The acid-dissociation constant for benzoic acid C6H5COOH is 6.3 * 10^-5. Calculate the equilibrium concentrations of H3O+, C6H5COO-, and C6H5COOH in the solution if the initial concentration of C6H5COOH is 0.050 M.

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1. Write the balanced chemical equation for the dissociation of benzoic acid: \( \text{C}_6\text{H}_5\text{COOH} \rightleftharpoons \text{C}_6\text{H}_5\text{COO}^- + \text{H}_3\text{O}^+ \).
2. Set up the expression for the acid-dissociation constant \( K_a \) using the equation: \( K_a = \frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{COOH}]} \).
3. Define the initial concentrations: \([\text{C}_6\text{H}_5\text{COOH}]_0 = 0.050 \text{ M}\), \([\text{C}_6\text{H}_5\text{COO}^-]_0 = 0\), \([\text{H}_3\text{O}^+]_0 = 0\).
4. Assume \( x \) is the change in concentration at equilibrium: \([\text{C}_6\text{H}_5\text{COOH}] = 0.050 - x\), \([\text{C}_6\text{H}_5\text{COO}^-] = x\), \([\text{H}_3\text{O}^+] = x\).
5. Substitute the equilibrium concentrations into the \( K_a \) expression and solve for \( x \): \( 6.3 \times 10^{-5} = \frac{x^2}{0.050 - x} \).