Question

1.24 grams of magnesium phosphate tribasic dissolved in 1 L of lemon juice. What is the Ksp of the magnesium phosphate tribasic in lemon juice at room temperature?
a. 2.52e^-10
b. 7.56e^-9
c. 1.26e^-11
d. 6.42e^-10

Answer 1

Answer: The solubility product of magnesium phosphate tribasic is $2.52\times 10^{-10}$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

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:

$\text{Molarity of solution}=\frac{1.24g}{262.85g/mol\times 1}\\\\\text{Molarity of solution}=4.72\times 10^{-3}M$

The equation for the ionization of the magnesium phosphate is given as:

$Mg_3(PO_4)_2\leftrightharpoons 3Mg^{2+}+2PO_4^{3-}$

Expression for the solubility product of $Mg_3(PO_4)_2$ will be:

$K_{sp}=[Mg^{2+}]^3[PO_4^{3-}]^2$

We are given:

$[Mg^{2+}]=(3\times 4.72\times 10^{-3})=1.416\times 10^{-2}M$

$[PO_4^{3-}]=(2\times 4.72\times 10^{-3})=9.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(1.416\times 10^{-2})^3\times (9.44\times 10^{-3})^2=2.52\times 10^{-10}$

Hence, the solubility product of magnesium phosphate tribasic is $2.52\times 10^{-10}$