Answer
$[H_3O^+] = 0.048M$
$pH = 1.32$
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.
-$[H_3O^+] = [H_2P{O_4}^-] = x$
-$[H_3PO_4] = [H_3PO_4]_{initial} - x = 0.350 - x$
For approximation, we consider: $[H_3PO_4] = 0.350M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][H_2P{O_4}^-]}{ [H_3PO_4]}$
$Ka = 7.5 \times 10^{- 3}= \frac{x * x}{ 0.35}$
$Ka = 7.5 \times 10^{- 3}= \frac{x^2}{ 0.35}$
$ 2.625 \times 10^{- 3} = x^2$
$x = 5.123 \times 10^{- 2}$
Percent dissociation: $\frac{ 5.123 \times 10^{- 2}}{ 0.35} \times 100\% = 14.64\%$
%dissociation > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 7.5 \times 10^{- 3}= \frac{x^2}{ 0.35- x}$
$ 2.625 \times 10^{- 3} - 7.5 \times 10^{- 3}x = x^2$
$ 2.625 \times 10^{- 3} - 7.5 \times 10^{- 3}x - x^2 = 0$
Bhaskara:
$\Delta = (- 7.5 \times 10^{- 3})^2 - 4 * (-1) *( 2.625 \times 10^{- 3})$
$\Delta = 5.625 \times 10^{- 5} + 1.05 \times 10^{- 2} = 1.056 \times 10^{- 2}$
$x_1 = \frac{ - (- 7.5 \times 10^{- 3})+ \sqrt { 1.056 \times 10^{- 2}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 7.5 \times 10^{- 3})- \sqrt { 1.056 \times 10^{- 2}}}{2*(-1)}$
$x_1 = - 5.5 \times 10^{- 2} (Negative)$
$x_2 = 4.8 \times 10^{- 2}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.048)$
$pH = 1.32$
