Question

Calculate the pH of a 0.200 M solution of C₅H₅NHF.

Short answer

The pH of the given solution is 4.2.

Step-by-step solution

 Calculation of Ka for Hydrolysis

Species present are C₅H₅NH⁺, F^-, and H₂O.

Reaction involving:

$$\begin{gathered} {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{{\text{NH}}^{\text{+}}}_{\left(aq\right)}\rightleftarrows {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}_{\left(aq\right)}\text{+}{{\text{H}}^{\text{+}}}_{\left(aq\right)} \\ {\text{K}}_{\text{a}}{\text{= 5.88×10}}^{\text{- 6}} \end{gathered}$$

$$\begin{gathered} {{\text{F}}^{\text{-}}}_{\left(aq\right)}{\text{+H}}_{\text{2}}{\text{O}}_{{}_{}}\rightleftarrows {\text{HF}}_{{}_{\left(aq\right)}}\text{+}{{\text{OH}}^{\text{-}}}_{{}_{\left(aq\right)}} \\ {\text{K}}_b{\text{= 1.38×10}}^{\text{- 11}} \end{gathered}$$

$$\begin{gathered} {\text{H}}_{\text{2}}{\text{O +H}}_{\text{2}}\text{O}\rightleftarrows {\text{H}}_{\text{3}}{{\text{O}}^{\text{+}}}_{\left(aq\right)}\text{+}{{\text{OH}}^{\text{-}}}_{\left(aq\right)} \\ {\text{K}}_w{\text{= 1×10}}^{\text{- 14}} \end{gathered}$$

We can consider the overall reaction asfollows:

$$\begin{gathered} {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{{\text{NH}}^{\text{+}}}_{\left(aq\right)}+{{\text{F}}^{\text{-}}}_{\left(aq\right)}\rightleftarrows {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}_{\left(aq\right)}\text{+}{\text{HF}}_{\left(aq\right)} \\ k=\frac{\left[C_5{H}_5{N}_{\left(aq\right)}\right]{\displaystyle \left[H{F}_{\left(aq\right)}\right]}}{\left[C_5{H}_{5}N{H^{+}}_{\left(aq\right)}\right]\left[{F^{-}}_{\left(aq\right)}\right]} \\ k=\frac{{\displaystyle \left[H^{+}\right]\left[C_5{H}_5{N}_{\left(aq\right)}\right]}}{\left[C_5{H}_{5}N{H^{+}}_{\left(aq\right)}\right]{\displaystyle }}\times \frac{\left[H{F}_{\left(aq\right)}\right]}{\left[H^{+}\right]\left[{F^{-}}_{\left(aq\right)}\right]} \end{gathered}$$

$$\begin{gathered} \text{K}=K_a\left(C_5{H}_{5}N{H}^{+}\right)\times \frac{1}{K_a\left(H{F}_{\left(aq\right)}\right)} \\ \text{K}=\frac{{\text{5.88×10}}^{\text{- 6}}}{7.2\times {\text{10}}^{\text{- 4}}} \\ \text{K}=8.17\times {\text{10}}^{\text{- 3}} \end{gathered}$$

Calculation of pH  for the given salt

The initialconcentrationofC₅H₅NH⁺ = 0.200

The initialconcentrationofC₅H₅N= 0

The initialconcentrationofHF = 0

The initialconcentrationofF^- = 0.200

Attheequilibriumconcentration,C₅H₅NH⁺ = 0.200-x

Attheequilibriumconcentration,C₅H₅N= x

Attheequilibriumconcentration,HF = x

Attheequilibriumconcentration,F^- = 0.200-x

$$\begin{gathered} {\text{K}}_{\text{a}}=\frac{{\text{[x]}}^{\text{2}}}{{\text{[}0.2\text{- x]}}^{\text{2}}}=8.17\times {10}^{-3} \\ {\text{K}}_{\text{a}}=\frac{\text{x}}{0.2\text{- x}}=9.04\times {10}^{-2} \\ \text{x}=1.65\times {10}^{-2} \end{gathered}$$

$\text{[HF]}=x=1.65\times {10}^{-2}$${\text{[F}}^{\text{-}}\text{]}=0.2-x=0.1835$

$$\begin{gathered} {\text{K}}_a=\frac{{\text{[H}}^{\text{+}}{\text{][F}}^{\text{-}}\text{]}}{\text{[HF]}}=\frac{{\text{[H}}^{\text{+}}\text{](}0.1835)}{{1.65\times 10}^{-2}} \\ {\text{K}}_a=7.2\times {10}^{-4} \\ {\text{[H}}^{\text{+}}\text{]}=\frac{{7.2\times 10}^{-4}{\times 1.65\times 10}^{-2}}{0.1835} \\ {\text{[H}}^{\text{+}}\text{]}=6.48\times {10}^{-5} \end{gathered}$$

$$\begin{gathered} \text{pH}=-\log {\text{[H}}^{\text{+}}\text{]} \\ \text{pH}=-\log (6.48\times {10}^{-5}) \\ \text{pH}=4.2 \end{gathered}$$