Question

What is the pH of a 0.125 M solution of the weak base B if $\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{28}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{J}$ and $\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{175}\mathbf{}\mathit{J}\mathbf{/}\mathit{K}$for the following equilibrium reaction at 25°C?
$\mathit{B}\mathit{l}\left(aq\right)\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(l\right)\mathbf{\iff }\mathit{B}{\mathit{H}}^{\mathbf{+}}\left(aq\right)\mathbf{+}\mathit{O}{\mathit{H}}^{\mathbf{-}}\left(aq\right)$

Short answer

ThepH of a 0.125 M solution of the weak base B is $\mathbf{11}\mathbf{.}\mathbf{43}$.

Step-by-step solution

Step 1: Introduction to the Concept

• For a particular reaction, the Gibbs free energy change( $\mathit{\Delta }\mathit{G}$) is determined by the enthalpy change( $\mathit{\Delta }\mathit{H}$), entropy change ($\mathbf{\Delta S}$ ), and temperature(T).

$\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{-}\mathit{T}\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(1\right)$

The sign of$\mathit{\Delta }\mathit{G}$is determined by the sum of the two terms in the above equation.

Therefore for a reaction to be spontaneous, $\mathit{\Delta }\mathit{G}\mathbf{<}\mathbf{0}$

• The Gibbs-Helmholtz isotherm links the Gibbs free energy of a reaction at equilibrium to the equilibrium constant (K):

$\mathit{\Delta }{\mathit{G}}_{\mathbf{R}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathit{R}\mathit{T}\mathit{l}\mathit{n}\mathit{K}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\left(2\right)$

Where

$R=8.314J/molK$ and

$T=$temperature

Step 2: Determination of ΔGo and K

Here,

By equation (1),

$\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}\mathit{\Delta }{\mathit{H}}^{\mathbf{o}}\mathbf{-}\mathit{T}\mathit{\Delta }{\mathit{S}}^{\mathbf{o}}$

$=-28.0\text{kJ -}\left(298\text{K}\right)\times \left(-0.175kJ/J\right)$

$=24.15\text{kJ}$

By equation (2),

$\mathit{\Delta }{\mathit{G}}_{\mathbf{R}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathit{R}\mathit{T}\mathit{l}\mathit{n}\mathit{K}$

$K=e^{-\frac{\Delta G_R°}{RT}}$

$=e^{-\frac{24.15KJ/mol}{0.008314kJ/molK\times 298\text{K}}}$

$=5.84\times {10}^{-5}$

Step 3: Determination of pH

From the given reaction,

$B\left(aq\right)+H_{2}O\left(l\right)\iff B{H}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$

	B	${\text{BH}}^{\text{+}}$	${\text{OH}}^{\text{-}}$
Initial	$0.125M$	0	0
Change	$-x$	localid="1658768575254" $+x$	$+x$
Equilibrium	$\left(0.125-x\right)$	$x$	$x$

$K_{eq}=\frac{\left[B{H}^{+}\right]\times \left[O{H}^{-}\right]}{\left[B\right]}$

$=\frac{x^2}{0.125-x}$

$5.84\times {10}^{-5}=\frac{x^2}{0.125-x}$

$x=\left[O{H}^{-}\right]=0.0027\text{M}$

$pOH=-\log \left(0.0027\right)$

$=2.57$

$pH=14-pOH$

$=14-2.57$

$=11.43$

Therefore the pH is $\mathbf{11}\mathbf{.}\mathbf{43}$.