Question

In a titration, 25.0ml of .35 M hydrofluoric acid, HF solution is titrated with NaOH solution.

a) Calculate the solution ph when 13.0ml of .25 M NaOH is added to the acid.

b) Calculate the solution ph at the equivalence point.

Answer 1

Ka of HF = 6.6*10^-4

a)

Given:

M(HF) = 0.35 M

V(HF) = 25 mL

M(NaOH) = 0.25 M

V(NaOH) = 13 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.35 M * 25 mL = 8.75 mmol

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

mol(NaOH) = 0.25 M * 13 mL = 3.25 mmol

We have:

mol(HF) = 8.75 mmol

mol(NaOH) = 3.25 mmol

3.25 mmol of both will react

excess HF remaining = 5.5 mmol

Volume of Solution = 25 + 13 = 38 mL

[HF] = 5.5 mmol/38 mL = 0.1447M

[F-] = 3.25/38 = 0.0855M

They form acidic buffer

acid is HF

conjugate base is F-

Ka = 6.6*10^-4

pKa = - log (Ka)

= - log(6.6*10^-4)

= 3.18

use:

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

= 3.18+ log {8.553*10^-2/0.1447}

= 2.952

Answer: 2.95

b)

find the volume of NaOH used to reach equivalence point

M(HF)*V(HF) =M(NaOH)*V(NaOH)

0.35 M *25.0 mL = 0.25M *V(NaOH)

V(NaOH) = 35 mL

Given:

M(HF) = 0.35 M

V(HF) = 25 mL

M(NaOH) = 0.25 M

V(NaOH) = 35 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.35 M * 25 mL = 8.75 mmol

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

mol(NaOH) = 0.25 M * 35 mL = 8.75 mmol

We have:

mol(HF) = 8.75 mmol

mol(NaOH) = 8.75 mmol

8.75 mmol of both will react to form F- and H2O

F- here is strong base

F- formed = 8.75 mmol

Volume of Solution = 25 + 35 = 60 mL

Kb of F- = Kw/Ka = 1*10^-14/6.6*10^-4 = 1.515*10^-11

concentration ofF-,c = 8.75 mmol/60 mL = 0.1458M

F- dissociates as

F- + H2O -----> HF + OH-

0.1458 0 0

0.1458-x x x

Kb = [HF][OH-]/[F-]

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 ((1.515*10^-11)*0.1458) = 1.486*10^-6

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

so, x = 1.486*10^-6 M

[OH-] = x = 1.486*10^-6 M

use:

pOH = -log [OH-]

= -log (1.486*10^-6)

= 5.8278

use:

PH = 14 - pOH

= 14 - 5.8278

= 8.1722

Answer: 8.17