Question

A buffer is prepared by adding 300.0 mL of 0.050 M Ch3COONa to a 500.0 mL of 0.100 M CH3COOH. What is the pH of this buffer? Ka=1.8x10^-5

What is the pH of the buffer upon the addition of 10.0 mL of a 1.2x10^-3 M HCl solution?

Answer 1

Answer – Given, [CH₃COONa] = [CH₃COO^-] = 0.050 M , volume = 300.0 mL

[CH₃COOH] = 0.100 M , volume = 500.0 mL , Ka = 1.8*10^-5

We know the Henderson Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

now calculation of pKa

pKa = -log Ka

        = - log 1.8*10^-5

        = 4.74

pH = 4.74 + log 0.050 / 0.100

      = 4.44

Now pH when we added 10.0 mL of a 1.2x10^-3 M HCl solution

First we need to calculate the moles of each

Moles of CH₃COOH = 0.100 M * 0.500 L = 0.050 moles

Moles of CH₃COO^-= 0.050 M * 0.300 L = 0.0150 moles

Moles of HCl = 1.2*10^-3 M * 0.010 L = 1.2*10^-5 moles

We know when we added HCl there is moles of acid increase and moles of conjugate base decrease-

Moles of CH₃COOH = 0.050 moles + 1.2*10^-5 moles = 0.050012

Moles of CH₃COO^- = 0.0150 moles - 1.2*10^-5 moles = 0.014988 moles

Total volume = 300+500+10 = 810 mL

So, New molarity

[CH₃COOH] = 0.050012 moles / 0.810 L = 0.0617 M

[CH₃COO^-] = 0.014988 moles / 0.810 L = 0.0185 M

So,

pH = 4.74 + log 0.0185 M / 0.0617 M

      = 5.27

the pH of the buffer upon the addition of 10.0 mL of a 1.2x10^-3 M HCl solution is 5.27