Answer to Question #234628 in General Chemistry for Nana

Question

A 0.405 g sample of propanoic acid (CH₃CH₂COOH, K_a = 1.3. x 10^-5) was dissolved to a final volume of 50.0 cm³. This solution underwent titration with 0.21 M NaOH. Calculate the pH of solution at the equivalence point.

Answer 1

Molar mass of propanoic acid = 74.08

0.405/74.08 = 0.00547mol

= -log 1.3 × 10^-5 = 4.89

log (0.21/0.00547) = 1.58

= 4.89 + 1.58 = 6.47