Answer
$pH = 3.267$
Work Step by Step
1. Calculate the molar mass:
12.01* 3 + 1.01* 6 + 16* 3 = 90.09g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.056}{ 90.09}$
$n(moles) = 6.216\times 10^{- 4}$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 6.216\times 10^{- 4}}{ 0.25} $
$C(mol/L) = 2.486\times 10^{- 3}$
4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [C_3H_5{O_3}^-] = x$
-$[C_3H_6O_3] = [C_3H_6O_3]_{initial} - x = 2.486 \times 10^{- 3} - x$
For approximation, we consider: $[C_3H_6O_3] = 2.486 \times 10^{- 3}M$
5. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_3H_5{O_3}^-]}{ [C_3H_6O_3]}$
$Ka = 1.5 \times 10^{- 4}= \frac{x * x}{ 2.486\times 10^{- 3}}$
$Ka = 1.5 \times 10^{- 4}= \frac{x^2}{ 2.486\times 10^{- 3}}$
$ 3.73 \times 10^{- 7} = x^2$
$x = 6.107 \times 10^{- 4}$
Percent dissociation: $\frac{ 6.107 \times 10^{- 4}}{ 2.486\times 10^{- 3}} \times 100\% = 24.56\%$
%dissociation > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 1.5 \times 10^{- 4}= \frac{x^2}{ 2.486 \times 10^{- 3}- x}$
$ 3.73 \times 10^{- 7} - 1.5 \times 10^{- 4}x = x^2$
$ 3.73 \times 10^{- 7} - 1.5 \times 10^{- 4}x - x^2 = 0$
$\Delta = (- 1.5 \times 10^{- 4})^2 - 4 * (-1) *( 3.73 \times 10^{- 7})$
$\Delta = 2.25 \times 10^{- 8} + 1.492 \times 10^{- 6} = 1.514 \times 10^{- 6}$
$x_1 = \frac{ - (- 1.5 \times 10^{- 4})+ \sqrt { 1.514 \times 10^{- 6}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 1.5 \times 10^{- 4})- \sqrt { 1.514 \times 10^{- 6}}}{2*(-1)}$
$x_1 = - 6.903 \times 10^{- 4} (Negative)$
$x_2 = 5.403 \times 10^{- 4}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 5.403 \times 10^{- 4})$
$pH = 3.267$
