Question

A beaker with 115mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 4.70mL of a 0.470MHCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

first we need to calculate acid and conjugate base millimoles individually

(acid + conjugate base) molarity = 0.1 M

volume of buffer = 115ml

(acid + conjugate base) millimoles = 0.1 x 115

                                                    = 11.5

conjugate base millimoles = 11.5 -acid millimoles

for aicdic buffer

pH = pKa + log (conjugate base / acid)

5.000 = 4.740 + log (11.5-acid millimoles / acid millimoles)

0.26 = log (11.5-acid millimoles / acid millimoles)

11.5-acid millimoles / acid millimoles = 1.8197

11.5-acid millimoles = 1.8197 x acid millimoles

11.5 = 2.8197 acid millimoles

acid millimoles = 11.5/2.8197

acid millimoles = 4.078------------------------------(1)

conjugate base millimoles = 11.5 -acid millimoles

                                          = 11.5 -4.078

conjugate base millimoles       = 7.422 ----------------------------(2)

now additional added HCl millimoles (C) = molarity x volume (ml)

                                                              = 0.470 x 4.7

                                                                 = 2.209 ---------------------------(3)

on addition of