If K, for Mg-(PO4)2 is 5.2x 104 at 25°C what is the molar solubility of Mg (PO4)2 in 0.25M Na:PO4?

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If K, for Mg-(PO4)2 is 5.2x 104 at 25°C what is the molar solubility of Mg (PO4)2 in 0.25M Na:PO4?
Transcribed Image Text:If K, for Mg-(PO4)2 is 5.2x 104 at 25°C what is the molar solubility of Mg (PO4)2 in 0.25M Na:PO4?
Expert Solution
Step 1

Given: Ksp Mg3(PO4)2 = 5.2 * 10-24

         Concnentration of Na3PO4= 0.25 M

To Calculate: Molar solubility of Mg3(PO4)2 in 0.25 M Na3PO4

 

Step 2

Na3POionize as follows

Na3PO---------> 3 Na+ + PO43- 

So 0.25 M  Na3PO will give 0.25 M PO43-  ions.

 

Mg3(PO4)disociates as follows:

Mg3(PO4)2 <--------->3Mg2+          + 2PO43-  

The solubility product constant, Ksp is the equilibrium constant for a solid substance dissolving in an aqueous solution.

For the given reaction Ksp = [3 Mg2+]3 [2PO43- ]2

                                          Mg3(PO4)2 <--------->3Mg2+          + 2PO43-  

                                                                                3x                   2x+0.25M

Ksp = [3 x]3 [2x+0.25]2

5.2 * 10-24 = [3 x]3 [2x+0.25]

Since Ksp is small, so x will be small, we can neglect 2x in comparison to 0.25

So, 

5.2 * 10-24 = [3 x]3 [0.25]

5.2 * 10-24 = 1.6875 * x3

3.081 * 10-24 =  x

x =1.455 * 10-8  

 

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