The pH of a \(0.20 - M\) solution of \(HF\) is \(1.92.\) Determine \({K_a}\) for \(HF\) from these data.

We can compute the concentration of \({{\bf{H}}^{\bf{ + }}}\) ions since we know the pH of the solution:

\(\begin{aligned}\;\;\;{\bf{pH = - log}}\left( {{{\bf{H}}^{\bf{ + }}}} \right)\\\left( {{{\bf{H}}^{\bf{ + }}}} \right){\bf{ = 1}}{{\bf{0}}^{{\bf{ - pH}}}}\;{\bf{M}}\\\;\;\;\;\;\;\;\;{\bf{ = 1}}{{\bf{0}}^{{\bf{ - 1}}{\bf{.92}}}}\;{\bf{M}}\\\;\;\;\;\;\;\;\;{\bf{ = 0}}{\bf{.01}}\;{\bf{M}}.\end{aligned}\)

The dissociation process of hydrofluoric acid is \(HF(aq) \to {H^ + }(aq) + {F^ - }(aq).\)

\(HF(aq) \to {H^ + }(aq) + {F^ - }(aq)\)

So the \({K_a}\) value equals:

\({K_a} = \frac{{\left( {{H^ + }} \right){\bf{ \times }}\left( {{F^ - }} \right)}}{{\left( {HF} \right)}}.\)

The concentrations of \({H^ + }\) and \({F^ - }\) are assumed to be \(0.01 M,\) the same. We only have the \(HF\) in a concentration of \(0.20 M\) at the start of the reaction. The concentration of \(HF\) lowers by \(x,\) while the concentrations of \({H^ + }\) and \({F^ - }\) rise by \(x\) until equilibrium is established.

We know the concentrations of \({H^ + }\) and \({F^ - }\) because we know the concentrations of \({H^ + }\)and \({F^ - }.\) As a result, the equilibrium concentration of \(HF\) is as follows:

\(\begin{aligned}\left( {HF} \right) &= 0.20M - x\\(HF) &= (0.20 - 0.01) M\\\left( {HF} \right) &= 0.19 M.\end{aligned}\)

We can compute the \({K_a}\) value now that we have all of the equilibrium concentrations:

\(\begin{aligned}{K_a} &= \frac{{\left( {{H^ + }} \right) \times \left( {{F^ - }} \right)}}{{\left( {HF} \right)}}\\\;\;\;\;\; &= \frac{{0.01 M \times 0.01 M}}{{0.19 M}}\\\;\;\;\;\; &= 5.26 \times 1{0^{ - 4}}.\end{aligned}\)

A hydrofluoric acid has a \({K_a}\) value of \(5.26 \times 1{0^{ - 4}}.\)