The solubility of a slightly soluble compound can be greatly affected by the add

Question

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 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 0 190 M NaOH? Express your answer with the appropriate units.

Explanation / Answer

Ksp of Mg(OH)2 = 1.8 x 10-11 (literature value)

1) Dissociation equation:

MgOH (s) Mg2+ (aq) + 2OH-(aq)

2) Ksp expression:

Ksp = [Mg2+] [OH-]2

3) Let us substitue into the Ksp expression:

1.8 x 10-11 = [Mg2+] [OH-]2

4) Now, we have to reason out the values of the two terms on the right. The problem specifies that [OH-] is already 0.100. OH- get another '2x' amount from the dissolving Mg(OH)2. By the 1:2 stochiometry between magnesium ion and hydroxide ion, the [Mg2+] is 'x.' Substituting, we get:

1.8 x 10-11 = (x) (0.190 + 2x)2

4) The answer (after neglecting the + 2x in 0.190 + 2x):

1.8 x 10-11 = (x) (0.190)2

1.8 x 10-11 = (x) (0.0361)

x = 4.99 x 10-10 M

Molar solubility of Mg(OH)2 in 0.190 M NaOH is 4.99 x 10-10 moles/liter

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