Question

0.499g of dried KHP is dissolved in 35 mL of distilled water and titrated with KOH. If it took 24.40mL of KOH to reach an end point determine the concentration of KOH.

I got 0.1 M KOH.

But if you forgot to dry the above KHP sample and it was later determined to contain 5.00% water by weight what would be the actual concentration of KOH?

This part has me stumped. I would appreciate the help.

Answer 1

Molar Mass of KHP = 204.22 g / mol

Mass of KHP = 0.499 g

=> Moles of KHP = 0.499 / 204.22 = 0.00244 moles

Let Molarity of KOH = X

=> Moles of KOH = X * 0.0244

At end point,

Moles of KHP = Moles of KOH

=> X * 0.0244 = 0.00244

=> X = 0.1 M = Conc. of KOH solution

Case 2

Mass of water in KHP = 0.05 x 0.499 = 0.02495 g

Mass of KHP (dry) = 0.95 x 0.499 = 0.47405 g

=> Moles of KHP = 0.47405 / 204.22 = 0.00232

=> X * 0.0244 = 0.00232

=> X = 0.095 M = Actual Conc. of KOH