Question

Ka for hypochlorous acid, HClO is 3.0x10^-8. Calculate the pH after 10.0, 20.0, 30.0 and 40.0 mol 0.100 M NaOH have been added to 40.0 mL of 0.100 M HClO.

Answer 1

first convert the ka in to pKa

pKa = -log(Ka) = -log(3.0x10^-8) = 7.52

no of moles of HClO = molarity M x volume of the solution in liters

= 0.1M x 0.04L = 0.004 mol

Part A

basic reaction is

HClO + NaOH <=====> NaClO + H2O

after 10 mL of NaOH moles = 0.1M x 0.01 = 0.001 mol

0.001 mol of NaOh will react with 0.001 mol of HClO

no of moles of HClO remaining = 0.004-0.001 = 0.003

use the handerson equation

pH = pKa + log(NaClO/HClO)

here the concentration term is in ration so we can take directly moles in stead of Molarity

pH = 7.52 + log(0.001 / 0.003)

pH = 7.52 -0.48 = 7.04

Part B

20 mL NaOH

no of moles of NaOH = 0.1M x 0.02 = 0.002

moles of NaClO = 0.002

moles of HCL remaining = 0.002

pH = 7.52 + log(0.002 / 0.002)

pH = 7.42 + 0 = 7.52

Part 3

30 mL NaOH

no of moles of NaOH = 0.1M x 0.03L = 0.003 mol

moles of NaClO = 0.003

moles of HCO remaining = 0.001

pH = 7.52 + log(0.003 / 0.001)

pH = 7.99

Part -4

40 mL NaOH

no of moles of NaOH = 0.1 M x 0.04L = 0.004 mol

means exactly at equilent point

no fo moles of NaClO = 0.004 mol

total volume = 40 + 40 = 80 mL

concentration of NaClO = 0.004 mol / 0.08 = 0.05 M

convert the Ka in to Kb using the formula

Ka x Kb = 1.0 x 10^-14

Kb = 1.0 x 10^-14 / 3.0 x 10^-8

Kb = 3.3 x 10^-7

now

NaClO + H2O <-----> HClO + NaOH

cinstruct the ICE table and write the expression for Kb

Kb = [OH-][HClO] / [ClO-] = [OH-]^2 / [ClO-]

(since 1 ClO- gives equal amounts of OH- and HClO)

So [OH-] = (Kb*[ClO-])^0.5 --> (3.33*10^-7 * 0.05)^0.5 = 1.29*10^-4 M

pOH = -log[OH-] = 3.89
pH = 14 - pOH = 14 - 3.89 = 10.11