Question

A beaker with 115 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 8.00 mL of a 0.470 MHCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

CH3COOH = a = acid

CH3COO- = b = conjugate base

buffer total millimoles = 115 x 0.1 = 11.5

a + b = 11.5 ---------------------> 1

pH = 5.000

pKa = 4.740

pH = pKa + log [b/a]

5.00 = 4.740 + log (b/a)

b / a = 1.8197

b = 1.8197 a ---------------->2

from 1 & 2

a = 4.078

b = 7.422

strong acid HCl millimoles = 8 x 0.870 = 3.76 = C

new pH

pH = pKa + log [b -C / a + C]

pH = 4.74 + log (7.422 -3.76 / 4.078 + 3.76)

pH = 4.410

change in pH = final - initial

= 4.410 - 5.000

= - 0.5904