Question

A 1.458-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 40.0 mL of this solution was titrated with 0.07942-M NaOH. The pH after the addition of 16.55 mL of base was 2.94, and the equivalence point was reached with the addition of 46.63 mL of base.

a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution.

b) What is the molar mass of the acid?

c) What is the pK_a of the acid?

Answer 1

Given:

Mass of monoprotic acid= 1.458 g

Volume of solution = 100.0 mL

Volume of acid used for titration = 40.0 mL

[NaOH]= 0.07942 M

pH = 2.94 when 16.55 mL of NaOH is added.

At equivalence point volume of NaOH = 46.63 mL

Solution:

Then write reaction between NaOH and weak monoprotic acid.

By using moles of NaOH in the titration we find moles of acid.

From the pH of the solution we get an idea about the concentration [H3O+]

Then we find molar mass of acid by using moles and its mass.

ka expression is used to find out pka as we know pka = - log ka

Solution

Reaction

NaOH (aq)   + HA (aq) (weak monoprotic acid ) --- > NaA (aq) + H2O (l)

Here mole ratio between NaOH and HA is 1:1

Calculation of moles of of NaOH at equilvalence point.

n NaOH = volume of NaOH x molarity

= 0.04663 L x 0.07942 M= 0.003703 mol

n HA = 0.003703 mol NaOH x 1 mol HA / 1 mol NaOH

= 0.003703 mol HA

This much moles are present in 40.0 mL

So we calculate moles of acid in 100.0 mL

n HA in 100.0 mL

= 100.0 mL x 0.003703 mol HA / 40.0 mL

=0.00923 mol HA

Mili moles of HA

= 0.00923 M x 1000 mili moles / 1 M

= 9.23 mili moles HA

a). Answer : milimoles of Acid = 9.23

n HA = mass in g / molar mass

Molar mass of HA = mass in g / moles

= 1.458 g / 0.00923 mol

= 157.48 g/mol

b). Molar mass of HA = 157.48 g /mol

c).

calculation of moles of NaOH when 16.55 mL of NaOH is added

Moles of NaOH = 0.01655 L x 0.07942 M = 0.001341 mol NaOH

Moles of HA reacted = 0.001314   mol   (1:1 ratio)

Excess moles of HA = Original – reacted ) mol

= 0.00923 mol HA – 0.001314 mol HA

= 0.007944 mol HA

Calculation of concentration of HA in this solution

Total volume = 40.0 mL + 16.55 mL = 56.55 mL = 0.05655 L

[HA]= 0.007944 mol / 0.05655 L = 0.1405 M

Dissociation reaction of HA

            HA (aq) + H2O (l) --- > A- (aq) + H3O + (aq)

I       0.1405                                0                   0

C         -x                                 +x                    +x       

E    (0.1405-x)                         x                      x

pH for this solution = 2.94

[H3O+] = Antilog (-pH )

= Antilog (-2.94)

[H3O+]= 0.001145 M

Concentrations at equilibrium

x = [H3O+]= 0.001145 M

x = [A-]

[HA]= 0.1405 – x = 0.1405 – 0.001145

= 0.1393 M

Ka for HA = [A-] [H3O+]/[HA]

=(0.001145)² / (0.1393)

= 6.46 x 10^-6

Pka = - log ( 6.46E-6)

= 5.02

Pka of acid = 5.02