91Ó°ÊÓ

The propionic acid dissociation looks like this:

\(\begin{aligned}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{H(aq)}} \to {{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{CO}}_{\bf{2}}^{\bf{ - }}{\bf{(aq) + }}{{\bf{H}}^{\bf{ + }}}{\bf{(aq)}}\\{{\bf{K}}_{\bf{a}}}{\bf{ = }}\frac{{\left( {{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{CO}}_{\bf{2}}^{\bf{ - }}} \right){\bf{ \times }}\left( {{{\bf{H}}^{\bf{ + }}}} \right)}}{{{\bf{(}}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{CO}}{{\bf{O}}_{\bf{2}}}{\bf{H)}}}}\end{aligned}\)

We may assume that the concentrations of \({C_2}{H_5}CO_2^ - \) and \({H^ + }\)are the same, and we can designate them as \(x.\) We can also assume that the change in the concentration of \({C_2}{H_5}CO{O_2}H\) is small.

\(\begin{aligned}1.34 \times 1{0^{ - 5}} &= \frac{{{x^2}}}{{0.698}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;x &= 3.06 \times 1{0^{ - 3}}\\\;\;\;\;\;\;\;\;\;\left( {{H^ + }} \right) &= 3.06 \times 1{0^{ - 3}} M.\end{aligned}\)

The concentration of the hydronium ions in this solution is \(3.06 \times 1{0^{ - 3}} M.\)

Thus, the final solution is \(\left( {{H^ + }} \right) = 3.06 \times 1{0^{ - 3}} M.\)