Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solution at equilibrium. Ka(HF) = 7.2×10-4; Ka(HOC6H5) = 1.6×10-10. pH = [OC6H5-] = M

Explanation:

The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.

1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

                    HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹:       2.7                   0       0

C/mol·L⁻¹:      -x                   +x      +x

E/mol·L⁻¹:   2.7 - x                 x        x

2. Calculate the pH

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

     2.7                         x        0.0441

im not exactly sure but i think its b!