what is the pH of a solution made by mixing 60cm3 1.0M NaCH3COO(aq) with 40cm3 0.50 M HCl (aq)?

Question

what is the pH of a solution made by mixing 60cm^3 1.0M NaCH3COO(aq) with 40cm^3 0.50 M HCl (aq)?

Ka of CH3COOH is 1.8 x 10^-5.

Answer 1

CH₃COONa + HCl ==> NaCl + CH₃COOH

moles CHCOO^- present = 60 ml x 1 L / 1000 ml x 1.0 mol / L = 0.06 moles

moles H⁺ added = 40 ml x 1 L / 1000 ml x 0.50 mol/L = 0.02 moles H⁺

CH₃COONa + HCl ==> NaCl + CH₃COOH

0.06 ..............0.02.........................0............Initial

-0.02..............-0.02......................+0.02......Change

0.04...................0..........................0.02.......Equilibrium

This creates a buffer of a weak acid (CH₃COOH) and the conjugate base (CH₃COO-). Use the Henderson Hasselbalch equation pH = pKa + log (conj.base/acid)

pKa = -log Ka = -log 1.8x10^-5

pKa = 4.74

pH = 4.74 + log (0.04/0.02) = 4.74 + 0.30

pH = 5.04