Question

Part A

A beaker with 135 mL 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 4.30 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740. Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

Answer 1

millimoles of acid + conjugate base =135 x 0.1 = 13.5 ------------->1

pH = pKa + log [conjugate base / acid ]

5.00 = 4.74 + log [conjugate base / acid ]

1.82 = conjugate base / acid

conjugate base = 1.82 acid -------------> 2

from 1 and 2

1.82 acid + acid = 13.5

2.82 acid = 13.5

acid = 4.79

conjugate base = 8.71

millimoles of strong acid = C = 4.30 x 0.4 = 1.72

on additon of C millimoles of acid to buffer

pH = pKa + log [conjugate base - C / acid + C]

pH = 4.74 + log (8.71 - 1.72 / 4.79 + 1.72)

pH = 4.77

pH chnage = 4.77 - 5.000

= -0.23

pH chnage  = - 0.23