Determine the pH of a 0.461 mol L-1 C6H5CO2H solution if the Ka of C6H5CO2H is 6.5 × 10-5.

To determine the pH of the solution, we can use the formula for the ionization constant of water (Kw), which is the product of the concentrations of H+ and OH- ions. Since we know the concentration of the acid (C6H5CO2H) and its Ka, we can find the concentration of H+ ions and then calculate the pH. Here are the steps:

1. Write the ionization equation for the acid: C6H5CO2H ⇌ H+ + C6H5CO2-

2. Write the expression for the acid ionization constant (Ka): Ka = [H+][C6H5CO2-] / [C6H5CO2H]

3. Since the acid is weak, we can assume that only a small fraction of it ionizes, so [C6H5CO2H] in the denominator is approximately equal to the initial concentration of the acid, which is 0.461 mol/L.

4. Let x be the concentration of H+ ions, which is also the concentration of C6H5CO2- ions. Then we can write: 6.5 × 10^-5 = x^2 / 0.461

5. Solve this equation for x: x = sqrt(6.5 × 10^-5 * 0.461) = 0.0061 M

6. The pH is the negative logarithm (base 10) of the H+ concentration: pH = -log10(0.0061) = 2.21

So, the pH of the solution is approximately 2.21.