Question

Calculate the pH of a 0.200 M solution of C₅H₅NHF.

Short answer

pH of the given solution comes out to be 4.2.

Step-by-step solution

 Calculation of Ka for Hydrolysis

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

Reaction involving:

$$\begin{gathered} C_5{H}_{5}N{H^{+}}_{\left(aq\right)}\rightleftarrows C_5{H}_5{N}_{\left(aq\right)}+{H^{+}}_{\left(aq\right)} \\ {K}_a=5.88\times {10}^{-6} \\ {F^{-}}_{\left(aq\right)}+H_{2}O\rightleftarrows H{F}_{\left(aq\right)}+O{H^{-}}_{\left(aq\right)} \\ {K}_b=1.38\times {10}^{-11} \\ {H}_{2}O+H_{2}O\rightleftarrows H_3{O^{+}}_{\left(aq\right)}+O{H^{-}}_{\left(aq\right)} \\ {K}_w=1\times {10}^{-14} \end{gathered}$$

We can here consider the overall reaction as :

$$\begin{gathered} C_5{H}_{5}N{H^{+}}_{\left(aq\right)}+{F^{-}}_{\left(aq\right)}\rightleftarrows C_5{H}_5{N}_{\left(aq\right)}+{HF}_{\left(aq\right)} \\ K=\frac{\left[C_5{H}_5{N}_{\left(aq\right)}\right]\left[H{F}_{\left(aq\right)}\right]}{\left[C_5{H}_{5}N{H^{+}}_{\left(aq\right)}\right]\left[{F^{-}}_{\left(aq\right)}\right]} \\ K=\frac{\left[H^{+}\right]\left[C_5{H}_5{N}_{\left(aq\right)}\right]}{\left[C_5{H}_{5}N{H^{+}}_{\left(aq\right)}\right]}\times \frac{\left[H{F}_{\left(aq\right)}\right]}{\left[H^{+}\right]\left[{F^{-}}_{\left(aq\right)}\right]} \\ K=K_a\left(C_5{H}_{5}N{H^{+}}_{\left(aq\right)}\right)\times \frac{1}{K_a\left(H{F}_{\left(aq\right)}\right)} \\ K=\frac{5.88\times {10}^{-6}}{7.2\times {10}^{-4}} \\ K=8.17\times {10}^{-3} \end{gathered}$$

Calculation of pH  for Given Salt

Initial Concentration C₅H₅NH⁺ = 0.200

Initial Concentration C₅H₅N= 0

Initial Concentration HF = 0

Initial Concentration F^- = 0.200

At equilibrium Concentration C₅H₅NH⁺ = 0.200-x

At equilibrium Concentration C₅H₅N= x

At equilibrium Concentration HF = x

At equilibrium Concentration F^- = 0.200-x

$$\begin{gathered} K_a=\frac{{\left[x\right]}^2}{{\left[0.2-x\right]}^2=8.17\times {10}^{-3}} \\ {K}_a=\frac{x}{0.2-x}=9.04\times {10}^{-2} \\ x=1.65\times {10}^{-2} \\ \left[F^{-}\right]=0.2-x=0.1835 \\ \left[HF\right]=x=1.65\times {10}^{-2} \\ {K}_a=\frac{\left[H^{+}\right]\left[F^{-}\right]}{\left[HF\right]}=\frac{\left[H^{+}\right]\left(0.1835\right)}{1.65\times {10}^{-2}} \\ {K}_a=7.2\times {10}^{-4} \\ \left[H^{+}\right]=\frac{7.2\times {10}^{-4}\times 1.65\times {10}^{-2}}{0.1835} \\ \left[H^{+}\right]=6.48\times {10}^{-5} \\ pH=-\log \left[H^{+}\right] \\ pH=-\log \left(6.48\times {10}^{-5}\right) \\ pH=4.2 \end{gathered}$$