Question

Calculate the pH of the acetate buffer after 28.00 mL 0.100 M HCl has been added.

There is 10 mL of the acetate buffer (0.16 M NaC2H3O2 / 0.2 M HC2H3O2) mixed with 30 mL of water in the buffer solution.

The Pka of the buffer is 4.74. Ka = 1.76 x 10^-5

Please help, I've made ICE tables and tried using ph = pka +log (base/acid) and the pH I get is bigger than it should be. As more HCl is added to the solution, the pH continued to decrease and become more acidic, so my experimental value was around 2.52.

Answer 1

Volume of buffer = 10 mL

moles of sodium acetate = 0.16 M * 10 mL =1.6 mmol

moles of acetic acid = 0.2 M*10 mL = 2.0 mmol

moles of HCl added = molarity of HCl*volume of HCl

= 0.100 M * 28.00 mL

= 2.8 mmol

HCl reacts with sodium acetate to form acetic acid.

Now, here moles of HCl is more than moles of sodium acetate.

So, excess HCl = 2.8 mmol - 1.6 mmol = 1.2 mmol

Total volume of solution after addition of HCl = 10 mL (buffer) + 30 mL (water) + 28 mL (HCl) = 68 mL

Concentration of excess HCl = 1.2 mmol/68 mL = 0.0176 M

[H⁺] = 0.0176 M

pH = -log(0.0176)

= 1.75

Answer : pH of solution is 1.75. It is no longer a buffer as whole of the conjugate base has exhausted.