🤓 Based on our data, we think this question is relevant for Professor Davis' class at UCF.

**Step 1. Calculate [H ^{+}]:**

**C _{6}H_{5}COOH ⇌ C_{6}H_{5}COO^{- }+ H**

$\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{{\mathbf{a}}}\mathbf{=}\frac{\left[{H}^{+}\right]\left[{C}_{6}{H}_{5}{\mathrm{COO}}^{-}\right]}{\left[{C}_{6}{H}_{5}\mathrm{COOH}\right]}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{3}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{\left[}\mathbf{x}\mathbf{\right]}}{[0.40]}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{3}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\phantom{\rule{0ex}{0ex}}\sqrt{{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\sqrt{(6.3\times {10}^{-5})(0.40)}$

**x = 5.020x10 ^{-3 }M = **

What is the pH of a 0.40 M solution of benzoic acid, C_{6}H_{5}COOH? (The K_{a} value for benzoic acid is 6.3 x 10^{-5}.)

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Based on our data, we think this problem is relevant for Professor Davis' class at UCF.