Question

A beaker with 1.50×10² 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 8.50 mL of a 0.370 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Answer 1

the buffer equation

pH = pKA + log(conjugate/acid)

then

pH = 5 and pKa = 4.75

[A-] + [HA] = 0.1 M

V 150 ml

Solve for concentrations initallly

5 = 4.75 + log([A-]/[HA])

then

[A-]/[HA] = 10^(5-4.75) = 1.7782794

now we have 2 equyations relatin A- and HA

1.7782794 = [A-]/[HA]

and

[A-] + [HA] = 0.1

then

Solve for A-

[A-] = 0.1 - [HA]

then susbtitute in ratio

1.7782794 = (0.1 - [HA])/[HA]

1.7782794 = 0.1/[HA] - 1

2.7782794 = 0.1/[HA]

[HA] = 0.1/2.7782794 = 0.035993

mol of HA = M*V = 150*0.035993 = 5.39895 mmol of HA

then;

[A-] = 0.1 - 0.035993 = 0.064007

mol of A- = M*V = 150*0.064007= 9.60105 mmol of A-

This is initially;

now when addition of Hcl:

V = 8.5 ml of M = 0.37 M of HCl

mol of H+ added = M*V = 8.5*0.37 = 3.145 mmol of H+

Then:

Acid after addition:

mol HA = 5.39895 +3.145 = 8.54395

Conjguate after addition

mol A- =9.60105 -3.145 = 6.45605

then

substitut epH

pH = pKa + log(conjugate/acid)

pH = 4.75+ log(6.45605/8.54395) = 4.62830

change in pH = 5-4.62830 = 0.3717

there is a decrease of 0.3717 due to acid