Answer
$pH = 2.681$
Work Step by Step
1. Calculate the molar mass:
1.01* 1 + 12.01* 9 + 1.01* 7 + 16* 4 = 180.17g/mol
2. Calculate the number of moles
- Note, each tablet has 0.325g, and we are dissolving 2 tablets, so the total mass is 0.325 * 2 = 0.65g
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.65}{ 180.17}$
$n(moles) = 3.608\times 10^{- 3}$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 3.608\times 10^{- 3}}{ 0.237} $
$C(mol/L) = 0.01522$
4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [C_9H_7{O_4}^-] = x$
-$[HC_9H_7O_4] = [HC_9H_7O_4]_{initial} - x = 0.01522 - x$
For approximation, we consider: $[HC_9H_7O_4] = 0.01522M$
5. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_9H_7{O_4}^-]}{ [HC_9H_7O_4]}$
$Ka = 3.3 \times 10^{- 4}= \frac{x * x}{ 0.01522}$
$Ka = 3.3 \times 10^{- 4}= \frac{x^2}{ 0.01522}$
$ 5.023 \times 10^{- 6} = x^2$
$x = 2.241 \times 10^{- 3}$
Percent dissociation: $\frac{ 2.241 \times 10^{- 3}}{ 0.01522} \times 100\% = 14.72\%$
%dissociation < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 3.3 \times 10^{- 4}= \frac{x^2}{ 0.01522- x}$
$ 5.023 \times 10^{- 6} - 3.3 \times 10^{- 4}x = x^2$
$ 5.023 \times 10^{- 6} - 3.3 \times 10^{- 4}x - x^2 = 0$
$\Delta = (- 3.3 \times 10^{- 4})^2 - 4 * (-1) *( 5.023 \times 10^{- 6})$
$\Delta = 1.089 \times 10^{- 7} + 2.009 \times 10^{- 5} = 2.02 \times 10^{- 5}$
$x_1 = \frac{ - (- 3.3 \times 10^{- 4})+ \sqrt { 2.02 \times 10^{- 5}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 3.3 \times 10^{- 4})- \sqrt { 2.02 \times 10^{- 5}}}{2*(-1)}$
$x_1 = - 2.412 \times 10^{- 3} (Negative)$
$x_2 = 2.082 \times 10^{- 3}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 2.082 \times 10^{- 3})$
$pH = 2.681$
