Question

A beaker with 2.00×102 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 7.20 mL of a 0.420 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

a = acid = acetic acid

b = conjugate base = acetate

millimoles of buffer = 2.00 x10^2 x 0.1 = 20

a + b = 20 --------------------> 1

pH = pKa + log [b/a]

5.00 = 4.74   + log [b /a]

b / a = 1.297

b = 1.297 a ----------------->2

form 1 & 2

a = 8.707

b = 11.293

now strong acid C millimoles of HCl added = C = 7.20 x 0.420 = 3.024

new pH

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

pH = 4.74 + log [11.293 -3.024 / 8.707 + 3.024]

pH = 4.588

change in pH = final - initial

                      = 4.588 - 5.000

                      = - 0.412