Answer
$[H_3O^+] = 0.12M$
$pH = 0.93$
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.
-$[H_3O^+] = [HC_2{O_4}^-] = x$
-$[H_2C_2O_4] = [H_2C_2O_4]_{initial} - x = 0.350 - x$
For approximation, we consider: $[H_2C_2O_4] = 0.35M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][HC_2{O_4}^-]}{ [H_2C_2O_4]}$
$Ka = 6.0 \times 10^{- 2}= \frac{x * x}{ 0.35}$
$Ka = 6.0 \times 10^{- 2}= \frac{x^2}{ 0.35}$
$ 2.1 \times 10^{- 2} = x^2$
$x = 1.449 \times 10^{- 1}$
Percent dissociation: $\frac{ 1.449 \times 10^{- 1}}{ 0.35} \times 100\% = 41.4\%$
%dissociation > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 0.06= \frac{x^2}{ 0.35- x}$
$ 0.021 - 0.06x = x^2$
$ 0.021 - 6 \times 10^{- 2}x - x^2 = 0$
Bhaskara:
$\Delta = (- 6 \times 10^{- 2})^2 - 4 * (-1) *( 2.1 \times 10^{- 2})$
$\Delta = 3.6 \times 10^{- 3} + 8.4 \times 10^{- 2} = 8.76 \times 10^{- 2}$
$x_1 = \frac{ - (- 6 \times 10^{- 2})+ \sqrt { 8.76 \times 10^{- 2}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 6 \times 10^{- 2})- \sqrt { 8.76 \times 10^{- 2}}}{2*(-1)}$
$x_1 = - 1.8 \times 10^{- 1} (Negative)$
$x_2 = 1.2 \times 10^{- 1}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.12)$
$pH = 0.93$
- We don't need to consider the second constant dissociation, because it has a very small value compared to the hydronium concentration.
