Solution Stoichiometry: Acid/Base

Question

When aqueous solutions of hydroiodic acid and barium hydroxide are mixed, the following reaction occurs:

2HI + Ba(OH)2 BaI2 + 2 H2O

If 29.1 mL of a 7.25E-2 M Ba(OH)2 solution are required to react with 28.7 mL of HI, what is the molarity of the HI solution?

Answer 1

QUESTION

When aqueous solutions of hydroiodic acid and barium hydroxide are mixed, the following reaction occurs.

2HI_(aq) + Ba (OH)_2(aq) =======> BaI_2(ppt) + 2 H₂O_(l)

If 29.1 mL of a 7.25 x 10² M Ba (OH)_2(aq solution are required to react with 28.7 mL of HI, what is the molarity of the HI solution?

SOLUTION

GIVEN DATA: Volume of Ba (OH)_2(aq) = V₁= 29.1 mL

Volume of HI = V₂ = 28.7 mL

Molarity of Ba (OH)_2(aq) = M₁= 7.25 x 10² M

Molarity of HI = M_{2 = ?}

According to Dilution formula

FORMULA: M₁V₁ = M₂ V₂

M₂ = M₁V₁ / V₂ = 7.25 x 10² M x 29.1 mL / 28.7 mL = 7.35 x 10² M

Thus, the molarity of the HI solution is 7.35 x 10² M

Note: The unit of molarity is mol / L and it is denoted by M.