Question

1) What volume of a 0.370 M hydroiodic acid solution is required to neutralize 22.0 mL of a 0.171 M potassium hydroxide solution?

2) An aqueous solution of hydroiodic acid is standardized by titration with a 0.171 M solution of calcium hydroxide.

If 22.0 mL of base are required to neutralize 28.0 mL of the acid, what is the molarity of the hydroiodic acid solution?

Answer 1

HCl (aq) + KOH (aq) -----------> KCl (aq) + H2O (l)

From the equation n1 = n2 balcing coefficients of HCl and KOH

M1V1 = M1V2

M1 = molarity of HCl solution

V1 = volume of HCl solution

M2 = Molarity of KOH solution

V2 = volume of KOH solution

0.37 * V1 = 0.171 * 22

V1 = 10.1676 mL

2 HCl (aq) + Ca(OH)2 (aq) ---------------> CaCl2 (aq) + 2 H2O (l)

Here For HCl n1 = 2 and for Ca(OH)2 n2 = 1, balancing coefficients

M1V1/n1 = M2V2.n2

M1 * 28/2 = 0.171*22/1

M1 = 0.2687 mol/L