Question

A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.905 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 31.0 mL of calcium hydroxide to reach the endpoint.

A. What is the molarity of the calcium hydroxide solution?

This calcium hydroxide solution is then used to titrate an unknown solution of hydrochloric acid.

B. If 25.0 mL of the calcium hydroxide solution is required to neutralize 19.3 mL of hydrochloric acid, what is the molarity of the hydrochloric acid solution?

Answer 1

a)

mol of acid = 0.905 /204.22 = 4.43 x 10^-3 mol

2 KHC8H4O4   + Ca(OH)2   --------------> salt   +   water

1 mol Ca(OH)2   ---------------> 2 mol acid

moles of Ca(OH)2 = 4.43 x 10^-3 / 2 = 2.216 x 10^-3 mol

Volume of Ca(OH)2 = 31.0 mL

Molarity of Ca(OH)2 = 2.216 x 10^-3 / 0.031 = 0.0715

Molarity of Ca(OH)2 = 0.0715 M

B)

Volume of base used = 25.0 mL

Volume of HBr = 19.3

1 mol base ------------> 2 mol acid

moles of base = 0.0715 x 25 / 1000 = 1.79 x 10^-3

moles of acid = 2 x 1.79 x 10^-3 = 3.57 x 10^-3

Volume of acid = 19.3 mL

Molarity of HBr = 3.57 x 10^-3 / 0.0193 = 0.185 M

Molarity of HBr = 0.185 M