Question

Consider the titration of 42.0 mL of 0.270 M HF with 0.215 M NaOH. Calculate the pH at each of the following points.

a) How many milliliters of base are required to reach the equivalence point?

b) Calculate the pH after the addition of 10.5 mL of base

c) Calculate the pH at halfway to the equivalence point.

d) Calculate the pH at the equivalence point.

e) Calculate the pH after the addition of 84.0 mL of base.

Answer 1

We know that for HF; Ka=7.2 x 10^-4

HF_(aq) --> H⁺_(aq) + F^-_(aq)

NaOH (Aq) + HF   à H2O + NaF

[HF] initial = 0.27 mole/L = x moles/ 0.042 L ==> x moles = 0.011134 moles HF

Now to 42 ml of HF solution 10.5ml of 0.215M NaOH is added

NaOH moles= 0.0105 * 0.215 = 0.0022575 moles

When we add base moles of HF will reduce but at the same time F- will be added to solution.

Moles of HF depleted = 0.011134 - 0.0022575 = 0.0088765 moles of HF.

And Moles of F- generated = 0.0022575

Now total volume of solution is 42+10.5 = 52.5 ml

[HF] final = 0.0088765 moles HF/0.052 L = 0.17 M
[F-] final = 0.0022575 moles F-/0.052 L = 0.0434 M

We know that ; PKa = -log Ka

PKa = - log(7.2 x 10^-4)

PKa = 3.14

pH = pKa + log[F-]/[HF] = 3.14 + log(0.0434/0.17) = 2.55

when 10.5 ml of base is added pH = 2.55

when 84.0 ml of base is added, no. of moles of base = 0.084L ×0.215 mols/L = 0.01806 moles of OH-

initial moles of acid present= 0.011134

therefore excess of OH- ion in the solution= 0.01806 - 0.011134 = 0.006926 mols

pOH = -log [OH-]

pOH = 2.15

pH = 14-2.15= 11.84

So pH of the solution when 84 ml of base is added = 11.84

Now initial moles of acid present is 0.011134 and to reach equivalence point equal number of moles of base should be added as the ration between base and acid is 1:1.

So 0.011134 number of moles will be present in how much ml of base,

0.215 moles in 1000 ml ,

so 0.011134 moles in ‘X’ ml

X = 0.011134/0.215= 0.0517 L = 51.7 ml

51.7 ml of base should be added to reach equivalence point.

At equivalence point solution is compose of salt and weak acid

Total volume of the solution at equivalence point is 42+51.7 = 93.7 ml

Molarity of sodium acetate = initial moles of acid/final volume of solution = 0.011134/ 93.7 = 1.188 × 10^-4 M

Calculate Kb for NaF salt,

Kw = KaKb

Kb = 1.00 x 10^-14/ 7.2 x 10^-4 = 0.133 × 10^-10 = 1.388 × 10^-11

1.388 × 10^-11= [(OH-) (OH-)] / 1.188 × 10^-4 M

[OH-] = 3.97 × 10^-8

pOH = 7.40

pH at equivalence point = 14-7.4 = 6.59

pH = pKa at the half-equivalence point.

So at half way to equivalence point pH =Pka= 3.14