Question

A 230.0 mL buffer solution is 0.200 M in acetic acid and 0.200 M in sodium acetate.

What is the pH after addition of 0.0150 mol of HCl?   

What is the pH after addition of 0.0150 mol of NaOH?

Answer 1

pKa of CH3COOH = 4.76

a)

adding 0.0150 mol HI will increase number of moles of CH3COOH by 0.0150 mol and decrease mol of CH3COONa by0.0150 mol

mol of CH3COOH = 0.200*0.230 + 0.0150 = 0.061 mol

mol of CH3COONa = 0.200*0.230 – 0.0150 = 0.031 mol

pH = pKa + log {[ CH3COONa]/[ CH3COOH]}

    = 4.76 + log (0.031/0.061)

    = 4.47

b)

adding 0.0150 mol NaOH will increase number of moles of CH3COONa by 0.0150 mol and decrease mol of CH3COOH by0.0150 mol

mol of CH3COOH = 0.200*0.230 - 0.0150 = 0.031 mol

mol of CH3COONa = 0.200*0.230 + 0.0150 = 0.061 mol

pH = pKa + log {[ CH3COONa]/[ CH3COOH]}

    = 4.76 + log (0.061/0.031)

    = 5.05