Answer to Question #276059 in Chemistry for khryst

Question

 How many grams of CaCO₃ (K_sp = 5.0 × 10^–9) is soluble in 500.0 mL of water at 25 °C.

Answer 1

Solution:

CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃^2-(aq)

The K_sp expression for CaCO₃(s) is:

K_sp = [Ca²⁺][CO₃^2-] = 5.0×10^–9

CaCO₃(s) ⇌ Ca²⁺(aq) + CO₃^2-(aq)

S(CaCO₃) = [Ca²⁺] = [CO₃^2-] = S

K_sp = [Ca²⁺][CO₃^2-] = [S] × [S] = S²

Calculate solubility (S) of CaCO₃(s):

S = (K_sp)^0.5 = (5.0×10^–9)^0.5 = 7.07×10^–5

S(CaCO₃) = 7.07×10^–5 mol/L

The molar mass of CaCO₃ is 100.087 g/mol

V = 500.0 mL = 0.5 L

Calculate the mass of CaCO₃:

m(CaCO₃) = S(CaCO₃) × M(CaCO₃) × V

m(CaCO₃) = (7.07×10^–5 mol/L) × (100.087 g/mol) × (0.5 L) = 0.00354 g CaCO₃ = 3.54×10^–3 g CaCO₃

m(CaCO₃) = 3.54×10^–3 g

Answer: 3.54×10^–3 grams of CaCO₃