Question

Need help with this problem.

Calculate the pH during the titration of 10.00 mL of 0.400 M hypochlorous acid with 0.500 M NaOH. First what is the initial pH (before any NaOH is added)?
The K_a for HOCl is 3.0 x 10^-8 M.

A.What is the pH after 0.90 mL of NaOH are added?]

B.What is the pH after 7.50 mL of NaOH are added?

C.What is the pH after 9.30 mL of NaOH are added?

D.What is the pH after 14.90 mL of NaOH are added?

Answer 1

1)when 0.9 mL of NaOH is added

Given:

M(HOCl) = 0.4 M

V(HOCl) = 10 mL

M(NaOH) = 0.5 M

V(NaOH) = 0.9 mL

mol(HOCl) = M(HOCl) * V(HOCl)

mol(HOCl) = 0.4 M * 10 mL = 4 mmol

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

mol(NaOH) = 0.5 M * 0.9 mL = 0.45 mmol

We have:

mol(HOCl) = 4 mmol

mol(NaOH) = 0.45 mmol

0.45 mmol of both will react

excess HOCl remaining = 3.55 mmol

Volume of Solution = 10 + 0.9 = 10.9 mL

[HOCl] = 3.55 mmol/10.9 mL = 0.3257M

[OCl-] = 0.45/10.9 = 0.0413M

They form acidic buffer

acid is HOCl

conjugate base is OCl-

Ka = 3*10^-8

pKa = - log (Ka)

= - log(3*10^-8)

= 7.523

use:

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

= 7.523+ log {4.128*10^-2/0.3257}

= 6.626

Answer: 6.63

2)when 7.5 mL of NaOH is added

Given:

M(HOCl) = 0.4 M

V(HOCl) = 10 mL

M(NaOH) = 0.5 M

V(NaOH) = 7.5 mL

mol(HOCl) = M(HOCl) * V(HOCl)

mol(HOCl) = 0.4 M * 10 mL = 4 mmol

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

mol(NaOH) = 0.5 M * 7.5 mL = 3.75 mmol

We have:

mol(HOCl) = 4 mmol

mol(NaOH) = 3.75 mmol

3.75 mmol of both will react

excess HOCl remaining = 0.25 mmol

Volume of Solution = 10 + 7.5 = 17.5 mL

[HOCl] = 0.25 mmol/17.5 mL = 0.0143M

[OCl-] = 3.75/17.5 = 0.2143M

They form acidic buffer

acid is HOCl

conjugate base is OCl-

Ka = 3*10^-8

pKa = - log (Ka)

= - log(3*10^-8)

= 7.523

use:

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

= 7.523+ log {0.2143/1.429*10^-2}

= 8.699

Answer: 8.70

3)when 9.3 mL of NaOH is added

Given:

M(HOCl) = 0.4 M

V(HOCl) = 10 mL

M(NaOH) = 0.5 M

V(NaOH) = 9.3 mL

mol(HOCl) = M(HOCl) * V(HOCl)

mol(HOCl) = 0.4 M * 10 mL = 4 mmol

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

mol(NaOH) = 0.5 M * 9.3 mL = 4.65 mmol

We have:

mol(HOCl) = 4 mmol

mol(NaOH) = 4.65 mmol

4 mmol of both will react

excess NaOH remaining = 0.65 mmol

Volume of Solution = 10 + 9.3 = 19.3 mL

[OH-] = 0.65 mmol/19.3 mL = 0.0337 M

use:

pOH = -log [OH-]

= -log (3.368*10^-2)

= 1.4726

use:

PH = 14 - pOH

= 14 - 1.4726

= 12.5274

Answer: 12.53

4)when 14.9 mL of NaOH is added

Given:

M(HOCl) = 0.4 M

V(HOCl) = 10 mL

M(NaOH) = 0.5 M

V(NaOH) = 14.9 mL

mol(HOCl) = M(HOCl) * V(HOCl)

mol(HOCl) = 0.4 M * 10 mL = 4 mmol

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

mol(NaOH) = 0.5 M * 14.9 mL = 7.45 mmol

We have:

mol(HOCl) = 4 mmol

mol(NaOH) = 7.45 mmol

4 mmol of both will react

excess NaOH remaining = 3.45 mmol

Volume of Solution = 10 + 14.9 = 24.9 mL

[OH-] = 3.45 mmol/24.9 mL = 0.1386 M

use:

pOH = -log [OH-]

= -log (0.1386)

= 0.8584

use:

PH = 14 - pOH

= 14 - 0.8584

= 13.1416

Answer: 13.14