Question

A solution of an unknown acid is prepared by dissolving 0.250 g of the unknown in water to produce a total volume of 100.0 mL. Half of this solution is titrated to a phenolphthalein endpoint, requiring 12.2 mL of 0.0988 M KOH solution. The titrated solution is re-combined with the other half of the un-titrated acid and the pH of the resulting solution is measured to be 4.02. What is are the Ka value for the acid and the molar mass of the acid?

Answer 1

Half of the solution is titrated to a phenolphthalein endpoint and re-combined with the other half of the un-titrated acid. This implies that this solution is equivalent to a solution which has reached half equivalence point in an acid-base titration.

At half equivalence point,

pH = pK_a

or, pK_a = 4.02

or, - logK_a = 4.02

or, logK_a = - 4.02

or, K_a = 9.55 x 10^-5

Hence, the K_a value for the acid = 9.55 x 10^-5

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Mass of the acid dissolved = 0.250 g

Total volume of the acid solution = 100 mL

Now, half of this solution i.e. 50 mL of this solution which is equivalent to 0.250 g/2 = 0.125 g of acid is titrated with KOH.

Moles of KOH required for titration = 12.2 mL x 0.0988 M = 1.21 mmol = 1.21 x 10^-3 mol

Now, if the acid is monoprotic then 1.21 x 10^-3 mol of the base titrated 1.21 x 10^-3 mol of the acid.

Now,

1.21 x 10^-3 mol of the acid = 0.125 g of the acid

Therefore, 1 mol of the acid = (0.125 g x 1 mol)/(1.21 x 10^-3 mol) = 103 g of the acid

Hence, the molar mass of the acid = 103 g/mol