Question Video: Calculating the Solubility Product of a Salt Given Its Solubility | Nagwa Question Video: Calculating the Solubility Product of a Salt Given Its Solubility | Nagwa

# Question Video: Calculating the Solubility Product of a Salt Given Its Solubility Chemistry • Third Year of Secondary School

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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 × 10⁻⁷ mol⋅dm⁻³. What is the solubility product of nickel(II) phosphate? Give your answer in scientific notation to 2 decimal places.

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### 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.

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