Answer
$$pH = 2.66 $$
Work Step by Step
$ HC_9H_7O_4 $ : ( 1.008 $\times$ 8 )+ ( 12.01 $\times$ 9 )+ ( 16.00 $\times$ 4 )= 180.15 g/mol
2 tablets, each has 500 mg: 2 x 500 mg = 1000 mg = 1.00 g
- Calculate the amount of moles:
$$ 1.00 \space g \times \frac{1 \space mol}{ 180.15 \space g} = 5.55 \times 10^{-3} \space mol$$
- Calculate the molarity:
$$ \frac{ 5.55 \times 10^{-3} \space mol}{ 0.325 \space L} = 0.0171 \space M $$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HC_9H_7O_4 ]& [ C_9H_7{O_4}^- ]& [ H_3O^+ ]\\
Initial& 0.0171 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.0171 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_9H_7{O_4}^- ][ H_3O^+ ]}{[ HC_9H_7O_4 ]}$$
$$K_a = \frac{(x)(x)}{[ HC_9H_7O_4 ]_{initial} - x}$$
3. Assuming $ 0.0171 \gt\gt x:$
$$K_a = \frac{x^2}{[ HC_9H_7O_4 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HC_9H_7O_4 ]_{initial}} = \sqrt{ 3.3 \times 10^{-4} \times 0.0171 }$$
$x = 2.4 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 2.4 \times 10^{-3} }{ 0.0171 } \times 100\% = 14.0 \%$$
The percent is greater than 5%, therefore, the approximation is invalid.
5. Return for the original expression and solve for x:
$$K_a = \frac{x^2}{[ HC_9H_7O_4 ]_{initial} - x}$$
$$K_a [ HC_9H_7O_4 ] - K_a x = x^2$$
$$x^2 + K_a x - K_a [ HC_9H_7O_4 ] = 0$$
$$x_1 = \frac{- 3.3 \times 10^{-4} + \sqrt{( 3.3 \times 10^{-4} )^2 - 4 (1) (- 3.3 \times 10^{-4} ) ( 0.0171 )} }{2 (1)}$$
$$x_1 = 2.2 \times 10^{-3} $$
$$x_2 = \frac{- 3.3 \times 10^{-4} - \sqrt{( 3.3 \times 10^{-4} )^2 - 4 (1) (- 3.3 \times 10^{-4} )( 0.0171 )} }{2 (1)}$$
$$x_2 = -2.5 \times 10^{-3} $$
- The concentration cannot be negative, so $x_2$ is invalid.
$$x = 2.2 \times 10^{-3} $$
6. $$[H_3O^+] = x = 2.2 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 2.2 \times 10^{-3} ) = 2.66 $$
