Can you please explain the computation using pH= pKa + log (base/acid)

Question

50.0mL of 0.20M of lactic acid is added with;

1.) 100.0mL

2.)110.0mL

of 0.10M NaOH. Calculate the pH at each addition of NaOH

Answer 1

The equation that you refer to is called the Henderson Hasselbalch equation, in case you want to look it up and read more about it. The computation is exactly as it says: the pH of the buffer solution will be equal to the pKa of the weak acid + log of the ratio of the conjugate base to the weak acid. This ratio is also sometimes written log [salt]/[acid] since the salt is the conjugate base. With lactic acid, the conjugate base (salt) is lactate. CH3CH(OH)COOH is lactic acid. The conjugate base is CH3CH(OH)COO^-.

If you are given the Ka for the acid, then take the negative log to get the pKa. For lactic acid,

the Ka = 1.5x10^-4 so pKa = -log 1.5x10^-4 = 3.8

Look at the reaction between lactic acid and NaOH:

CH3CH(OH)COOH + NaOH ==> CH3CH(OH)COO^-Na⁺ + H2O

Initial moles lactic acid = 50.0 ml x 1L/1000 ml x 0.20 mol/L = 0.01 moles

After addition of 100.0 ml of 0.10 M NaOH, you have added 100.0 ml x 1L/1000 ml x 0.1 mol/L = 0.01 mol OH^- and that reacts with the lactic acid to produce the salt sodium lactate (the conjugate base).

In this case all of the lactic acid has been neutralized to produce 0.01 moles of the conjugate base, sodium lactate.

In this case, you DO NOT use the Henderson Hasselbalch equation because there is no weak acid left. It's all been converted to the conjugate base. If you want to know how to calculate the pH of this new solution, please submit a separate question, as this involves hydrolysis reactions and not pH = pKa + log [base]/[acid]