Question

a) Calculate the pH of the solution after the addtion of 0.010 mol NaOH to 1.00L of a 0.100 M acetic acid/ 0.100 M sodium acetate buffer solution ( Ka= 1.8 x 10^-5)

b) What is the neutralization reaction that happens?

c) Is the system still a buffer after neutralization?

P.S : If you can pls explain question C more specific.

Answer 1

a)

moles of CH3COOH = 1.00 x 0.1 = 0.1

moles of CH3COONa = 1.00 x 0.1 = 0.1

moles of NaOH = 0.01 = C

on addition of NaOH CH3COONa concentration will increases and CH3COOH concentration will decreases

pKa = -log Ka = 4.74

pH = pKa + log [CH3COONa - C / CH3COOH + C]

pH = 4.74 + log (0.1 + 0.01 / 0.1 - 0.01)

pH = 4.83

b)

netralization reaction :

CH3COOH + NaOH ------------------> CH3COONa + H2O

c)

yes it is still buffer .

the reason is that the buffer only fails when moles of strong acid or base added to more than CH3COOH or CH3COONa which are already present in buffer

here NaOH moles = 0.01

CH3COOH moles = 0.1

if we add more than 0.1 moles of NaOH then buffer will fail . so here buffer will not fail in this case