Question

What is the initial pH, at half way and after 12.5 mL of NaOH 0.3235 M is used to titrate 65.00 mL of 0.1237 M HAc?    Ka=1.75 x 10^-5

Answer 1

            CH3COOH + H2O -----> CH3COO- + H3O+

Initial      0.1237 M ......................0.................0

At equili 0.1237 -x .....................x................x

Ka = [CH3COO-][H3O+]/[CH3COOH]

1.75 x 10^-5 = x² / 0.1237-x

1.75 x 10^-5 = x² / 0.1237

x² = 1.75 x 10^-5 x 0.1237 = 1.47 x 10^-3 M

[H3o+] = 1.47 x 10^-3 M

Initial pH = -log(1.47 x 10^-3 ) = 2.83

At half way

[CH3COOH] = [CH3COO-]

So [H3O+] = 1.75 x 10^-5

pH = -log (1.75 x 10^-5) = 4.76

after 12.5 ml of 0.3235 M NaOH is added

NaOH + CH3COOH ---> CH3COONa + H2O

Moles of NaOH = 0.0125 L x 0.3235 M = 0.00404 moles

Moles of CH3COOH = 0.065L x 0.1237 M = 0.00804 moles

Moles of CH3COOH left unreacted = 0.00804 -0.00404 = 0.004 moles

pH = pKa + log[Salt]/[Acid]

     = -log(1.75 x 10^-5 ) + log 0.00404/0.004

     = 4.76 + 0.0043 = 4.7643