When 1.95 g of co (no3) 2 is dissolved in 0.350 l of 0.220 m koh, what are [ co2+], [ co (oh) 42-], and [ oh-] if kf of co (oh) 42 - = 5.00 * 109?

Mass of Co (NO₃) ₂ = 1.95 g

V KOH = 0.350 L

[KOH] = 0.220 M

Kf = 5.0 x 10⁹

molar mass of Co (NO₃) ₂ = 182.943 g/mol

so [Co (NO₃) ₂] = 1.95 / (0.350 * 182.943) = 0.03045 M

[Co²⁺] = 0.03045 M

[OH⁻] = 0.22 M

chemical reaction:

Co²⁺ (aq) + 4 OH⁻ ⇄ Co (OH) ₄²⁻

I (M) 0.03045 0.22 0

C (M) - 0.03045 - 4 (0.03045) 0.03045

E (M) - x 0.22 - 4 (0.03045) 0.03045

= 0.0982

Kf = [Co (OH) ₄²⁻] / [Co⁺²][OH⁻]⁴

5.0 x 10⁹ = (0.03045) / x (0.0982) ⁴

x = 6.5489 x 10⁻⁸

at equilibrium:

[Co²⁺] = 6.54 x 10⁻⁸

[OH⁻] = 0.0982 M

[Co (OH) ₄²⁻] = 0.03045 M