Question

# A 1.54 gram sample of an unknown monoprotic acid is dissolved in 30.0 mL of water...

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 Ka for the acid?

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 Ka 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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