**Step 1**. Calculate the **initial amounts** of C_{6}H_{5}COOH and NaOH.

• **30.0mL of 0.10M C _{6}H_{5}COOH**

$\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}\mathbf{COOH}\mathbf{=}(30.0\overline{)\mathrm{mL}}\times \frac{{10}^{-3}\overline{)L}}{1\overline{)\mathrm{mL}}})\frac{\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}\mathbf{COOH}}{\overline{)\mathbf{L}}}$

**mol C _{6}H_{5}COOH = 0.003 mol C_{6}H_{5}COOH**

Calculate the pH of a solution prepared by mixing 15.0mL of 0.10M NaOH and 30.0mL of 0.10M benzoic acid solution. (Benzoic acid is monoprotic; its dissociation constant is 6.5 x 10^{-5}.)

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