Question

Determine the pH at the equivalence point in the titration of 47 mL of 0.26 M HF (aq) with 0.26 M NaOH (aq). The Ka of HF is 7.4 x 10^-4.

Please show detailed process. Answer should be 8.12

Answer 1

mmol of acid = MV = 0.26*47 = 12.22

mmol of base = 12.22

Vbase = 12.22/0.26 = 47 mL

Total V in equivalence = 47+47 = 94 mL

now..

[F-] = mmol/V = 12.22 / 94 = 0.13

now..

Let HA --> HF and A- = F- for simplicity

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

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

Ka can be calculated as follows:

Kb = Kw/Ka = (10^-14)/(7.4*10^-4) = 1.35*10^-11

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

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

Kb= x*x/(M-x)

1.35*10^-11= x*x/(0.13-x)

solve for x

x^2 + Kb*x - M*Kb = 0

solve for x with quadratic equation

x = OH- =0.00000132475

[OH-]  0.00000132475

pOH = -log(OH-) = -log(0.00000132475 = 5.88

pH = 14-5.88= 8.12

pH = 8.12