Question

Hypochlorous acid has a pKa of 7530. If a 66.0 mL solution of o.180 M is titrated with o.150 M NaOH. Calculate the pH of the solution at each of the following points on the titration curve. Consult Textbook Numerical Answer a) Before the addition of any NaOH pH- b) After adding 11.40 mL of NaOH pH- c) At the equivalence point pH- d) After adding 82.2 mL of NaOH pH-

Answer 1

use:

pKa = -log Ka

7.53 = -log Ka

Ka = 2.951*10^-8

1)when 0.0 mL of NaOH is added

HClO dissociates as:

HClO -----> H+ + ClO-

0.18 0 0

0.18-x x x

Ka = [H+][ClO-]/[HClO]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.951*10^-8)*0.18) = 7.288*10^-5

since c is much greater than x, our assumption is correct

so, x = 7.288*10^-5 M

use:

pH = -log [H+]

= -log (7.288*10^-5)

= 4.1374

Answer: 4.14

2)when 11.4 mL of NaOH is added

Given:

M(HClO) = 0.18 M

V(HClO) = 66 mL

M(NaOH) = 0.15 M

V(NaOH) = 11.4 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.18 M * 66 mL = 11.88 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.15 M * 11.4 mL = 1.71 mmol

We have:

mol(HClO) = 11.88 mmol

mol(NaOH) = 1.71 mmol

1.71 mmol of both will react

excess HClO remaining = 10.17 mmol

Volume of Solution = 66 + 11.4 = 77.4 mL

[HClO] = 10.17 mmol/77.4 mL = 0.1314M

[ClO-] = 1.71/77.4 = 0.0221M

They form acidic buffer

acid is HClO

conjugate base is ClO-

Ka = 2.951*10^-8

pKa = - log (Ka)

= - log(2.951*10^-8)

= 7.53

use:

pH = pKa + log {[conjugate base]/[acid]}

= 7.53+ log {2.209*10^-2/0.1314}

= 6.756

Answer: 6.76

3)

find the volume of NaOH used to reach equivalence point

M(HClO)*V(HClO) =M(NaOH)*V(NaOH)

0.18 M *66.0 mL = 0.15M *V(NaOH)

V(NaOH) = 79.2 mL

Given:

M(HClO) = 0.18 M

V(HClO) = 66 mL

M(NaOH) = 0.15 M

V(NaOH) = 79.2 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.18 M * 66 mL = 11.88 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.15 M * 79.2 mL = 11.88 mmol

We have:

mol(HClO) = 11.88 mmol

mol(NaOH) = 11.88 mmol

11.88 mmol of both will react to form ClO- and H2O

ClO- here is strong base

ClO- formed = 11.88 mmol

Volume of Solution = 66 + 79.2 = 145.2 mL

Kb of ClO- = Kw/Ka = 1*10^-14/2.951*10^-8 = 3.389*10^-7

concentration ofClO-,c = 11.88 mmol/145.2 mL = 0.0818M

ClO- dissociates as

ClO- + H2O -----> HClO + OH-

0.0818 0 0

0.0818-x x x

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

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((3.389*10^-7)*8.182*10^-2) = 1.665*10^-4

since c is much greater than x, our assumption is correct

so, x = 1.665*10^-4 M

[OH-] = x = 1.665*10^-4 M

use:

pOH = -log [OH-]

= -log (1.665*10^-4)

= 3.7786

use:

PH = 14 - pOH

= 14 - 3.7786

= 10.2214

Answer: 10.22

4)when 82.2 mL of NaOH is added

Given:

M(HClO) = 0.18 M

V(HClO) = 66 mL

M(NaOH) = 0.15 M

V(NaOH) = 82.2 mL

mol(HClO) = M(HClO) * V(HClO)

mol(HClO) = 0.18 M * 66 mL = 11.88 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.15 M * 82.2 mL = 12.33 mmol

We have:

mol(HClO) = 11.88 mmol

mol(NaOH) = 12.33 mmol

11.88 mmol of both will react

excess NaOH remaining = 0.45 mmol

Volume of Solution = 66 + 82.2 = 148.2 mL

[OH-] = 0.45 mmol/148.2 mL = 0.003 M

use:

pOH = -log [OH-]

= -log (3.036*10^-3)

= 2.5176

use:

PH = 14 - pOH

= 14 - 2.5176

= 11.4824

Answer: 11.48