Question

18. The molar solubility of M(OH)3 in 0.4 M M(NO3)3 solution interms of solubility product of M(OH)3
1) (Ksp / 10.8)^1/3

2) (Ksp / 3.6)^1/3

3) (Ksp / 10.8)^1/4

4) (Ksp / 0.4)^1/3


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plz solve urgently plz plz plz I will mark you as a brainlist plz I would be very thankful to you
plz answer th right one plz!​

Answer 1

Answer:

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Explanation:

I don't know answer

Answer 2

The molar solubility of M(OH)3 in 0.4 M M(NO3)3 solution in terms of the solubility product of M(OH)3 is equal to  (Ksp / 10.8)^1/3.

• One mole of M(OH)3 will dissociate to produce one mole of M(3+) and three moles of OH(-) ions.
• So, if the solubility is taken to be S, then the concentration of OH(-) is equal to 3S while the concentration of M(3+) will be equal to (S+0.4).
• S is contributed by the M(OH)3 and 0.4 is the contribution from M(NO3)3, since nitrates are completely soluble in water.
• Now, Ksp = [M(3+)].[OH(-)]³ = (3S)³.(S+0.4) = 27S⁴ + 10.8S³ ≈ 10.8S³(since higher powers of S will be neglected as the value of S will be very small).
• Therefore, S =  (Ksp / 10.8)^1/3.