Question

For the titration of 25.0 mL of 0.20 M hydrofluoric acid with 0.20 M sodium hydroxide , determine the volume of based added when pH is a) 2.85 b) 3.15 c)11.89

Answer 1

Let the volume of sodium hydroxide added = V mL
Before the end point at v < 25 ml

The following reaction takes place

HF = H+ + F-

Moles of HF at starting point = molarity x volume

= 0.2 mol/L x 25/1000 L

= 0.005 moles
Moles of NaOH added = molarity x volume

= 0.2 mol/L x v/1000 L

Moles of HF left = initial moles of HF present - moles of HF reacted

moles of HF reacted = moles of NaOH reacted

Moles of HF left = 0.005 - 0.2 v/1000

moles of F- formed = moles of NaOH reacted

= 0.2 x v/1000

Concentration of F-
[F-] = (0.2v/1000) x 1000/25

= 0.008v

Concentration of HF

[HF] = (0.005 - 0.2v/1000) x 1000/25

= 0.2 - 0.008v

Equilibrium constant expression for weak acid HF

K = [H+][F-]/[HF]

Part a

When pH = 2.85

[H+] = 10^-pH = 10^-2.85

K = 10^-2.85 x 0.008v/(0.2 - 0.008v)

v = 0.2K/[0.008(10^-2.85 + K)]

For HF,
K = 6.7 x 10^-4

v = (0.2 x 6.7 x 10^-4) /[0.008(10^-2.85 + (6.7 x 10^-4))]

v = 8.3 mL

Part b

When pH = 3.15

[H+] = 10^-pH = 10^-3.15

K = 10^-3.15 x 0.008v/(0.2 - 0.008v)

v = 0.2K/[0.008(10^-3.15 + K)]

For HF,
K = 6.7 x 10^-4

v = (0.2 x 6.7 x 10^-4) /[0.008(10^-3.15 + (6.7 x 10^-4))]

v = 12.5 mL

Part C

After the end point the pH is determined by the excess base added. If

pH = 11.89

pOH = 14 - 11.89 = 2.11

[OH-] = 10^-pOH

= 7.7625 x 10^-3 M

Let the volume of sodium hydroxide added after end point = V mL

total volume added = 50 + V

moles of OH- present =( 7.7625 x 10^-3) x (50 + v)/1000

moles of NaOH present = 0.2v/1000 ml

Now Solve for V

V = 2 ml

total volume added = 25 + 2 = 27 ml