You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with KOH(aq) according to the following balanced chemical equation:

Question

You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with KOH(aq) according to the following balanced chemical equation:

HA(aq)+KOH(ap)---->KA(aq)+H2O(l)

If 12.15 mL of 0.490 M KOH is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid?

What is the molar mass of HA?

Answer 1

HA(aq) + KOH(ap) ----> KA(aq) + H2O(l) ... balanced equation

Find moles KOH needed to reach equivalence:

12.15 mls x 1 L / 1000 mls x 0.490 mol/L = 0.005954 mols KOH

Since equivalence means when moles of base (KOH) equals mols of acid (HA), we now know there are

0.005954 mols of HA present in the original 20.00 mls.

Concentration of HA = 0.005954 mols / 20.00 mls x 1000 ml / L = 0.298 mol / L = 0.298 M

Molar mass HA = g / mol = 1.153 g / 0.005954 mols = 194 g / mol