Question

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

for 10ml

Answer 1

Start with the equation and the see the moles of acid and salt left after the addition of the base:
CH3CH2CH2COOH + NaOH -----> CH3CH2CH2COONa + H2O
NaOH reacts with C3H7COOH in 1:1 mol ratio

moles of acid= 20.0mL x 0.2480M =4.96 mmol =0.00496 mol

Moles of base= 10.00mL x 0.2480M =2.48 mmol =0.00248 mol

When these react you will produce 0.00248 mol C3H7COONa and there will be 0.00496-0.00248 = 0.00248 mol C3H7COOH left unreacted
The final volume of solution = 30.00mL
[salt] = 0.00248/.030 = 0.0826M
[acid] = 0.00248/0.030 = 0.0826M

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

Use Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid]
pH = 4.81 + log ( 0.0826/0.0826)
pH = 4.81 + log 1
pH = 4.81