Question

Find the pH during the titration of 20.00 mL of 0.2480 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 ✕ 10-5), with 0.2480 M NaOH solution after the following additions of titrant. for 15ml

Answer 1

Lets call Butanoic acid BH.

milimoles of butanoic acid =0.2480M*20mL =4.96mmol

milimoles of NaOH = 0.2480M*15mL = 3.72mmol

So the reaction taking place after adding NaOH is-

BH + OH^- <---> B^- + H2O

Initial 4.96 0 0

Adding 3.72

Change -3.72 -3.72 +3.72

Equilbm 1.24 0 3.72

So,after addition of 15 mlNaOH,

[BH] = 1.24mmoles

[B^-] =3.72mmoles

Now using henderson-hasslebach equation-

pH = pKa + log[conjugate base/acid]

Ka = 1.54*10^-5

pKa = -log(Ka) = -log(1.54*10^-5) = 4.81

pH = 4.81 +log(3.72/1.24)

pH = 4.81 + 0.477

pH = 5.29

answered by: anonymous