Answer to Question #161940 in Chemistry for Chinny

Question

A 0.2025g sample consisting of only BaCl2 and KCl required 20.25ml of 0.1200M AgNO3 solution for the quantitative precipitation of chloride. Calculate the %Ba and %K in the sample

Answer 1

Solution:

let x = mass of KCl, y = mass of BaCl₂

Thus: x + y = 0.2025 g; → x = 0.2025 - y

The molar mass of BaCl₂ is 208.23 g mol^-1

The molar mass of KCl is 74.55 g mol^-1

Moles of AgNO₃ = Molarity of AgNO₃ × Solution volume

Moles of AgNO₃ = 0.1200 M × 0.02025 L = 0.00243 mol

Moles of AgNO₃ = 0.00243 mol

Balanced chemical equations:

KCl + AgNO₃ → AgCl + KNO₃

BaCl₂ + 2AgNO₃ → 2AgCl + Ba(NO₃)₂

According to these equations above:

Moles AgNO₃ from KCl = (x g KCl × 1 mol AgNO₃) / (74.55 g mol^-1 KCl) = 0.01341x mol AgNO₃

Moles AgNO₃ from BaCl₂ = (y g BaCl₂ × 2 mol AgNO₃) / (208.23 g mol^-1 BaCl₂l) = 0.00960y mol AgNO₃

0.01341x + 0.00960y = 0.00243

0.01341 × (0.2025 - y) + 0.00960y = 0.00243

0.002716 - 0.01341y + 0.00960y = 0.00243

-0.0038y = -0.00027

y = 0.07506 g - mass of BaCl₂

x = 0.2025 - y = 0.2025 - 0.07506 = 0.12744

x = 0.12744 g - mass of KCl

Moles of BaCl₂ = Mass of BaCl₂ / Molar mass of BaCl₂

Moles of of BaCl₂ = 0.07506 g / 208.23 g mol^-1 = 0.0003605 mol BaCl₂

BaCl₂ → Ba²⁺ + 2Cl^-

Hence,

Moles of Ba²⁺ = Moles of BaCl₂ = 0.0003605 mol

Mass of Ba²⁺ = Moles of Ba²⁺ × Molar mass of Ba²⁺

The molar mass of Ba is 137.33 g mol^-1

Hence,

Mass of Ba²⁺ = 0.0003605 mol × 137.33 g mol^-1 = 0.04951 g

%Ba = (0.04951 g / 0.2025 g) × 100% = 24.45%

%Ba = 24.45%

Moles of KCl = Mass of KCl / Molar mass of KCl

Moles of of KCl = 0.12744 g / 74.55 g mol^-1 = 0.001709 mol KCl

KCl → K⁺ + Cl^-

Hence,

Moles of K⁺ = Moles of KCl = 0.001709 mol

Mass of K⁺ = Moles of K⁺ × Molar mass of K⁺

The molar mass of K is 39.098 g mol^-1

Hence,

Mass of K⁺ = 0.001709 mo × 39.098 g mol^-1 = 0.06682 g

%K = (0.06682 g / 0.2025 g) × 100% = 32.99% = 33.00%

%K = 33.00%

Answer:

%Ba = 24.45%;

%K = 33.00%.