Question

The Ka for HOBr is 2.8 x 10 ^-9. You titrate 20.0 mL of a 0.400 M HOBr solution using a 0.400 M NaOH. What is the pH after 10.0 mL of NaOH have been added (1/2 equivalence pt) and after 22.0 mL of NaOH have been added (overshot endpoint)?

pH after 10.0 mL of 0.400 M NaOH =

pH after 22.0 mL of 0.400 M NaOH = (total volume = 42.0 mL)

Answer 1

1)when 10.0 mL of NaOH is added

Given:

M(HOBr) = 0.4 M

V(HOBr) = 20 mL

M(NaOH) = 0.4 M

V(NaOH) = 10 mL

mol(HOBr) = M(HOBr) * V(HOBr)

mol(HOBr) = 0.4 M * 20 mL = 8 mmol

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

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

We have:

mol(HOBr) = 8 mmol

mol(NaOH) = 4 mmol

4 mmol of both will react

excess HOBr remaining = 4 mmol

Volume of Solution = 20 + 10 = 30 mL

[HOBr] = 4 mmol/30 mL = 0.1333M

[OBr-] = 4/30 = 0.1333M

They form acidic buffer

acid is HOBr

conjugate base is OBr-

Ka = 2.8*10^-9

pKa = - log (Ka)

= - log(2.8*10^-9)

= 8.553

use:

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

= 8.553+ log {0.1333/0.1333}

= 8.553

Answer: 8.55

2)when 22.0 mL of NaOH is added

Given:

M(HOBr) = 0.4 M

V(HOBr) = 20 mL

M(NaOH) = 0.4 M

V(NaOH) = 22 mL

mol(HOBr) = M(HOBr) * V(HOBr)

mol(HOBr) = 0.4 M * 20 mL = 8 mmol

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

mol(NaOH) = 0.4 M * 22 mL = 8.8 mmol

We have:

mol(HOBr) = 8 mmol

mol(NaOH) = 8.8 mmol

8 mmol of both will react

excess NaOH remaining = 0.8 mmol

Volume of Solution = 20 + 22 = 42 mL

[OH-] = 0.8 mmol/42 mL = 0.019 M

use:

pOH = -log [OH-]

= -log (1.905*10^-2)

= 1.7202

use:

PH = 14 - pOH

= 14 - 1.7202

= 12.2798

Answer: 12.28