Respuesta :
Answer: The solubility product of magnesium phosphate tribasic is [tex]2.52\times 10^{-10}[/tex]
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
Given mass of magnesium phosphate = 1.24 g
Molar mass of magnesium phosphate = 262.85 g/mol
Volume of solution = 1 L
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{1.24g}{262.85g/mol\times 1}\\\\\text{Molarity of solution}=4.72\times 10^{-3}M[/tex]
The equation for the ionization of the magnesium phosphate is given as:
[tex]Mg_3(PO_4)_2\leftrightharpoons 3Mg^{2+}+2PO_4^{3-}[/tex]
Expression for the solubility product of [tex]Mg_3(PO_4)_2[/tex] will be:
[tex]K_{sp}=[Mg^{2+}]^3[PO_4^{3-}]^2[/tex]
We are given:
[tex][Mg^{2+}]=(3\times 4.72\times 10^{-3})=1.416\times 10^{-2}M[/tex]
[tex][PO_4^{3-}]=(2\times 4.72\times 10^{-3})=9.44\times 10^{-3}M[/tex]
Putting values in above expression, we get:
[tex]K_{sp}=(1.416\times 10^{-2})^3\times (9.44\times 10^{-3})^2=2.52\times 10^{-10}[/tex]
Hence, the solubility product of magnesium phosphate tribasic is [tex]2.52\times 10^{-10}[/tex]
