Consider the dissolution of AB(s): AB(s)⇌A+(aq)+B−(aq) Le Châtelier's principle tells us that an increase in either [A+] or [B−] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A+ or B− ions. This is an example of the common-ion effect. The generic metal hydroxide M(OH)2 has Ksp = 8.45×10−12. (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.) Part A What is the solubility of M(OH)2 in pure water? Express your answer with the appropriate units. View Available Hint(s) nothing nothing Part B What is the solubility of M(OH)2 in a 0.202 M solution of M(NO3)2? Express your answer with the appropriate units. View Available Hint(s) nothing nothing

A. 0.000128 M is the solubility of M(OH)2 in pure water.

B. is the solubility of in a 0.202 M solution of .

Explanation:

A

Solubility product of generic metal hydroxide =

                      S         2S

The expression of a solubility product is given by :

Solving for S:

0.000128 M is the solubility of M(OH)2 in pure water

B

Concentration of = 0.202 M

Solubility product of generic metal hydroxide =

                   S          2S

So,

The expression of a solubility product is given by :

Solving for S:

is the solubility of in a 0.202 M solution of .

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explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.