# Problem: The solubility of zinc(II) phosphate, Zn3(PO4)2, in pure water is 1.5×10–7 moles per liter. Calculate the value of Ksp for zinc(II) phosphate from this data. a. 2.3×10–14b. 5.1×10–28c. 2.7×10–33d. 8.2×10–33e. 7.6×10–35

###### FREE Expert Solution

Recall that Ksp is an equilibrium constant related to slightly soluble ionic compounds

The dissociation of slightly soluble ionic compounds in solution is as follows:

MnNm(s)  n Mm+(aq) + m Nn–(aq)

The Ksp expression then is:

$\overline{){{\mathbf{K}}}_{{\mathbf{sp}}}{\mathbf{=}}\frac{\mathbf{products}}{\overline{)\mathbf{reactants}}}{\mathbf{=}}{\mathbf{\left[}}{\mathbf{nx}}{{\mathbf{\right]}}}^{{\mathbf{n}}}{\mathbf{\left[}}{\mathbf{mx}}{{\mathbf{\right]}}}^{{\mathbf{m}}}}$

where x = molar solubility of the compound.

Solve Ksp for each:

Zn3(PO4)2; molar solubility = 1.5×10–7

MX(s)  M+(aq) +X(aq)

96% (47 ratings) ###### Problem Details

The solubility of zinc(II) phosphate, Zn3(PO4)2, in pure water is 1.5×10–7 moles per liter. Calculate the value of Ksp for zinc(II) phosphate from this data.

a. 2.3×10–14

b. 5.1×10–28

c. 2.7×10–33

d. 8.2×10–33

e. 7.6×10–35