Question

Question: Calculate the pH of a solution prepared by mixing equal volumes of 1.0x10^-4 MNH₃ and 1.0x10^-4 MHCl.

Short answer

Answer:The pH of the solution is 6.72.

Step-by-step solution

Determination of equilibrium

When NH₃ reacts with HCl, it produces NH₄⁺.

The equilibrium chemical reaction is,

$\mathit{N}{\mathit{H}}_{\mathbf{4}}^{\mathbf{+}}\mathbf{\to }\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{+}{\mathit{H}}^{\mathbf{+}}$

The ICE table for the above reaction is,

	NH4+	H+	NH3
Initial	5.0x10-5	0	0
Change	-x	+x	+x
Equilibrium	5.0x10-5 - x	x	x

Using the values in K_a expression,

$$\begin{gathered} \frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}\mathbf{=}\frac{\left[H^{+}\right]\left[N{H}_3\right]}{\left[N{H}_4^{+}\right]} \\ \frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{-}\mathbf{x}} \\ \mathbf{5}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{-}\mathbf{x}} \\ x=1.7\times {10}^{-7} \end{gathered}$$

Calculation of [H+]

The value of [H⁺] is,

$$\begin{gathered} \left[H^{+}\right]\mathbf{=}\sqrt{\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}{\left(HA\right)}_{\mathbf{0}}\mathbf{+}{\mathbf{K}}_{\mathbf{w}}} \\ \left[H^{+}\right]\mathbf{=}\sqrt{\left(5.6\times {10}^{-10}\right)\left(5.0\times {10}^{-5}\right)\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}} \\ \left[H^{+}\right]\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathit{M} \end{gathered}$$

Calculation of pH

The pH of the solution is,

$$\begin{gathered} \mathit{p}\mathit{H}\mathbf{=}\mathbf{-}\mathit{l}\mathit{o}\mathit{g}\left[H^{+}\right] \\ \mathbf{=}\mathbf{-}\mathit{l}\mathit{o}\mathit{g}\left(1.9\times {10}^{-7}\right) \\ \mathbf{=}\mathbf{6}\mathbf{.}\mathbf{72} \end{gathered}$$

Thus, the pH of the solution is 6.72.