### Video Transcript

Nickel(II) phosphate, Ni3(PO4)2, is
only slightly soluble, dissolving to form multiple nickel and phosphate ions. In a laboratory, the solubility of
nickel(II) phosphate is found to be 2.13 times 10 to the negative seven moles per
decimeter cubed. What is the solubility product of
nickel(II) phosphate? Give your answer in scientific
notation to two decimal places.

We often think of substances as
being either insoluble or soluble in water. However, even substances that we
think of as being insoluble may still be soluble, just in very small amounts. So this is a more realistic and
accurate way to think about solubility. Nickel(II) phosphate is only
slightly soluble.

We can express the dissolution of a
slightly soluble metal salt, let’s call it MA, as being in equilibrium with its
component ions according to this equation. M is a metal, and it forms a cation
in solution. A is the nonmetal component. It could be one element or several
elements in a polyatomic ion. This portion forms an anion in
solution.

For this general dissolution
equation of a metal salt, we can write a corresponding equation for the solubility
product constant 𝐾 sp. 𝐾 sp is equal to the
multiplication product of the molar concentrations of the ions in the solution. However, what happens if our solid
does not dissolve into a one-to-one molar ratio of ions? In this case, the concentration of
each ion is raised to the power of its respective stoichiometric coefficient.

Now that we know how to write the
𝐾 sp expression for any general metal salt, let’s look at the metal salt in
question, nickel(II) phosphate. The formula for the metal salt is
Ni3(PO4)2. So let’s substitute this for
MA. The metal ion is Ni2+. Let’s substitute. For every one formula unit of the
metal salt that dissolves, three nickel cations will be formed in solution. So we substitute three for small
m. The anion will be PO4 3‒, the
phosphate anion. And we can substitute this as the
anion. And for every formula unit of the
salt that dissolved, two anions go into solution. We can substitute two for small
a.

We get this solubility product
expression. Now that we have the solubility
expression, we can solve this question. We are told the solubility of
nickel(II) phosphate, in other words, the concentration of this metal salt which
dissolves. It is 2.13 times 10 to the negative
seven moles per decimeter cubed. And we know that for every one mole
of a salt which dissolves, three nickel ions and two phosphate ions go into
solution. So the concentration of nickel ions
in solution at equilibrium must be three times this concentration value. We can put this into the
expression. And the concentration of phosphate
ions which go into solution is twice the concentration of the metal salt which
dissolves. And we can put this into the
equation too.

Now we can solve. And we get a value of 4.74 times 10
to the negative 32 for 𝐾 sp. We were asked to give the answer in
scientific notation to two decimal places. We have expressed the answer
accordingly. But what about the units? Often, 𝐾 sp values are expressed
without units. If we choose to include the units,
we must remember to multiply the unit by the sum of the powers, which is five. This gives a final unit of moles to
the power five decimeters to the power negative 15.

Finally, what is the solubility
product of nickel(II) phosphate? The answer is 4.74 multiplied by 10
to the negative 32 moles to the five decimeters to the negative 15.