posted by Debbie on .
5 drops of 0.10M NaOH were added to 20mL of the buffer problem, the pHwent from 3.40 to 3.43. Write a net ionic equiation to explain why the pH did not go up to about 10, as it would have if that amount of NaOH were added to distilled water or to 20mL 0.00040.
What buffer did you have and what was its concn? It will be the acid which I'll call HA.
5 drop will be approximately 0.25 mL x 0.1M = 0.025 millimoles.
.........HA + OH^- ==> A^- + H2O