Question

Question: Calculate the pH of a solution initially with 0.10 MNaHSO₄ and 0.10 MNH₃.

Short answer

Answer:The pH of the solution is 5.59.

Step-by-step solution

Determination of value of equilibrium

The chemical reaction is,

${\text{NH}}_{\text{3}}\left(\text{aq}\right){\text{+ HSO}}_{\text{4}}^{\text{-}}\left(\text{aq}\right)\to N{H}_4^{+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)$

The value of equilibrium is,

$$\begin{gathered} {\text{K}}_{\text{a}}=\frac{\left[H^{+}\right]\left[S{O}_4^{2-}\right]}{\left[HS{O}_4^{-}\right]} \\ {K}_b=\frac{\left[O{H}^{-}\right]\left[N{H}_4^{+}\right]}{\left[N{H}_3\right]} \\ \frac{K_a K_b}{K_w}=\frac{\frac{\left[H^{+}\right]\left[S{O}_4^{2-}\right]}{\left[HS{O}_4^{-}\right]}.\frac{\left[O{H}^{-}\right]\left[N{H}_4^{+}\right]}{\left[N{H}_3\right]}}{\left[H^{+}\right]\left[O{H}^{-}\right]} \\ =\frac{\left(1.2\times {10}^{-2}\right)\left(1.8\times {10}^{-5}\right)}{\left(1.0\times {10}^{-14}\right)} \\ =2.2\times {10}^7 \end{gathered}$$

The ICE table for the reaction is,

	NH3	HSO4-	NH4+	SO42-
Initial	0.10	0.10	0	0
Change	-x	-x	+x	+x
Equilibrium	0.10-x	0.10-x	x	x

Using the above value, calculate H⁺,

Determination of pH

The pH for the solution is,

$$\begin{gathered} {\text{K}}_{\text{a}}=\frac{\left[H^{+}\right]\left[S{O}_4^{2-}\right]}{\left[HS{O}_4^{-}\right]} \\ {K}_b=\frac{\left[O{H}^{-}\right]\left[N{H}_4^{+}\right]}{\left[N{H}_3\right]} \\ \frac{K_a K_b}{K_w}=\frac{\frac{\left[H^{+}\right]\left[S{O}_4^{2-}\right]}{\left[HS{O}_4^{-}\right]}.\frac{\left[O{H}^{-}\right]\left[N{H}_4^{+}\right]}{\left[N{H}_3\right]}}{\left[H^{+}\right]\left[O{H}^{-}\right]} \\ =\frac{\left(1.2\times {10}^{-2}\right)\left(1.8\times {10}^{-5}\right)}{\left(1.0\times {10}^{-14}\right)} \\ =2.2\times {10}^7 \end{gathered}$$

Thus, the pH of the solution is 5.59.