Question

The solubility of Mg3(PO4)2 in 's' mol L - 1 The solubility product is given by the relation
(A) S^5
(B) 12S ^3
(C) 6S ^ 5
(D) 108 S^5

with explanation wrong answers are reported​

Answer 1

Answer:

(d)

Explanation:

Hope this will help you.

Answer 2

Answer:

The given reaction's solubility product, $K_{sp}$, calculated is $108s^{5}$.

Therefore, option d) $108s^{5}$ is correct.

Explanation:

Given data,

The given solubility of $Mg_{3}(PO_{4})_{2}$ = s mol/L

The solubility product, $K_{sp}$ =?

The reaction of dissociation of $Mg_{3}(PO_{4})_{2}$:

• $Mg_{3}(PO_{4})_{2} \rightarrow 3Mg^{2+} + 2PO_{4} ^{3-}$
•     s                    3s            2s

As we know,

• The solubility product is an equilibrium constant for the dissociation of the compounds.

The solubility product of the above reaction is given by:

• $K_{sp}$ = $[Mg^{2+}]^{3} \times [PO_{4} ^{3-}]^{2}$
• $K_{sp}$ = $[3s]^{3} \times [2s]^{2}$
• $K_{sp}$ = $27s^{3} \times 4s^{2}$
• $K_{sp}$ = $108 ( s^{3}\times s^{2})$             ( $a^{m} \times a^{n} = a^{m+n}$ )
• $K_{sp}$ = $108 s^{5}$

Hence, the solubility product of the given reaction, $K_{sp}$, calculated is $108s^{5}$.