Answer to Question #310214 in General Chemistry for Thel

Question

If 0.590% solution of Na²CO3 will be used to precipitate Ca+² from solution, determine the amount of solution, in grams, needed to precipitate 5.00×10²mL of 0.01120 M Ca+²?

Answer 1

CaCO_3(s) = Ca²⁺_(aq) + CO₃^2-_(aq)

K_sp = 4.8 × 10^-9

n(Ca²⁺) = 0.01120 mol/l × 5.00×10²×10^-3 l = 0.0056 mol

0.590 g Na₂CO₃ in 100 g solution

n(Na₂CO₃) = m(Na₂CO₃) / Mr(Na₂CO₃) = 0.590 g / 106 g/mol = 0.005566 mol

V(solution) = 100 g / 1 g/ml = 100 ml = 0.1 l

M = 0.005566 mol / 0.1 l = 0.05566 mol/l

K_sp = [Ca²⁺] × [CO₃^2-]

4.8×10^-9 = 0.0056 mol × [CO₃^2-]

[CO₃^2-] = 4.8×10^-9 / 0.0056 = 8.57×10^-7 mol

V(solution) = 8.57×10^-7 mol / 0.05566 mol/l = 1.54×10^-5 l = 0.0154 ml

m(solution) = 0.0154 ml × 1 g/ml = 0.0154 g