Calculate the molar solubility in water Mg(OH)_2 is a sparingly soluble compound

Question

Calculate the molar solubility in water Mg(OH)_2 is a sparingly soluble compound, in this case a base, with a solubility product, K_sp, of 5.61 times 10^-11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams. Based on the given value of the K_sp, what is the molar solubility of Mg(OH)_2 in pure H_2O? Express your answer with the appropriate units. molar solubility = Value Units The solubility of a slightly soluble compound can be greatly affected by the addition of a soluble compound with a common ion, that is, with one of the ions in the added soluble compound being identical to one of the ions of the slightly soluble compound. The general result of the addition of the common ion is to greatly reduce the solubility of the slightly soluble compound. In other words, the addition of the common ion results in a shift in the equilibrium of the slightly soluble compound. Calculate the molar solubility in NaOH Based on the given value of the K_sp, what is the molar solubility of Mg(OH)_2 in 0.200 M NaOH? Express your answer with the appropriate units. molar solubility = Value Units

Explanation / Answer

A)

Mg(OH)2 ---> Mg2+ + 2 OH-

s 2s

Ksp = [Mg2+] [OH-]^2

5.61*10^-11= s*(2s)^2

5.61*10^-11 = 4*s^3

s = 2.41*10^-4 M

Answer: 2.41*10^-4 M

B)

IN NaOH, initial [OH-] = 0.200M

Mg(OH)2 ---> Mg2+ + 2 OH-

s 2s+0.200

Ksp = [Mg2+] [OH-]^2

5.61*10^-11= s*(2s+0.2)^2

Since Ksp is small, s will be small and can be ignored in comparison to 0.200

so, above expression becomes,

5.61*10^-11= s*(0.2)^2

s = 1.4*10^-9 M

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