Question

Calculate the pH at 25 degree C of a solution formed by combining and 0.10 M HC2H3O2(aq) and 0.10 M NaC2H3O2(aq) Calculate the new pH formed after addition of 0.10 mol of HCl (aq) to 1.0 L of the solution formed in part (a)

Answer 1

Using Henderson-Hesselbalach equation

pH = pKa + log { [salt] / [acid] }

pKa of CH₃COOH = 4.76

So, pH = pKa + log { [CH₃COONa] / [CH₃COOH] }

= 4.76 + log ( 0.1 / 0.1 )

= 4.76 + log (1)

= 4.76 + 0

= 4.76

0.10 M CH₃COOH(aq) and 0.10 M CH₃COONa(aq) are present in 1.0 L.

So, moles of CH₃COOH = 0.10 M x 1 L = 0.1 mole

moles of CH₃COONa = 0.10 M x 1 L = 0.1 mole

When you add HCl, it will react with CH₃COONa to form CH₃COOH.

So, 0.1 mole of HCl will react with 0.1 mole of CH₃COONa to produce 0.1 mole of CH₃COOH. All CH₃COONa are now formed to CH₃COOH.

So, the total moles of CH₃COOH present after HCl addition = 0.1 mol + 0.1 mol = 0.2 mol

Volume = 1 L

[CH₃COOH] = 0.2 mol / 1 L = 0.2 M

Now, CH₃COOH is a weak acid.

CH₃COOH   CH₃COO^- + H⁺

Ka of CH₃COOH = 1.8 x 10^-5

Now,

Ka = [CH₃COO^-] [H⁺] / [CH₃COOH]

1.8 x 10^-5 = x² / 0.2

x² = 3.6 x 10^-6

x = 1.90 x 10^-3

So, [H⁺] = x = 1.90 x 10^-3

pH = - log[H⁺]

= - log(1.90 x 10^-3)

= 2.72