# 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

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Ksp concept. You can view video lessons to learn Ksp. Or if you need more Ksp practice, you can also practice Ksp practice problems.