Answer
4. %ionization: $\frac{1.796\times10^{-3}}{0.25} \times 100$% = 0.7184%
Work Step by Step
1. Write the Ka formula:
$K_a = \frac{[C_2H_5COO^-][H^+]}{[C_2H_5COOH] - [H^+]}$
2. Since $[H^+] = [C_2H_5COO^-]:$
$K_a = \frac{x^2}{[C_2H_5COOH]- x}$
$1.3 \times 10^{-5} = \frac{x^2}{0.25-x}$
$3.25 \times 10^{-6} - 1.3\times10^{-5}x = x^2$
$3.25 \times 10^{-6} - 1.3\times10^{-5}x - x^2 = 0$
3. Find the value of x:
$x_1 = \frac{-b - \sqrt {b^2-4ac}}{2a}$
$x_1 = \frac{1.3 \times 10^{-5} - \sqrt[] {(1.3\times10^{-5})^2 - 4 \times (-1) \times 3.25 \times 10^{-6}}}{-2}$
$x_1 = \frac{1.3\times10^{-5}-\sqrt {1.3 \times 10^{-5}}}{-2}$
$x_1 \approx 1.796 \times 10^{-3}$
* If we used $x_1 = \frac{-b + \sqrt {b^2-4ac}}{2a}$ the result would be negative.
4. %ionization: $\frac{1.796\times10^{-3}}{0.25} \times 100$ %= 0.7184%
