When 7.5 x 10-4 g of KOH is dissolved in 1.00 L of 1.0 x 10-8 M

Question

When 7.5 x 10^-4 g of KOH is dissolved in 1.00 L of 1.0 x 10^-8 M Cu(NO₃)₂, a precipitate of Cu(OH)₂ is formed. True or False? (K_sp of Cu(OH)₂ is 2.2 x 10^-20)

The answer is true

Answer 1

To answer this question, we will calculate Q and compare it to the given value of Ksp. If Q > Ksp, then a precipitate will form. If Q < Ksp, no precipitate will form. But before we can do this, we must first write the balanced equation.

Looking at the reaction in question:

2KOH(aq) + Cu(NO₃)₂(aq) ==> Cu(OH)₂(s) + 2KNO₃(aq)

moles KOH present = 7.5x10^-4 g x 1 mol/56.1 g = 1.33x10^-5 moles KOH = 1.3x10^-5 M OH-

moles Cu(NO₃)₂ present = 1.0x10^-8 mol/L x 1 L = 1.0x10^-8 moles Cu(NO₃)₂ = 1x10^-8 M Cu²⁺

Cu(OH)₂(s) ==> Cu²⁺(aq) + 2OH^-(aq)

Q = [Cu²⁺][OH-]²

Q = (1x10^-8)(1.3x10^-5)²

Q = 1.3x10^-18

Q > Ksp so TRUE, a precipitate will form