Question

A solution was made by dissolving 0.339 g of an acid in water, then adding a few drops of indicator. This solution was titrated with 0.204 M KOH, and 19.96 mL were required to reach the endpoint. Assuming the acid is monoprotic, what is its molar mass? Report your answer to the tenths place.

Answer 1

Concentration of KOH = 0.204 M = 0.204 mol/L

Volume of KOH solution = 19.96 ml = 19.96 L / 1000 = 0.01996 L

Number of moles of KOH = 0.204 mol/L * 0.01996 L

= 0.004072 mol

Reaction of monoprotic acid(HA) and KOH

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

From reaction, 1.0 mol of KOH required 1.0 mol of HA so 0.004072 mol of KOH will required 0.004072 mol of HA.

Mass of HA in gram = 0.339 g

Molar mass = mass in gram / number of moles = 0.339 g / 0.004072 mol = 83.25 g/mol

Molar mass of monoprotic acid = 83.25 g/mol