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.

[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