Question

142. Consider a weak acid HX. If a 0.10 M solution of HX has a pH of 5.83 at 25°C, what is $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$for the acid's dissociation reaction at 25°C?

Short answer

The value of$\Delta G$for an acid's dissociation reaction at 25°C is$61kJ$

Step-by-step solution

Step 1: Introduction to the Concept

The acid dissociation constant expression at equilibriumis as follows:

$K_a=\frac{ProductConcentration}{ReactantsConcentration}$

Step 2: Determination of Dissociation constant

To get the value of, we must firstdetermine the dissociation constant of a weak acid:

We must draw an ICE table to determine the dissociation constant.Initial Change and Equilibrium is the full version of the ICE table. The ICE table is a collection of product and reactant concentrations at various phases of a reaction that is used to calculate the equilibrium constant:

$HX\left(aq\right)\stackrel{}{\rightleftarrows }{\text{H}}^{+}\left(aq\right)+X^{-}$

Initial	$0.10\text{M}$	0	0
Change	$-x\text{M}$	$+x\text{M}$	$+x\text{M}$
Equilibrium	$(0.10-x)\text{M}$	$x\text{M}$	$x\text{M}$

Here, the dissociation constant is:

$K_a=\frac{\left[H^{+}\right]\times \left[X^{-}\right]}{\left[HX\right]}$

$K_a=\frac{\left[x\right]\times \left[x\right]}{[0.10-x]}-----\left(1\right)$

Step 3: Determination of pH value

From the given,

$pH=5.83$

Be familiar with,

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

$\left[H^{+}\right]={10}^{-pH}$

$\left[H^{+}\right]=1.48\times {10}^{-6}$

Let’ssubstitute the values in equation (1),

$K_a=\frac{[1.48\times {10}^{-6}]\times [1.48\times {10}^{-6}]}{[0.10-1.48\times {10}^{-6}]}$

$K_a=\frac{2.19\times {10}^{-12}}{0.0999}$

$=2.19\times {10}^{-11}$

Step 4: Determination of ΔGvalue

We know that,

$\Delta G=-RTln{K}_a$

Where,

$R=8.314\text{J}\times {\text{mol}}^{-1}\times {\text{K}}^{-1}$

$T={25}^{\circ }C$

$=(25+273)\text{K}$

$=298\text{K}$

$K_a=2.19\times {10}^{-11}$

Let’s substitute all the values,

$\begin{array}{l}=-(8.314\text{J}\times {\text{mol}}^{-1}\times {\text{K}}^{-1}\times \frac{1\text{kJ}}{1000\text{J}})\times \left(298\text{K}\right)\times \ln \times (2.19\times {10}^{-11})\\ =60.8\text{kJ}\\ \approx \text{61kJ}\end{array}$

Therefore, the value of$\Delta G$ is$61kJ$.