Question

A 5.65 −g sample of a weak acid with Ka=1.3×10−4 was combined with 5.20 mLof 6.10 M NaOH and the resulting solution was diluted to 750 mL. The measured pH of the solution was 4.20.

What is the molar mass of the weak acid?

Answer 1

A 5.65 −g sample of a weak acid with Ka=1.3×10−4 was combined with 5.20 mLof 6.10 M NaOH and the resulting solution was diluted to 750 mL. The measured pH of the solution was 4.20.

What is the molar mass of the weak acid?

Solution :-

Lets first calculate the moles of NaOH

Moles = molarity * volume in liter

Moles of NaOH = 6.10 mol per L * 0.00520 L = 0.03172 mol NaOH

pH = 4.20

using the Henderson equation lets calculate the ratio of the conjugate base to acid

pH= pka + log (A-/HA)

pka = -log ka

pka = - log 1.3*10^-4

pka= 3.87
4.20 = pKa + log(A-/HA)

4.20= 3.89 + log(A-/HA)

4.20 = 3.89 + log (A-/HA)
0.33 = log(A-/HA)
(A-/HA) = antilog 0.33

               = 2.14

HA + NaOH --> A^- + Na^+ + H2O
A^-1 = 0.03172 mol

Therefore
HA = X mol - 0.03172 mol
now lets find the value of the x using the ratio as follows

0.03172 / (X - 0.03172) = 2.14

2.14x – 0.06788 = 0.03172

2.14x= 0.03172+0.06788

2.14x= 0.0996

x= 0.0996 / 2.14

x= 0.04654

So the moles of the Acid before any NaOH added are 0.04654 moles

Now lets calculate the molar mass of the acid

Molar mass = mass / moles

                     = 5.65 g / 0.04654 mol

                    = 121.4 g per mol

So the molar mass of the acid = 121.4 g per mol