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 8.40 mL of a 0.350 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

V = 2*10^2 = 200 ml

pH = 5

[H+] = 10^-pH = 10^-5

[A-] + [HA] = 0.1

V = 8.4 ml M = 0.35 HCl

pH change?

pKa = 4.74

apply

pH = pKa +log(base/acid)

initially

[A-] + [HA] = 0.1

note that

5 = 4.74 + log(base/acid)

base/acid = 10^(5-4.74) = 1.819700

now we have 2 euqations

[A-] + [HA] = 0.1

[A-]/[HA] = 1.819700

[A-] = 1.8197*[HA]

1.8197*[HA] + [HA] = 0.1

2.8197(HA) = 0.1

[HA] = 0.1/2.8197 = 0.03546

[A-] + [HA] = 0.1

[A] =0.1-0.03546 = 0.06454

then

the changes:

acid added = MV = 8.4*0.35 = 2.94 mmol of acid

[HA] = MV + 2.94 = 200*0.03546 + 2.94 = 10.032

[A-] = MV - 2.94 = 200*0.06454 - 2.94 = 9.968

new pH

pH = pKa + log(A-/HA) = 4.74 + log(9.968/10.032)

pH = 4.7372