Question

# A 1.224-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...

A 1.224-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 55.0 mL of this solution was titrated with 0.08096-M NaOH. The pH after the addition of 12.88 mL of base was 7.00, and the equivalence point was reached with the addition of 41.81 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 pKa of the acid? pKa =

Solution:

(a) Let us assume the acid be HA

HA + NaOH => NaA + H2O

mm of NaOH added at equivalence point = volume x concentration

= 41.81 x 0.08096 = 3.38 mmol

mm of HA in 55.0 mL of solution = Millimoles of NaOH added at equivalence point

= 3.38 mmol

mmoles of HA in original sample = 100.0/55.0 x 3.38 = 6.145 mmol

(b) Molar mass of HA = mass of HA/moles of HA

= 1.224/6.145 x 10-3 = 199.19 g/mol

(c) At pH = 7.00

Millimoles of NaOH added = volume x concentration

= 12.88 x 0.08096 = 1.04 mmol

Millimoles of NaA formed = Millimoles of NaOH added = 1.04 mmol

Millimoles of HA left = 3.38 - 1.04 = 2.34 mmol

Henderson-Hasselbalch equation:

pH = pKa + log([NaA]/[HA])

= pKa + log(mmol of NaA/mmol of HA) because the volume is the same for both

7.00 = pKa + log(1.04/2.34)

pKa = 7.35

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