Chemistry 30

3.3 The Solubility Product Constant, K_sp

We will now return to an important mathematical relationship we first learned about in our unit on Equilibrium (Unit 3, Section 2-1), the equilibrium constant expression.

Recall that for any general reaction:

aA + bB

cD + dD

an equilibrium constant expression can be defined as:

Keq = [C]c × [D]d [A]a × [B]b

Since saturated solutions are equilibrium systems, we can apply this mathematical relationship to solutions. We will refer to our equilibrium constant as K_sp, where "sp" stands for "solubility product"

For our silver sulfate saturated solution,

Ag2SO4 (s)

2Ag+(aq) + SO42-(aq)

we can write our solubility product constant expression as

Ksp

=

[Ag+]2[SO42-] [Ag2SO4]

But remember from our earlier introduction to the equilibrium constant expression that the concentrations of solids and liquids are NOT included in the expression because while their amounts will change during a reaction, their concentrations will remain constant.

(Unless otherwise indicated, our examples will always involve dissolving a solid to produce aqueous ions.)

Therefore, we will write our solubility product constant expression for our saturated silver sulfate solution as:

K_sp= [Ag⁺]²[SO₄^2-]

Please note that it is VERY IMPORTANT to include the ion charges when writing this equation.

Try this example:

Write the expression for the solubility product constant, K_sp, for Ca₃(PO₄)₂.

Step 1: Begin by writing the balanced equation for the reaction. Remember that polyatomic ions remain together as a unit and do not break apart into separate elements.

Ca3(PO4)2 (s)

3 Ca2+(aq) + 2 PO43-(aq)

Step 2: Write the expression for K_sp:

K_sp= [Ca²⁺]³[PO_4^3-]²

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Solubility Product Tables that give K_sp values for various ionic compounds are available (print this out and keep it handy). Because temperature affects solubility, values are given for specific temperatures (usually 25°C).

Recall what we learned about K_eq

If Keq is very large, the concentration of the products must be much greater than the concentration of the reactants. The reaction essentially "goes to completion", which all - or most of - the reactants being used up to form the products.

If Keq is very small, the concentration of the reactants is much greater than the concentration of the products. The reaction does not occur to any great extent - most of the reactants remain unchanged, and there are few products produced.

When Keq is not very large or very small, then roughly equal amounts of reactants and products are present at equilibrium.

K_sp, which again is just a special case of K_eq, provides us with the same useful information:

A low value of Ksp indicates a substance that has a low solubility (it is generally insoluble); for ionic compounds this means that there will be few ions in solution.

Iron(II) sulfide, FeS, is an example of a low Ksp : Ksp = 4 ×10-19 In a saturated solution of FeS there would be few Fe2+ or S2- ions.

A large value of Ksp indicates a soluble substance; for ionic compounds it tells us that there will be many ions in solution.

An example of a relatively large Ksp would be for lead(II) chloride, PbCl2 which has a Ksp of 1.8 ×10-4 A saturated solution of PbCl2 would have a relatively high concentration of Pb 2+ and Cl - ions.

Calculations involving K_sp

There are several types of problems we can solve:

1.	Calculating Ksp when concentration of a saturated solution is known.
	Example	The concentration of a saturated solution of BaSO4 is 3.90 × 10-5M. Calculate Ksp for barium sulfate at 25°C
	Solution	Always begin problems involving Ksp by writing a balanced equation:
		BaSO4 (s) Ba2+(aq) + SO42- (aq)
		Next, write the Ksp expression:
		Ksp= [Ba2+][SO42-]
		The question provides us with the concentration of the solution, BaSO4 . We need to find the concentration of the individual ions for our equation. Recall from Section 2-5 - since 1 mole of BaSO4 produces 1 mole of Ba2+ and also 1 mole of SO42-, then . . . [BaSO4] = 3.90 × 10-5M [Ba2+] = 3.90 × 10-5M [SO42-] = 3.90 × 10-5M Substitute values into the Ksp expression and solve for the unknown:
		Ksp = [Ba2+][SO42-] = (3.90 × 10-5)(3.90 × 10-5) Ksp = 1.52 × 10-9 Answer
	Example	The concentration of lead ions in a saturated solution of PbI2 at 25°C is 1.3 × 10-3 M. What is its Ksp?
	Solution	Begin problems involving Ksp by writing a balanced equation:
		PbI2 (s) Pb2+(aq) + 2 I-(aq)
		Write the Ksp expression (be careful!):
		Ksp= [Pb2+][I-]2
		Determine the concentration of the ions. Take care to determine the concentration of I- ions: [PbI2] = 1.30 × 10-3M [Pb2+] = 1.30 × 10-3M [I-] = 2 × 1.30 × 10-3 = 2.60 × 10-3
		Substitute values into the Ksp expression and solve for the unknown:
		Ksp = [Pb2+][I-]2   = (1.30 × 10-3)(2.60 × 10-3)2 Watch Exponents!!! = (1.30 × 10-3)(6.76× 10-6) Ksp = 8.79 × 10-9M Answer
2.	Calculating ion concentrations when Ksp is known.
	Example	Ksp for MgCO3 at 25°C is 2.0 × 10-8. What are the ion concentrations in a saturated solution at this temperature?
	Solution	As always, begin with a balanced equation:
		MgCO3(s) Mg2+(aq) + CO32-(aq)
		Write the Ksp expression:
		Ksp= [Mg2+][CO32-]
		For this example, we are given the value for Ksp and need to find the ion concentrations. We will let our unknown ion concentrations equal x. The balanced equation tells us that both Mg2+ and CO32- will have the same concentration!
		Substitute values into the equation and solve for the unknown:
		Ksp = [Mg2+][CO32-] 2.0 × 10-8 = (x) (x) = x2 x2 = 2.0 × 10-8 x = √2.0 × 10-8 find the square root of 2.0 × 10-8 = 1.4× 10-4M x = [Mg2+] = 1.4× 10-4M Answer AND    x = [CO32-] = 1.4× 10-4M
	Example	Ksp for silver chromate, Ag2CrO4 , is 1.1 × 10-12. Calculate the molar concentration of silver chromate in a saturated solution.
	Solution
		Ag2CrO4 2 Ag+(aq)+ CrO42-(aq)
		Write the Ksp expression:
		Ksp= [Ag+]2[CrO42-]
		Read the question carefully. For this question we are given the value of Ksp but need to find the concentration of the Ag2CrO4 solution, not the concentration of the individual ions. But in order to solve our Ksp equation, we first must find the ion concentrations. The balancing coefficients from our equation tell us that the concentration of Ag+ ions will be twice as great as the concentration of CrO42-. The concentration of the Ag2CrO4 solution will equal the concentration of CrO42- ions. Let x = [CrO42-] Then [Ag+] = 2x
		Substitute values into the equation and solve for the unknowns:
		Ksp = [Ag+]2[CrO42-] 1.1 × 10-12 = (2x)2 (x) = 4 x3 Read this carefully! 4 x3 = 1.1 × 10-12 x3 = 2.75 × 10-13 = (cube root) √2.75 × 10-13 = 6.5 × 10-5 Find cube root of 2.75 × 10-13 x = [CrO42-] = 6.5 × 10-5M   2x = [Ag+] = 1.3× 10-4M [Ag2CrO4] = 6.5 × 10-5M FINAL Answer

Complete Assignment 4 before continuing to the next section.

 

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