Answer
(a) = 1.805%
(b) = 43.00%
Work Step by Step
Acid benzoic: Ka = $6.5 \times 10^{-5}$
$Ka = \frac{[H^+][A^-]}{[HA]}$
$[H^+] = [A^-] = x$
$[HA] = [HA]_{initial} - x$
(a)
1. Calculate $[H^+]$ and $[A^-]$
- Since the concentration is too large, compared to the Ka, we can consider, $[HA]_{initial} = [HA] = 0.20M$
Therefore:
$6.5 \times 10^{-5} = \frac{x * x}{0.20}$
$1.3 \times 10^{-5} = x^2$
$x = 3.61 \times 10^{-3}M$
$x = [H^+] = [A^-]$
2. Now, calculate the percent ionization.
$\% ionization = \frac{[H^+]}{[HA]} \times 100 \%$
$\% ionization = \frac{3.61 \times 10^{-3}}{0.20M} \times 100\% = 1.805 \%$
(b)
1. Calculate $[H^+]$ and $[A^-]$
$Ka = 6.5 \times 10^{- 5}= \frac{x * x}{ 2 \times 10^{- 4}- x}$
$Ka = 6.5 \times 10^{- 5}= \frac{x^2}{ 2 \times 10^{- 4}- x}$
$ 1.3 \times 10^{- 8} - 6.5 \times 10^{- 5}x = x^2$
$ 1.3 \times 10^{- 8} - 6.5 \times 10^{- 5}x - x^2 = 0$
$\Delta = (- 6.5 \times 10^{- 5})^2 - 4 * (-1) *( 1.3 \times 10^{- 8})$
$\Delta = 4.22 \times 10^{- 9} + 5.2 \times 10^{- 8} = 5.62 \times 10^{- 8}$
$x_1 = \frac{ - (- 6.5 \times 10^{- 5})+ \sqrt { 5.62 \times 10^{- 8}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 6.5 \times 10^{- 5})- \sqrt { 5.62 \times 10^{- 8}}}{2*(-1)}$
$x_1 = - 1.51 \times 10^{- 4} (Negative)$
$x_2 = 8.6 \times 10^{- 5}$
The concentration can't be negative, therefore: $x = [H^+] = 8.6 \times 10^{- 5}$
$\%ionization = \frac{8.6 \times 10^{- 5}}{0.0002} \times 100\% = 43 \%$
