Question

15.25 mL of 0.421 M of La(CH₃CO₂)₃ is combined with 85.25 mL of 0.536 M NaX (X= a halide, i.e. F^-, Cl^-, Br^-, or I^-; Ksp LaX₃ = 0.00074). Calculate Q, and determine if a precipitate will form.

Answer 1

Total volume of solution = 15.25 mL + 85.25 mL = 100.5 mL = 0.1005 L.

La(CH₃CO₂)₃ ------> La³⁺ + 3 CH₃CO₂^-

So, the number of moles of La(CH₃CO₂)₃ is equal to the number of moles of La³⁺ .

Moles of La³⁺ = 15.25 mL x 0.421 M = 6.42 mmol = 0.00642 mol

Concentration of La³⁺, [La³⁺ ] = moles of La³⁺ / total volume = 0.00642 mol / 0.1005 L = 0.0639 M

Nax   -----> Na⁺ + X^-

So, the number of moles of X^- equal to the number of moles of NaX.

Moles of X^- = 85.25 mL x 0.536 M = 45.7 mmol = 0.0457 mol

[X^-] = 0.0457 mol/0.1005 L = 0.455 M

LaX₃(s) <-------> La³⁺ (aq) + 3X^- (aq)

Reaction quotient (Q) for the above equilibrium is,

Q = [La³⁺][X^-]³ = (0.0639)x(0.455)³ = 0.00602

Since Q is greater than K_sp , precipitate will form.