3.364 g of hydrated barium chloride of BaCL2. xH2O was dissolved in water and made up to a total volume of 250.0 mL. 10.00 mL of this solution required 46.92 mL of 2.530 x 10-2 M silver nitrate for complete reaction. Calculate the value of x in the formula of hydrated barium chloride. given the net ionic equation for precipitation below.

Given:

Mass of hydrated barium chloride = 3.364 g

Total volume of barium chloride V (total) = 250 ml

Volume taken for titration V = 10 ml

Volume of AgNO3 consumed = 46.92 ml

Concentration of AgNO3 = 0.0253 M

To determine:

The value of x i. e. the water of hydration in BaCl2

Explanation:

The net ionic equation is-

Ag⁺ (aq) + Cl⁻ (aq) → AgCl (s)

Based on the reaction stoichiometry: Equal moles of Ag + and Cl - combine to form AgCl

Moles of Ag + consumed = moles of Cl - present

Moles of Ag + = V (AgNO3) * M (AgNO3) = 0.04692 * 0.0253 = 0.00119moles

Moles of Cl - present = 0.00119 moles

Thus, 0.00119 moles of Cl - are present in 10 ml of the solution

Therefore, number of moles of Cl - in 250 ml would be-

= 0.00119 * 250 / 10 = 0.02975 moles of cl-

Now:

2 moles of Cl - are present in 1 mole of BaCl2

Therefore, 0.02975 moles of Cl - correspond to - 0.02975 * 1/2 = 0.01488 moles of BaCl2

Molar mass of BaCl2 = 208.22 g/mol

Thus, mass of BaCl2 = 0.01488 moles * 208.22 g. mol-1 = 3.098 g

Mass of water of hydration = 3.364 - 3.098 = 0.266 g

# moles of water 'x' =.266/18 = 0.015 ≅ 1

Ans: Formula for hydrated barium chloride = BaCl2. 1H2O