Can you please help me with part C ± The Common-Ion Effect Part A - Calculate th

Question

Can you please help me with part C

± The Common-Ion Effect

Part A - Calculate the molar solubility in water

Mg(OH)2 is a sparingly soluble compound, in this case a base, with a solubility product, Ksp, of 5.61×1011. 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 Ksp, what is the molar solubility of Mg(OH)2 in pure H2O?

Express your answer with the appropriate units.

2.41×104 M

The common-ion effect and solubility

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.

Part B - Calculate the molar solubility in NaOH

Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in 0.100 M NaOH?

Express your answer with the appropriate units.

5.61×109 M

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 0.100 M NaOH solution.

Express your answer numerically to three significant figures.

=???

2.41×104 M

Explanation / Answer

Part C- Answer :-

Magnesium hydroxide is actually a base.

It,s KSp walue is 5. 61x10^-11

Mg(OH)2 solubility = 5.61x10^-9M in NaOH

Now we want to find the solubility in plaine water, use ICE method

In plain water.....
Mg(OH)2(s) <==> Mg2+ + 2OH- .......... Ksp = 5.61x10^-11
...........................0 .......... 0 ........ initial
.......................... +x ..... +2x ....... change
...........................x ......... 2x ....... equilibrium

Ksp = [Mg2+] [OH-]²
5.61x10^-11 = 4x³
x = 2.31x10^-4

The solubility of Mg(OH)2 in plaine water is 2.31x10^-4M

Now we want to find the solubility in NaOH, use ICE method

Mg(OH)2(s) <==> Mg2+ + 2OH- .......... Ksp = 5.61x10^-11
...........................     0 ......0.100M ........ initial
..........................    +x ..... +2x ............. change
...........................   x ....... 0.100+2x ..... equilibrium

Ksp = [Mg2+] [OH-]²
5.61x10^-11 = x (0.100+2x)²
x = 5.61x10^-9

Ratio = 2.31x10^-4M / 5.61x10^-9M

         = 41,200

So , Mg(OH)2 is over 41 thousand times more soluble in water than in 0.1M NaOH

Hope this will help u......

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