Answer
$pH = 3.197$
Work Step by Step
1. Calculate the molar mass:
12.01* 6 + 1.01* 8 + 16* 6 = 176.14g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.25}{ 176.14}$
$n(moles) = 1.419\times 10^{- 3}$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 1.419\times 10^{- 3}}{ 0.25} $
$C(mol/L) = 5.677\times 10^{- 3}$
4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [C_6H_7{O_6}^-] = x$
-$[C_6H_8O_6] = [C_6H_8O_6]_{initial} - x = 5.677 \times 10^{- 3} - x$
For approximation, we consider: $[C_6H_8O_6] = 5.677 \times 10^{- 3}M$
5. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_6H_7{O_6}^-]}{ [C_6H_8O_6]}$
$Ka = 8 \times 10^{- 5}= \frac{x * x}{ 5.677\times 10^{- 3}}$
$Ka = 8 \times 10^{- 5}= \frac{x^2}{ 5.677\times 10^{- 3}}$
$ 4.542 \times 10^{- 7} = x^2$
$x = 6.739 \times 10^{- 4}$
Percent dissociation: $\frac{ 6.739 \times 10^{- 4}}{ 5.677\times 10^{- 3}} \times 100\% = 11.87\%$
%dissociation < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 8 \times 10^{- 5}= \frac{x^2}{ 5.677 \times 10^{- 3}- x}$
$ 4.542 \times 10^{- 7} - 8 \times 10^{- 5}x = x^2$
$ 4.542 \times 10^{- 7} - 8 \times 10^{- 5}x - x^2 = 0$
$\Delta = (- 8 \times 10^{- 5})^2 - 4 * (-1) *( 4.542 \times 10^{- 7})$
$\Delta = 6.4 \times 10^{- 9} + 1.817 \times 10^{- 6} = 1.823 \times 10^{- 6}$
$x_1 = \frac{ - (- 8 \times 10^{- 5})+ \sqrt { 1.823 \times 10^{- 6}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 8 \times 10^{- 5})- \sqrt { 1.823 \times 10^{- 6}}}{2*(-1)}$
$x_1 = - 7.151 \times 10^{- 4} (Negative)$
$x_2 = 6.351 \times 10^{- 4}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 6.351 \times 10^{- 4})$
$pH = 3.197$
