Question

Determine the pH during the titration of 65.9 mL of 0.410 M hypochlorous acid (K_a = 3.5×10^-8) by 0.410 M NaOHat the following points.

(a) Before the addition of any NaOH-?

(b) After the addition of 16.0 mL of NaOH-?

(c) At the half-equivalence point (the titration midpoint) -?

(d) At the equivalence point -?

(e) After the addition of 98.9 mL of NaOH-?

Answer 1

a)when 0.0 mL of NaOH is added

HClO dissociates as:

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

0.41 0 0

0.41-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 ((3.5*10^-8)*0.41) = 1.198*10^-4

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

so, x = 1.198*10^-4 M

use:

pH = -log [H+]

= -log (1.198*10^-4)

= 3.9216

Answer:3.92

b)when 16.0 mL of NaOH is added

Given:

M(HClO) = 0.41 M

V(HClO) = 65.9 mL

M(NaOH) = 0.41 M

V(NaOH) = 16 mL

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

mol(HClO) = 0.41 M * 65.9 mL = 27.019 mmol

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

mol(NaOH) = 0.41 M * 16 mL = 6.56 mmol

We have:

mol(HClO) = 27.019 mmol

mol(NaOH) = 6.56 mmol

6.56 mmol of both will react

excess HClO remaining = 20.459 mmol

Volume of Solution = 65.9 + 16 = 81.9 mL

[HClO] = 20.459 mmol/81.9 mL = 0.2498M

[ClO-] = 6.56/81.9 = 0.0801M

They form acidic buffer

acid is HClO

conjugate base is ClO-

Ka = 3.5*10^-8

pKa = - log (Ka)

= - log(3.5*10^-8)

= 7.456

use:

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

= 7.456+ log {8.01*10^-2/0.2498}

= 6.962

Answer: 6.96

c)

At half equivalence point,

pH = pKa = 7.456

Answer: 7.46

d)

find the volume of NaOH used to reach equivalence point

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

0.41 M *65.9 mL = 0.41M *V(NaOH)

V(NaOH) = 65.9 mL

Given:

M(HClO) = 0.41 M

V(HClO) = 65.9 mL

M(NaOH) = 0.41 M

V(NaOH) = 65.9 mL

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

mol(HClO) = 0.41 M * 65.9 mL = 27.019 mmol

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

mol(NaOH) = 0.41 M * 65.9 mL = 27.019 mmol

We have:

mol(HClO) = 27.019 mmol

mol(NaOH) = 27.019 mmol

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

ClO- here is strong base

ClO- formed = 27.019 mmol

Volume of Solution = 65.9 + 65.9 = 131.8 mL

Kb of ClO- = Kw/Ka = 1*10^-14/3.5*10^-8 = 2.857*10^-7

concentration ofClO-,c = 27.019 mmol/131.8 mL = 0.205M

ClO- dissociates as

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

0.205 0 0

0.205-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 ((2.857*10^-7)*0.205) = 2.42*10^-4

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

so, x = 2.42*10^-4 M

[OH-] = x = 2.42*10^-4 M

use:

pOH = -log [OH-]

= -log (2.42*10^-4)

= 3.6162

use:

PH = 14 - pOH

= 14 - 3.6162

= 10.3838

Answer: 10.38

e)when 98.9 mL of NaOH is added

Given:

M(HClO) = 0.41 M

V(HClO) = 65.9 mL

M(NaOH) = 0.41 M

V(NaOH) = 98.9 mL

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

mol(HClO) = 0.41 M * 65.9 mL = 27.019 mmol

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

mol(NaOH) = 0.41 M * 98.9 mL = 40.549 mmol

We have:

mol(HClO) = 27.019 mmol

mol(NaOH) = 40.549 mmol

27.019 mmol of both will react

excess NaOH remaining = 13.53 mmol

Volume of Solution = 65.9 + 98.9 = 164.8 mL

[OH-] = 13.53 mmol/164.8 mL = 0.0821 M

use:

pOH = -log [OH-]

= -log (8.21*10^-2)

= 1.0857

use:

PH = 14 - pOH

= 14 - 1.0857

= 12.9143

Answer: 12.91