Question

A 35.00 mL solution of 0.2500 M HF is titrated with a standardized 0.1083 M solution of NaOH at 25 degrees C.

(a) What is the pH of the HF solution before titrant is added?

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

(c) What is the pH at 0.50 mL before the equivalence point?

(d) What is the pH at the equivalence point?

(e) What is the pH at 0.50 mL after the equivalence point?

Answer 1

(a) What is the pH of the HF solution before titrant is added?

First of all, volumes will change in every case (Volume of base + Volume of acid), therefore the molarity will change as well

M = moles / volume

pH = -log[H+]

a) HA -> H+ + A-

Ka = [H+][A-]/[HA]

a) no NaOH

Ka = [H+][A-]/[HA]

Assume [H+] = [A-] = x

[HA] = M – x

Substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

7.2*10^-4 = x*x/(0.25-x)

This is quadratic equation

x =0.01306

For pH

pH = -log(H+)

pH =-log(0.01306)

pH in a = 1.884

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

Vrequired:

mmol of acid present = MV = 0.25*35 = 8.75 mmol

therefore, we need

mmol of base = 8.75

M = mmol/mL

mL = mmol/M = 8.75/0.1083 = 80.79 mL required

V = 80.79 mL

(c) What is the pH at 0.50 mL before the equivalence point?

V = 80.79 -0.50 = 80.29 mL

mmol of acid present = MV = 0.25*35 = 8.75 mmol

mmol of base = MV = 0.1083*80.29 = 8.695407

mmol of acid left = 8.75 -8.695407 = 0.054593 mmol

mmol of conjugate (F-) formed = 8.695407

apply buffer equation

pH = pKa + log(F-/HF)

pH = 3.14 + log(8.695407/0.054593 )

pH = 5.34

(d) What is the pH at the equivalence point?

This will be Hydrolisis (equilibrium of acid-base) and the weak acid/base will form an equilibrium

We will need Kb

Ka*Kb = Kw

And Kw = 10^-14 always at 25°C for water so:

Kb = Kw/Ka = (10^-14)/(7.2*10^-4) = 1.388*10^-11

Now, proceed to calculate the equilibrium

H2O + A- <-> OH- + HA

Then K equilibrium is given as:

Kb = [HA][OH-]/[A-]

Assume [HA] = [OH-] = x

[A-] = M – x

Substitute in Kb

1.388*10^-11 = [x^2]/[M-x]

recalculate M:

mmol of conjugate = 8.75

Vnew = V1+V2 = 35+80.79= 115.79 mL

[M] = 8.75/115.79 = 0.07556 M

1.388*10^-11 = [x^2]/[0.07556-x]

x = 6.89*10^-7

[OH-] = 6.89*10^-7

Get pOH

pOH = -log(OH-)

pOH = -log ( 6.89*10^-7) = 6.16

pH = 14-pOH = 14-6.16= 7.84

e) What is the pH at 0.50 mL after the equivalence point?

mmol of acid present = MV = 0.25*35 = 8.75 mmol

mmol of base = MV = 0.1083*(81.29) = 8.803

mmol of OH- left = 8.75 -8.803= 0.053 mmol of OH-

Vtotal = V1+V2 = 35+81.29 = 116.29

[OH-] = 0.053 /116.29 = 0.00045 M

pOH = -log(0.00045) =

pH = 14-pOH = 14-3.3467 = 10.653