Question

Mg(OH ) 2 is a sparingly soluble compound, in this case a base, with a solubility product, K sp , of 5.61× 10 −11 . It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams.

Part C - Calculate how many times more soluble Mg(OH)2 is in pure water

Based on the given value of the Ksp , calculate the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.110 M NaOH solution.

Part D

What is the pH change of a 0.280 M solution of citric acid (pKa=4.77 ) if citrate is added to a concentration of 0.175 M with no change in volume?

Answer 1

Part. C :-

Let solubility of Mg(OH)2 in pure water = S mol/L

Partial Dissociation is :

Mg(OH)2 (s) <----------------------------> Mg²⁺.....................+.................2OH^-

............................................................S mol/L.....................................2 S mol/L ( at equilibrium)

Expression of Ksp is :

K_sp = [Mg²⁺].[OH^-]²

5.61 x 10^-11 = 4S³

S = 2.41 x 10^-4 mol/L

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Let Solubility of Mg(OH)2 in 0.110 M NaOH = S' mol/L

Total concentration of common ion i.e. OH^- is = [OH^-]_Total = (2S' + 0.110 ) M

Expression of Ksp is :

K_sp = [Mg²⁺].[OH^-]_Total²

5.61 x 10^-11 = S'(2S'+0.110)

2S'<<<0.110, so neglect 2S'

We have,

5.61 x 10^-11 = S'(0.110)

S' = 5.1 x 10^-10 mol/L

So,

S/S' = 2.41 x 10^-4 mol/L/5.1 x 10^-10 mol/L = 4.73 x 10⁵

So,

S = 4.73 x 10⁵ times of S'

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Part.D :-

According to Henderson-Hasselbalch equation, we have

pH = pKa + log [Citrate]/[Citric acid]

pH = 4.77 + log 0.175 M / 0.280 M

pH = 4.77 - 0.204

pH = 4.57