Answer
$[H_3O^+] = 2.5 \times 10^{-3} \space M$
$pH = 2.59$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HC_7H_5O_2 ]& [ C_7H_5O{_2}^- ]& [ H_3O^+ ]\\
Initial& 0.100 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.100 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_7H_5O{_2}^- ][ H_3O^+ ]}{[ HC_7H_5O_2 ]}$$
$$K_a = \frac{(x)(x)}{[ HC_7H_5O_2 ]_{initial} - x}$$
3. Assuming $ 0.100 \gt\gt x$:
$$K_a = \frac{x^2}{[ HC_7H_5O_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HC_7H_5O_2 ]_{initial}} = \sqrt{ 6.5 \times 10^{-5} \times 0.100 }$$
x = $2.5 \times 10^{-3}$
4. Test if the assumption was correct:
$$\frac{ 2.5 \times 10^{-3} }{ 0.100 } \times 100\% = 2.5 \% \lt 5\%$$
5. Thus, it is correct to say that $x = 2.5 \times 10^{-3}$
6. $$[H_3O^+] = x = 2.5 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( \sqrt{ 6.5 \times 10^{-5} \times 0.100 } ) = 2.59 $$
