Question

A certain weak acid, HA , with a Ka value of 5.61×10−6 , is titrated with NaOH .

Part A: A solution is made by titrating 8.00 mmol (millimoles) of HA and 2.00 mmol of the strong base. What is the resulting pH?

Part B: More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 71.0 mL ?

Express the pH numerically to two decimal places.

Answer 1

Given that;

Ka value of 5.61×10−6

So

pKa = - log K a= 5.25

And the recation is as follows:

HA + OH- = A- + H2O

mmol HA in excess = 8.00 - 2.00 = 6.00
mmoles A- formed = 2.00
pH= 5.25 + log 2.00/ 6.00

=5.25 + log 0.333

=4.77


at the equivalence point mmoles A- formed = 8.00 => 8.00 x 10^-3 mol
[A-]= 8.00 x 10^-3 mol/ 0.0710 L=0.113 M

A- + H2O <=> HA + OH-

Kb = Kw/Ka = 1.0 x 10^-14 / 5.61 x 10^-6= x^2/ 0.113-x

1.78*10^-9 = x^2/ 0.113-x

Due to small value of Kb we can write 0.113-x = 0.113
1.78*10^-9 = x^2/ 0.113

X = 1.42*10^-5
x = [OH-]= 1.42 x 10^-5 M

pOH = - log [OH-] = 4.85

And pH+ pOH = 14

So pH = 14-4.85

pH = 9.15