Question

A tritration is carried out where 50.0 m of .300M HF is tritrated with a 0.190 M NaOh solution. Determine the pH for each of the following points in the tritration. Ka (HF)=7.2 x 10^-4

A) 0.0 mL of NaOH added

B) 30.0mL of NaOH added

C) At the equivalence point

Answer 1

a)

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/(0.3-x)

This is quadratic equation

x^2 + (7.2*10^-4)x -(0.3)(7.2*10^-4) = 0

x = 0.01434

For pH

pH = -log(H+)

pH =-log(0.01434)

pH in a = 1.84

b)

mmol of HF = MV = 0.3*50 = 15

mmol of NaOH = MV = 0.19*30 = 5.7

after reaction

mmol of HF left = 15-5.7 = 9.3

mmol of F- formed = 5.7

this is a buffer

pH = pKa + log(F-/HF)

pH = -log(7.2*10^-4) + log(5.7 /9.3)

pH = 2.930

c)

in equivalence

Total V

Vbase = (50*0.3)/(0.190) = 78.9 mL

Vtotal = 50+78.9 = 128.9 mL

[F-] = mmol/V = 15/128.9 = 0.1163 M

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

Kb = [HF][OH-]/[F-]

(10^-14)/(7.2*10^-4) = (x*x)/(0.1163-x)

x^2 + 1.388*10^-11 -0.1163* 1.388*10^-11 = 0

x = 0.000001270

pOH = -log(0.000001270) = 5.89

pH = 14-pOH = 14-5.89

pH = 8.11