Question

A **1.54** gram sample of an unknown monoprotic
acid is dissolved in **30.0** mL of water and titrated
with a a **0.225** M aqueous **potassium
hydroxide** solution. It is observed that after
**19.3** milliliters of **potassium
hydroxide** have been added, the pH is
**7.171** and that an additional **34.5**
mL of the **potassium hydroxide** solution is required
to reach the equivalence point.

(1) What is the molecular weight of the
acid? g/mol

(2) What is the value of K_{a} for the acid?

Answer #1

1) What is the molecular weight of the acid? g/mol

First calculate the number of moles of base as follows:

Number of moles = molarity * volume in L

= 0.225* 34.5 ml/1000 ml*1.00 L

= 0.00776 Moles

At equivalence point number of moles of acid = number moles of base

Mole of acid = 0.00776 Moles

Molar mass – amount in g / number of moles

= **1.54** gram /0.00776 Moles

= 198.45 g/ mole

(2) What is the value of K_{a} for the acid?

When one-half of the volume of added base, called the half-equivalence point, at this point; half of the acid reacted to form A– , the concentrations of A– and HA at the half-equivalence point are the same. Therefore, at the half-equivalence point, the pH is equal to the pKa.

Thus pH or pKa = approximately 7.171

Ka = 10^-pKa

= 10^-7.171

= 6.745*10^-8

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