {\text{1}}{{\text{0}}^{ - {\text{3}}}}{\text{mol/L}} \\\ \Rightarrow \alpha = \dfrac{{{\text{1}} \times {\text{1}}{{\text{0}}^{ - {\text{3}}}}{\text{mol/L}}}}{{0.050}} \\\ \Rightarrow \alpha = 0.02 \\\$$

Substitute the value of the degree of dissociation in the acid constant expression and calculate the value of the acid dissociation constant.

$${{\Rightarrow \text{K}}_a} = 0.0025{\alpha ^3} \\\ {{\Rightarrow \text{K}}_a} = 0.0025 \times {\left( {0.02} \right)^3} \\\ {{\Rightarrow \text{K}}_a} = 0.0025 \times 8 \times {10^{ - 6}} \\\ {{\therefore \text{K}}_a} = 2 \times {10^{ - 8}} \\\$$

Hence, the value of the equilibrium constant ${{\text{K}}_a}$ for the dibasic acid is $2 \times {10^{ - 8}}$ .

Note: For a dibasic acid, the normality is twice its molar concentration. In other words, the molar concentration is one half its normality. In general, for an acid, the molarity is equal to the product of normality and basicity. A dibasic acid gives two protons per molecule upon dissociation.