Question

Using the appropriate Ksp values, find the concentration of K+ ions in the solution at equilibrium after 600 mL of 0.45 M aqueous Cu(NO3)2 solution has been mixed with 450 mL of 0.25 M aqueous KOH solution. (Enter in M.) (Ksp for Cu(OH)2 is 2.6x10-19).
Now find the concentration of OH? ions in this solution at equilibrium.

Answer 1

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