Question

Calculate the hydronium ion concentration and the pH of the solution that results when 11.8 mL of 0.22 M acetic acid,  (), is mixed with 5.9 mL of 0.20 M .

Answer 1

[The question is missing a part..I assume it is mixed with 5.9 mL of 0.20 M NaOH]

Moles acetic acid = 0.0118 L x 0.22 mol/L = 0.0026
Moles NaOH = 0.0059 L x 0.20 mol/L = 0.00012 moles
Moles of acetic acid un-neutralized = 0.0026 - 0.00012  = 0.00248
Total volume = 11.8 ml + 5.9 ml = 17.7 mL = 0.0177 L
[acetic acid] = 0.00248 mol/0.0177 L = 0.140 M

Resulting solution is 0.140 M CH3COONa

CH3COO^-1+ H2O <--> CH3COOH + OH^-1

K_a for acetic acid=1.8 x 10^-5 .

Therefore K_b = 1 x 10^-14 / 1.8 x 10^-5 = 5.6 x 10^-10

5.6x10^-10 = [CH3COOH][OH-] / [CH3COO-]

[CH3COOH]=[OH-] = x   

5.6x10^-10 = x²/0.140-x , and assuming x is small
5.6x10^-10 = x²/0.140

=> x =8.85 x 10^-6

[OH-] =8.85 x 10^-6
[H+] = 1x10^-14 / 8.85 x 10^-6
[H+] = 1.2x10^-9
pH = -log (1.2x10^-9 ) = 8.9