Question

A student performs a titration of 50.0mL o 0.100M hydrofluoric acid, using 0.050 M sodium hydroxide. The Ka for hydrofluoric acid is 8.8x10^-5

a) What is the pH of the solution after the addition of 20.0 mL of sodium hydroxide solution?

b) How many milliliters of NaOh are required to reach the equivalence point?

c) Write the chemical reaction that will determine the pH of the solution at the equivalence point.

d) What is the pH at the equivalence point

Answer 1

a)

Given:

M(HF) = 0.1 M

V(HF) = 50 mL

M(NaOH) = 0.05 M

V(NaOH) = 20 mL

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

mol(HF) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.05 M * 20 mL = 1 mmol

We have:

mol(HF) = 5 mmol

mol(NaOH) = 1 mmol

1 mmol of both will react

excess HF remaining = 4 mmol

Volume of Solution = 50 + 20 = 70 mL

[HF] = 4 mmol/70 mL = 0.0571M

[F-] = 1/70 = 0.0143M

They form acidic buffer

acid is HF

conjugate base is F-

Ka = 8.8*10^-5

pKa = - log (Ka)

= - log(8.8*10^-5)

= 4.056

use:

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

= 4.056+ log {1.429*10^-2/5.714*10^-2}

= 3.453

Answer: 3.45

b)

find the volume of NaOH used to reach equivalence point

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

0.1 M *50.0 mL = 0.05M *V(NaOH)

V(NaOH) = 100 mL

Answer: 100 mL

c)

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

d)

Given:

M(HF) = 0.1 M

V(HF) = 50 mL

M(NaOH) = 0.05 M

V(NaOH) = 100 mL

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

mol(HF) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.05 M * 100 mL = 5 mmol

We have:

mol(HF) = 5 mmol

mol(NaOH) = 5 mmol

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

F- here is strong base

F- formed = 5 mmol

Volume of Solution = 50 + 100 = 150 mL

Kb of F- = Kw/Ka = 1*10^-14/8.8*10^-5 = 1.136*10^-10

concentration ofF-,c = 5 mmol/150 mL = 0.0333M

F- dissociates as

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

0.0333 0 0

0.0333-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 ((1.136*10^-10)*3.333*10^-2) = 1.946*10^-6

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

so, x = 1.946*10^-6 M

[OH-] = x = 1.946*10^-6 M

use:

pOH = -log [OH-]

= -log (1.946*10^-6)

= 5.7108

use:

PH = 14 - pOH

= 14 - 5.7108

= 8.2892

Answer: 8.29