What is the pH at the equivalence point in the titration of 100.0 mL of 0.0500M HOCl (Ka=3.5×10−8) ( K a = 3.5 × 10 − 8 ) with 0.450M NaOH?

Question

What is the pH at the equivalence point in the titration of 100.0 mL of 0.0500M HOCl (Ka=3.5×10−8)(Ka=3.5×10−8) with 0.450M NaOH? 

Express your answer to two decimal places.

Answer 1

HClO + NaOH ==> NaClO + H2O ... balanced equation

At the equivalence point the moles of HClO = moles NaOH

moles HClO present = 100.0 ml x 1 L / 1000 ml x 0.0500 mol / L = 0.00500 mols HClO

moles NaOH needed = 0.00500 mols NaOH (1:1 mole ratio in the balanced eq.)

Volume of NaOH needed to supply 0.005 mols: 0.005 mols x 1 L / 0.450 mol = 0.0111 L = 11.11 mls

Total volume after equivalence is reached = 100.0 mls + 11.1 mls = 111.1 mls = 0.1111 L

All of the HClO has been converted to ClO^- (in the form of NaClO). This is 0.00500 mols

Final [HClO] = 0.00500 mols / 0.1111 L = 0.0450 M

Now we must look at the hydrolysis of the ClO^-

ClO^- + H2O ==> HClO + OH^-

ClO^- is acting as a base so we need the Kb for ClO^-

KaKb = 1x10^-14 and Kb = 1x10^-14 / 3.5x10^-8

Kb = 2.86x10^-7

Kb = 2.86x10^-7 = [HClO][OH-] / [HClO]

2.86x10^-7 = (x)(x) / 0.0450

x = 1.13x10^-4 M = [OH-]

pOH = -log 1.13x10^-4 M

pOH = 3.95

pH = 14 - 3.95

pH = 10.05