Answer
$pH \approx 2.00$
Work Step by Step
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x = 0.1M - x$
For approximation, we consider: $[HA] = [HA]_{initial} = 0.1M$
** HA = Acid, A- = Conjugate base.
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = 7.1 \times 10^{- 4}= \frac{x * x}{ 0.15}$
$Ka = 7.1 \times 10^{- 4}= \frac{x^2}{ 0.15}$
$ 1.06 \times 10^{- 4} = x^2$
$x = 1.03 \times 10^{- 2}$
5% test: $\frac{ 1.03 \times 10^{- 2}}{ 1.5 \times 10^{- 1}} \times 100\% = 6.87\%$
%ionization < 5% : Inappropriate approximation, so we have to consider: $[HA] = [HA]_{initial} - x$
$Ka = 7.1 \times 10^{- 4}= \frac{x^2}{ 0.15- x}$
$ 1.06 \times 10^{- 4} - 7.1 \times 10^{- 4}x = x^2$
$ 1.06 \times 10^{- 4} - 7.1 \times 10^{- 4}x - x^2 = 0$
Bhaskara:
$\Delta = (- 7.1 \times 10^{- 4})^2 - 4 * (-1) *( 1.06 \times 10^{- 4})$
$\Delta = 5.04 \times 10^{- 7} + 4.26 \times 10^{- 4} = 4.26 \times 10^{- 4}$
$x_1 = \frac{ - (- 7.1 \times 10^{- 4})+ \sqrt { 4.26 \times 10^{- 4}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 7.1 \times 10^{- 4})- \sqrt { 4.26 \times 10^{- 4}}}{2*(-1)}$
$x_1 = - 1.06 \times 10^{- 2} (Negative)$
$x_2 = 9.97 \times 10^{- 3}$
$[H_3O^+] = x = 9.97 \times 10^{- 3}$
3. Find the pH:
$pH = -log[H_3O^+]$
$pH = -log( 9.97 \times 10^{- 3})$
$pH = 2.00$
