Question

Calculate the solubility of Ca₃(PO4)₂ (K_sp = 1.3 x 10^-32) in a 0.083 M Ca(NO₃)₂ solution.

A. 2.4 x 10^-15 mol/L

B. 2.0 x 10^-16 mol/L

C. 4.8 x 10^-15 mol/L

D. 5.7 x 10^-30 mol/L

Answer 1

HAVE BEEN DOING SAME MODEL FROM SO LONG PLEASE PARDON

This is solved for 0.063M substitute ur value to get the perfect answer...

Ca3(PO4)2 (s) <==> 3 Ca++ (aq) + 2 PO4--- (aq)

Ksp = [Ca++]³ [SO4---]²

When x moles of Ca3(PO4)2 dissolve in a liter of water, [Ca++] = 3x and [SO4---] = 2x, so

Ksp = (3x)³ (2x)².

Since initially [Ca++] = 0.063 M, the equation becomes

Ksp = 1.3×10^–32 = (0.063 + 3x)³*4x².

This would be difficult to solve for x, the molar solubility, except we can make the assumption that because Ksp is so small, then x is very small. Therefore 0.063 >> x, and thus 0.063 + 3x ≈ 0.063. The equation becomes:

1.3×10^–32 = 0.063*4x².

The equation is easily solved to x = 2.3x10^-16 M.