Question

Will KClO4 precipitate when 20mL of a 0.050-M solution of K+ is added to 80mL of a 0.50-M solution of ClO4−? (Hint: Use the dilution equation to calculate the concentrations of potassium and perchlorate ions in the mixture.)

Answer 1

Moles of K+ = molarity x volume

= 0.050 mol/L x 20 mL x 1L/1000 mL

= 0.001 mol

Moles of ClO4- = molarity x volume

= 0.50 mol/L x 80 mL x 1L/1000 mL

= 0.04 mol

Total volume V = 20 + 80 = 100 mL x 1L/1000 mL = 0.100 L

Concentration of K+ = moles/V

= 0.001/0.100

= 0.010 M

Concentration of ClO4- = moles/V

= 0.04/0.100

= 0.40 M

Equilibrium constant expression of the reaction

Q = [K+] [ClO4-]

= (0.010) x (0.40)

= 4.0 × 10–3

Ksp = 1.07 × 10–2

Q < Ksp

KClO4 will not precipitate.