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2 years ago

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The generic metal hydroxide M(OH)2 has Ksp = 8.05×10−12. (NOTE: In this particular problem, because of the magnitude of the Ksp

and the stoichiometry of the compound, the contribution of OH− from water can be ignored. However, this may not always be the case.)
a) What is the solubility of M(OH)2 in pure water?
b) What is the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2?

1 answer:

vodka [1.7K]2 years ago

6 0

Answer:

a. in pure water Solubility (x) = 1.26 x 10⁻⁴M

b. in 0.202M M⁺² Solubility (x) = 9.963 x 10⁻¹²M

The large drop in solubility is consistent with the common ion effect.

Explanation:

a. Solubility in pure water

Given: M(OH)₂ ⇄ M⁺² + 2OH⁻

I --- 0 0

C --- x 2x

E --- x 2x

Ksp = [M⁺²][OH⁻]² = (x)(2x)² = 4x³ => x = CubeRt(Ksp/4)

solubility in pure water = x = CubeRt(8.05 x 10⁻¹²/4) = 1.26 x 10⁻⁴M

b. Solubility in presence of 0.202M M⁺² as common ion.

Given: M(OH)₂ ⇄ M⁺² + 2OH⁻

I --- 0.202M 0

C --- +x +2x

E --- 0.202M + x 2x

≈ 0.202M

Ksp = [M⁺²][2x]² = (0.202)(2x)² = (0.202)(4x²) = 8.05 x 10⁻¹²

=> x = (8.05 x 10⁻¹²)/(0.202)(4) = 9.963 x 10⁻¹²M

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The question is incomplete, here is the complete question:

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The image is attached below.

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<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

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For the given images:

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<u>For b:</u>

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$K_c=\frac{(4/1)^2}{(8/1)}\\\\K_c=2$

<u>For c:</u>

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