A reaction produced 0.40 mmol of H in a buffered solution which contained 300ml of 0.06M succinate buffer, pH is 4. What was the pH of the medium after the reaction was completed? The pKa of succinate buffer is 4.21

Question

A reaction produced 0.40 mmol of H in a buffered solution which contained 300ml of 0.06M succinate buffer, pH is 4. What was the pH of the medium after the reaction was completed? The pKa of succinate buffer is 4.21

Answer 1

Let's just call these acid(A) and base(B),

pH = 4.21 + log B/A
B/A = 0.6 but you should obtain the exact answer.
You know A + B = 0.06.
Solve the two equations simultaneously to find A and B. I obtained approximately 0.02 for B and 0.04 for A.

300 mL x 0.04 = approximately 12 mmols A.
300 mL x 0.02 = approx 6 mmols B.
...........B + H^+ ==> A
Initial....6...0.......12
add...........4..............
change...-4..-4.......+4
Equil......2..0........16

Substitute the E line (remember to use your answers instead of my approximations) into the HH equation and solve for the new pH. I obtained about 3.7 or so.