Misc Solubility Notes

Solubility notes:

Parts per million (ppm): This is a unit of concentration used for very dilute solutions - such as pollutant concentrations or the concentrations of solutions of insoluble salts. It's defined like this:

Remember that for pure water (and very dilute solutions in water), 1 g of solution is equivalent to 1 mL. The density of water at room temperature is approximately one gram per milliliter.

Let's make this unit a little simpler to remember by expressing it in terms of milligrams rather than grams.

So, for dilute aqueous solutions, parts per million is equivalent to milligrams per liter.

Solubility calculations:

For Ca₃(PO₄)₂, K_sp = 1x10^-26. Find the solubility of this salt in mg/L (ppm).

To find the solubility in milligrams per liter (ppm), write a solubility equilibrium and a solubility product constant expression:

Ca₃(PO₄)₂(s) <--> 3Ca²⁺(aq) + 2PO₄^3-(aq)

Note that we can solve this if we express the calcium and phosphate ion concentrations in terms of one variable. If we let x equal the amount of calcium phosphate that dissolves, then:

• [Ca²⁺] = 3x

• [PO₄^3-] = 2x

and

x = 2.4735x10 ^-6 = moles of Ca ₃ (PO ₄ ) ₂ per liter of solution.

Now all we need to do is convert this number of moles to milligrams:

So, the solubility of Ca ₃ (PO ₄ ) ₂ is 0.763 mg/L .

For AgCl, K_sp = 1.8x10^-10. Find the solubility of this salt in mg/L (ppm).

To find the solubility in milligrams per liter (ppm), write a solubility equilibrium and a solubility product constant expression:

AgCl(s) <--> Ag+(aq) + Cl-(aq)

Note that we can solve this if we express the silver and chloride ion concentrations in terms of one variable. If we let x equal the amount of silver chloride that dissolves, then:

• [Ag⁺] = x

• [Cl^-] = x

and

x = 1.34x10 ^-5 = moles of AgCl per liter of solution.

Now all we need to do is convert this number of moles to milligrams:

So, the solubility of AgCl is 1.92 mg/L .

The solubility of SrF₂ in water is 112 mg/L. Find the K_sp of this salt to two significant digits.

First, we write a solubility equilibrium and a solubility product constant expression:

SrF₂(s) <--> Sr²⁺(aq) + 2F^-(aq)

We then express the strontium and fluoride ion concentrations in terms of one variable as if we were going to solve for their concentrations. If we let x equal the amount of strontium fluoride that dissolves, then:

• [Sr²⁺] = x

• [F^-] = 2x

and

Now, we remember that we defined x as the amount of SrF₂ that dissolved - which we know from the problem statement. All we have to do is to convert mg/L to molarity.

So, x = 2.643x10^-5 M. Now, we can find K_sp from the expression we wrote before:

The K_sp of SrF₂ is 2.8x10^-9.

All original site content ©2008 Charles Taylor. Page updated: April 21, 2008.