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What is the solubility of M (OH)2 in a 0.202 solution of M(NO3)2 ? ksp =…

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Question “What is the solubility of M (OH)2 in a 0.202 solution of M(NO3)2 ? ksp =…”

What is the solubility of M (OH)2 in a 0.202 solution of M(NO3)2
?
ksp = 9.05*10^-18

Answer

Concepts and Reason

Molar solubility is the ability to dissolve a small number of moles in a solution to create saturated solution. This is called molar solubility.

Fundamentals

The molar solubility value can be used to calculate the solubility property. The compound’s molar solubility is determined as the sum of its dissociated compounds.

Example:


A, B AB+B -X X X K. =[AB][B] =(x)(x) {:x is molar solubility} =x


You can write the

as follows:

Example:


AgCl2 Ag* + CI Ksp =[Ag* ][cr]


Given that

м(OH),

solution is present in the

0.202 M

of

M(NO3)2



M(NO), A 0.202 M M + + 2NO, 0.202M 0.202M



is the common ion found in both solutions.

ion concentration will be before #media_tag_10$ dissociation.



M(OH), A Initial: Change: Equilibrium: M2+ + 2OH 0.202 0.0 +s +2s 0.202+s +2s



Below is the calculation of the solubility product.



Ksp = [M[ou] 9.05*10*1 = (0.202+s)(25) since s<<0.202, soscan be neglected 9.05x10-18 = 0.808 s?





9.05x10-18 SV0.808 S = 3.35x100M



Therefore, the Molar solubility

M(OH)2

is

3.35x10M

Ans:

Molar solubility of

M(OH)2

is

3.35x10°M

Conclusion

Above is the solution for “What is the solubility of M (OH)2 in a 0.202 solution of M(NO3)2 ? ksp =…“. We hope that you find a good answer and gain the knowledge about this topic of science.


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