Question

Using the dissociation constant, Kd = 3.4 × 10⁻¹⁵, calculate the equilibrium concentrations of Zn²⁺ and OH⁻ in a 0.0465-M solution of Zn(OH)₄²⁻.

a) [Zn²⁺] = 2.04 × 10⁻⁹ M, [OH⁻] = 2.04 × 10⁻⁹ M
b) [Zn²⁺] = 6.8 × 10⁻¹⁰ M, [OH⁻] = 6.8 × 10⁻¹⁰ M
c) [Zn²⁺] = 1.37 × 10⁻⁹ M, [OH⁻] = 1.37 × 10⁻⁹ M
d) [Zn²⁺] = 4.08 × 10⁻¹⁰ M, [OH⁻] = 4.08 × 10⁻¹⁰ M

Answer 1

The equilibrium concentrations of Zn²⁺ and OH⁻ in a 0.0465-M solution of Zn(OH)₄²⁻ are [Zn²⁺] = 2.04 × 10⁻⁹ M and [OH⁻] = 2.04 × 10⁻⁹ M.

Step-by-step explanation:

To calculate the equilibrium concentrations of Zn²⁺ and OH⁻ in a 0.0465-M solution of Zn(OH)₄²⁻, we can use the dissociation constant (Kd).

The dissociation constant is given as Kd = 3.4 × 10⁻¹⁵. The equation for the dissociation of Zn(OH)₄²⁻ in water is Zn(OH)₄²⁻ ⇌ Zn²⁺ + 4OH⁻.

Since Zn(OH)₄²⁻ dissociates into one Zn²⁺ ion and four OH⁻ ions, the equilibrium concentrations would be [Zn²⁺] = 2.04 × 10⁻⁹ M and [OH⁻] = 2.04 × 10⁻⁹ M.