Question

Show ALL your calculations and the respective dissociation equations for each step. Assignments without calculations or assignments that are submitted late are not going to be graded 1) Calculate the pH in the titration of 50,00 ml of 0.060 M hypochlorous acid (HCIO) with a 0.120 M sodium hydroxide, NaOH solution after the addition of the following volumes of base: Ka for hypochlorous acid = 2.9 x 10-8 A) 0 mL pH = B) 12.5 mL pH = C) 25.0 mL pH = *D) 40.0 mL pH =

Answer 1

A)when 0.0 mL of NaOH is added

HClO dissociates as:

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

6*10^-2 0 0

6*10^-2-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 ((2.9*10^-8)*6*10^-2) = 4.171*10^-5

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

so, x = 4.171*10^-5 M

use:

pH = -log [H+]

= -log (4.171*10^-5)

= 4.3797

Answer: 4.38

B)when 12.5 mL of NaOH is added

Given:

M(HClO) = 0.06 M

V(HClO) = 50 mL

M(NaOH) = 0.12 M

V(NaOH) = 12.5 mL

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

mol(HClO) = 0.06 M * 50 mL = 3 mmol

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

mol(NaOH) = 0.12 M * 12.5 mL = 1.5 mmol

We have:

mol(HClO) = 3 mmol

mol(NaOH) = 1.5 mmol

1.5 mmol of both will react

excess HClO remaining = 1.5 mmol

Volume of Solution = 50 + 12.5 = 62.5 mL

[HClO] = 1.5 mmol/62.5 mL = 0.024M

[ClO-] = 1.5/62.5 = 0.024M

They form acidic buffer

acid is HClO

conjugate base is ClO-

Ka = 2.9*10^-8

pKa = - log (Ka)

= - log(2.9*10^-8)

= 7.538

use:

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

= 7.538+ log {2.4*10^-2/2.4*10^-2}

= 7.538

Answer: 7.54

C)when 25.0 mL of NaOH is added

Given:

M(HClO) = 0.06 M

V(HClO) = 50 mL

M(NaOH) = 0.12 M

V(NaOH) = 25 mL

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

mol(HClO) = 0.06 M * 50 mL = 3 mmol

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

mol(NaOH) = 0.12 M * 25 mL = 3 mmol

We have:

mol(HClO) = 3 mmol

mol(NaOH) = 3 mmol

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

ClO- here is strong base

ClO- formed = 3 mmol

Volume of Solution = 50 + 25 = 75 mL

Kb of ClO- = Kw/Ka = 1*10^-14/2.9*10^-8 = 3.448*10^-7

concentration ofClO-,c = 3 mmol/75 mL = 0.04M

ClO- dissociates as

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

0.04 0 0

0.04-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 ((3.448*10^-7)*4*10^-2) = 1.174*10^-4

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

so, x = 1.174*10^-4 M

[OH-] = x = 1.174*10^-4 M

use:

pOH = -log [OH-]

= -log (1.174*10^-4)

= 3.9302

use:

PH = 14 - pOH

= 14 - 3.9302

= 10.0698

Answer: 10.07

D)when 40.0 mL of NaOH is added

Given:

M(HClO) = 0.06 M

V(HClO) = 50 mL

M(NaOH) = 0.12 M

V(NaOH) = 40 mL

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

mol(HClO) = 0.06 M * 50 mL = 3 mmol

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

mol(NaOH) = 0.12 M * 40 mL = 4.8 mmol

We have:

mol(HClO) = 3 mmol

mol(NaOH) = 4.8 mmol

3 mmol of both will react

excess NaOH remaining = 1.8 mmol

Volume of Solution = 50 + 40 = 90 mL

[OH-] = 1.8 mmol/90 mL = 0.02 M

use:

pOH = -log [OH-]

= -log (2*10^-2)

= 1.699

use:

PH = 14 - pOH

= 14 - 1.699

= 12.301

Answer: 12.30