Answer
The percent ionization of butanoic acid in that solution is equal to 0.018%.
Work Step by Step
1. Drawing the ICE table we get these concentrations at the equilibrium:
$HC_4H_7O_2(aq) + H_2O(l) \lt -- \gt C_4H_7O_2^-(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[HC_4H_7O_2] = 0.0075 M - x$
$[C_4H_7O_2^-] = 0.085M + x$
$[H_3O^+] = 0 + x$
2. Calculate 'x' using the $K_a$ expression.
$ 1.5\times 10^{- 5} = \frac{[C_4H_7O_2^-][H_3O^+]}{[HC_4H_7O_2]}$
$ 1.5\times 10^{- 5} = \frac{( 0.085 + x )* x}{ 0.0075 - x}$
Considering 'x' has a very small value.
$ 1.5\times 10^{- 5} = \frac{ 0.085 * x}{ 0.0075}$
$ 1.5\times 10^{- 5} \times \frac{0.0075}{0.085} = x$
$x = 1.32 \times 10^{-6}M$
Percent dissociation: $\frac{ 1.32\times 10^{- 6}}{ 0.0075} \times 100\% = 0.018\%$
