Calculate the pH of a solution formed by mixing 150.0 mL of 0.10 mol L^-1 C₆H₅COOH with 100.0 mL of 0.30 mol L^-1 C₆H₅COONa. The K_a for C₆H₅COOH is 6.5 × 10^-5. A) 4.19 B) 9.69 C) 4.49 D) 4.31 E) 10.51

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Step 1: Calculate the amount of moles of L^-1 C₆H₅COOH in the solution. n = CV n = 0.10 mol/L × 0.150 L n = 0.015 mol Step 2: Calculate the amount of moles of L^-1 C₆H₅COONa in the solution. n = CV n = 0.30 mol/L × 0.100 L n = 0.030 mol Step 3: Determine which species will contribute to the pH of the solution. Since L^-1 C₆H₅COOH is a weak acid and L^-1 C₆H₅COONa is the conjugate base, the pH of the solution will be determined by the acid dissociation of L^-1 C₆H₅COOH. Step 4: Write the equation for the dissociation of L^-1 C₆H₅COOH. L^-1 C₆H₅COOH + H2O ⇌ L^-1 C₆H₅COO- + H3O+ Step 5: Write the expression for the acid dissociation constant (K_a). K_a = [L^-1 C₆H₅COO-][H3O+] / [L^-1 C₆H₅COOH] Step 6: Substitute the given values into the expression for the acid dissociation constant. 6.5 × 10^-5 = [(0.015 - x)(x)] / [0.015 + 0.1] Step 7: Simplify the equation. 6.5 × 10^-5 = (0.015x - x2) / 0.115 Step 8: Rearrange the equation into a quadratic form. x2 - 0.015x - 8.93 × 10^-6 = 0 where x is the concentration of H3O+. Step 9: Solve for x. Using the quadratic formula, we get x = 4.38 × 10^-4 mol/L. Step 10: Calculate the pH of the solution. pH = -log[H3O+] pH = -log(4.38 × 10^-4) pH = 4.49 Therefore, the answer is C.

Calculate the PH of a Solution Formed by Mixing 150