Answer:
[K⁺] = 0.107 M
[OH⁻] = 1.13 × 10⁻⁹ M
Explanation:
600 mL of 0.45 M Cu(NO3)2 gives equal mole of Cu²⁺ and (NO₃)²⁻
⇒ 0.45 × 600 × 10⁻³
= 0.27 moles of Cu²⁺ and (NO₃)²⁻
450 mL of 0.25 M KOH gives equal moles of K⁺ and OH⁻
⇒ 0.25 × 450 × 10⁻³
= 0.1125 moles of K⁺ and OH⁻
Now after mixing 0.1125 moles of OH⁻ precipitates 0.05625 moles of Cu²⁺ (because 1 Cu²⁺ needs 2 OH⁻)
Therefore , moles of remaining Cu²⁺ = 0.27 - 0.05625
=0.21375 moles which is equal to :
⇒ 0.21375/(( 600+450))× 10⁻³
= 0.21375/1050 × 10⁻³
= 0.20357 M
Given that :
(Ksp for Cu(OH)2 is 2.6 × 10⁻¹⁹)
We know that , Ksp = [Cu²⁺][OH⁻]²
2.6 × 10⁻¹⁹ = 0.20357 × [OH⁻]²
[OH⁻]² = 2.6 × 10⁻¹⁹/0.20357
[OH⁻] = 1.13 × 10⁻⁹ M
[K⁺] = moles of K⁺ /total volume
[K⁺] = 0.1125 / 1050 × 10⁻³
[K⁺] = 0.107 M
