Question

A 1.491-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 35.0 mL of this solution was titrated with 0.08730-M NaOH. The pH after the addition of 19.69 mL of base was 4.56, and the equivalence point was reached with the addition of 39.30 mL of base.

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

mmol acid

b) What is the molar mass of the acid?

g/mol

c) What is the pK_a of the acid?

pK_a =

Answer 1

Mass of weak acid taken = 1.491 g

Volume of water = 100mL

Volume of weak acid used =35mL

On addition of base volume = 19.69 , the pH = 4.56

At this condition there must be formation of salt of weak acid, so this is buffer region

for buffers

pH = pKa + log [salt] / [acid]

Moles of [Salt] = volume of base X molarity of base = 19.69 X 0.08730 = 1.7189 millimoles

a) the moles of acid present initially = Volume of base used at equivalence point X molarity = 0.08730 X 39.30

moles of acid present initiaaly = 3.431 millmoles

b) Moles = Mass / Molecular mass

Molecular mass = Mass / Moles = 1.491 / 3.431 millimoles = 434.57 g / moles

c) moles of acid present at the point where pH of solution is 4.56 will be

Moles of acid present = Moles of acid initially present - Moles of salt formed = 3.431millimles - 1.7189 millimoles

Moles of acid present = 1.712millimoles

pH = pKa + log [salt] / [acid]

4.56 = pKa + log [1.719]/ [1.712]

pKa = 4.558

pKa = -logKa

Ka = antilog of (-4.558)

Ka = 2.77 X 10^-5