Chemistry 212 titration question. Please Help!

Question

Find the pH during the titration of 20.00 mL of 0.1860 M benzoic acid, C₆H₅COOH (K_a = 6.3  10^-5), with 0.1860 M NaOH solution after the following additions of titrant.

(a) 0 mL

2.47

(b) 10.00 mL

?

(c) 15.00 mL

?

(d) 20.00 mL

?

(e) 25.00 mL

?

Answer 1

I won't do all of them, but will guide you so you should be able to complete the problem.

C₆H₅COOH + NaOH ===> C₆H₅COONa + H2O

Initial moles C₆H₅COOH = 20.00 ml x 1 L / 1000 ml x 0.1860 mol/L = 0.00372 moles C₆H₅COOH

(a)

Ka = 6.3x10-5 = [H+][C₆H₅COO-] / [C₆H₅COOH] = (x)(x) / 0.1860

x2 = 1.17x10-5

x = 3.42x10-3 = [H+]

pH = -log 3.42x10-3

pH = 2.47

(b) After adding 10.00 ml of 0.1860 M NaOH:

moles NaOH added = 10.00 ml x 1 L / 1000 ml x 0.1860 mol/L = 0.00186 mols NaOH

C₆H₅COOH + NaOH ===> C₆H₅COONa + H2O

0.00372.........0.00186..............0................Initial

-0.00186......-0.00186.........+0.00186........Change

0.00186............0.................0.00186.........Equillibrium

pH = pKa + log [conj.base]/[acid] and pKa = -log Ka = 4.20

pH = 4.20

(c) After adding 15.00 ml of 0.1860 M NaOH

moles NaOH added = 15.00 ml x 1 L / 1000 ml x 0.1860 mol /L = 0.00279 mols NaOH

C₆H₅COOH + NaOH ===> C₆H₅COONa + H2O

0.00372.........0.00279...............0............Initial

-0.00279......-0.00279...........+0.00279...Change

0.00093............0...................0.00279....Equilibrium

pH = pKa + [conj.base]/[acid]

pH = 4.20 + log (0.00279/0.00093) = 4.20 + 0.48

pH = 4.68

(d) After adding 20.00 mls 0f 0.1860 M NaOH

Hint: you are now at the equivalence point and no longer have a buffer since ALL of the C₆H₅COOH and all of the NaOH have been converted to . To find the pH, use the hydrolysis of C₆H₅COO- and the Kb for C₆H₅COO-.

C₆H₅COO- + H2O ==> C₆H₅COOH + OH- and use Kb for C₆H₅COO-

(e) Here you will have excess OH since in (d) above you've already completely neutralized all of the C₆H₅COOH.