1 Ksp This is a special equilibrium constant known as the “ Product Constant”.

Ksp is used to relate the relative solubility of “insoluble” salts.

Generally, a salt solubility of 0.1 M or above is considered soluble.

How to Determine Solubility from K sp

Example: Calculate the molar solubility of Fe(OH)3 if its -39 Ksp = 1.6 x 10 .

Step 1 Write a balanced equation for the slight of the salt in water.

3  FeOH()3  Fe 3 OH

Step 2 Write the equilibrium expression.

3 3   KFeOHsp    Step 3 Make an ICE box

3  FeOH( )3  Fe  3 OH Solid 0 0 -X +X + 3 X X 3 X

2 Step 4 Substitute & Solve

3 3   KFeOHsp    3 1.6 1039 X  3X 

1.6 1039 27 X 4

X 8.8 1011

Step 5 Interpret the results and express as the molarity of the saturated solution.

+3 Since 1 mole Fe(OH)3  1 mole Fe

AND

Since [Fe+3] = 8.8 x 10-11 M

-11 Then; Fe(OH)3 solubility is 8.8 x 10 M

SOMETHING EXTRA

If we also wanted (or needed) to know the total ion concentration of this saturated salt, then we would do this:

(8.8 x 10-11) + (3)(8.8 x 10-11) = 3.52 x 10-10 M ions in solution

3 How to Determine K sp from Solubility

Example 1: If the concentration of a saturated CaSO4 solution is measured to be 0.955 g/L at 25°C, calculate its Ksp .

Step 1 Calculate the molarity if it is not given.

0.955g CaSO1mol CaSO 447.02 103M CaSO 1L 136.04gCaSO 4 4

Step 2 Write a balanced equation for the slight dissociation of the salt in water.

22 CaSO4  Ca SO4

Step 3 Write the equilibrium expression.

 22  KCaSOsp   4  Step 4 Identify the mole ratio in the balanced equation and use it to relate the number of FORMULA UNITS to the number of cations or anions. (Or make an ICE BOX if you need to)

2 In this reaction the ratio of CaSO4 : Ca is 1:1 +2 so [CaSO4] = [Ca ]; therefore the molar solubility of +2 CaSO4 = [Ca ]

4 Step 5 Substitute and solve. 2   KCaSOsp   4 

3 2 Ksp  7.02 10 M 5 Ksp 4.93 10

Example 2: If the concentration of a saturated Ag2SO4 solution is measured to be 4.49 g/L at 25°C, calculate its Ksp .

Step 1 Calculate the molarity if it is not given.

4.49gAgSO 1mol AgSO 24 24 1.44 102M Ag SO 1L 311.80gAgSO 24 24

Step 2 Write a balanced equation for the slight dissociation of the salt in water.

 2 Ag24 SO  2 AgSO 4

Step 3 Write the equilibrium expression.

2 KAgSO   2  sp  4  Step 4 Identify the mole ratio in the balanced equation and use it to relate the number of FORMULA UNITS to the number of cations or anions. (Or make an ICE BOX if you need to)

5

 2 Ag24 SO  2 AgSO  4 Solid 0 0 -X + 2 X +X 2 X X

Step 5 Substitute and solve. 2 KAgSO  2 sp 4

222 Ksp  2 1.44 10M 1.44 10 M 5 Ksp 1.19 10

s Can you compare salt Ksp to each other?

Well, let’s use the last two examples to see if we can.

CaSO4 vs Ag2SO4 Ag2SO4 is 4.1 -5 -5 Ksp 4.93x10 1.19x10 times smaller

Ag2SO4 is 2.0 -3 -2 Molar solubility 7.02x10 1.44x10 times bigger

Ag2SO4 is 4.7 Mass Solubility 0.955 g/L 4.49 g/L times bigger

In conclusion, you CAN’T compare Ksp values unless the formulas are the same in their RATIO of ions.

6 Using K sp

Predicting Precipitation from Solution

The basic premise is: determine the , Qsp and compare to Ksp

Logic Unsaturated Qsp < Ksp No ppt will form Saturated at Equilibrium Qsp = Ksp No ppt will form Saturated AND Beyond Qsp > Ksp ppt WILL form

Example Problem 1

-3 A 200. mL solution of 4.00x10 M BaCl2 is added to a 600. mL -3 solution of 8.00x10 M K2SO4. Assume volumes are additive. -10 Will BaSO4 precipitate? The Ksp of BaSO4 is 1.08x10

Step 1 Write a BALANCED NET IONIC equation for the possible formation of a precipitate.

BaCl224  K SO BaSO 4 2 KCl 2  2   Ba 2 Cl  2K  SO44 BaSO2 K  2Cl

22 Ba SO44 BaSO

7 Step 2 Find the concentration of each ion in the net equation. HINT: Don’t forget to combine volumes.

0.200LM 4.00 103 Ba23  1.00 10 M 0.200LL 0.600

0.600LM 8.00 103 SO23  6.00 10 M 4 0.200LL 0.600

Step 3 Write Qsp expression, substitute and solve.

QBaSO221.00 10  3 M 6.00 10  3 M 6.00 10  6 sp 4   

Step 4 Compare Qsp to Ksp and apply the logic from above.

610 QKsp6.00 10 and sp  1.08 10 QKspsp 

therefore BaSO4 WILL precipitate! Yea!

Example Problem 2

If 2.00 mL of 0.200 M NaOH is mixed with 1000. mL of 0.100M CaCl2, will the Ca(OH)2 that is produced in this reaction form a -6 precipitate? The Ksp for Ca(OH)2 is 5.02 x10

8

Step 1 Write a BALANCED NET IONIC equation for the possible formation of a precipitate.

2 NaOH  CaCl22 Ca ( OH )  2 NaCl 2     Ca 2 Cl  2Na  2()2OH Ca OH2 Na  2Cl

2 Ca2 OH Ca( OH )2

Step 2 Find the concentration of each ion in the net equation. HINT: Don’t forget to combine volumes.

1.00LM 0.100  Ca229.98 10 M 0.002LL 1.000 0.00200LM 0.200  OH3.99 10 4 M 0.00200LL 1.000

Step 3 Write Qsp expression, substitute and solve.

2 2 QCaOH22489.98 10  M 3.99 10  M 1.59 10  sp   

Step 4 Compare Qsp to Ksp and apply the logic from above.

86 QKsp1.59 10 and sp  5.02 10 QKspsp 

therefore Ca(OH)2 will NOT precipitate.

9 Fractional Precipitation (partial precipitation)

This process can be utilized to separate mixtures of ions. Sometimes the ions we want to separate are very similar to each other in their chemical behavior. Consider a mixture of halogen ions like, say, Cl- and Br- . These act very much alike and are very challenging to separate from a mixture. How is it done?

Example Problem 1

A solution contains 2.00 x10-2 M Cl- and 2.00 x10-2 M Br-. Solid AgNO3 is added very slowly while stirring the mixture. The solution volume is 100. mL.

a. What is the [Ag+] when AgBr begins to precipitate? b. What is the [Ag+] when AgCl begins to precipitate? c. What is the [Br-] when AgCl begins to precipitate? d. What mass of AgBr will precipitate before the AgCl begins to precipitate?

Additional information: -10 -13 AgCl Ksp = 1.77 x10 and AgBr Ksp = 5.35 x10

(a) The AgBr will begin to ppt just beyond the conditions + - + where [Ag ][Br ]=Ksp . So, if we want to know the [Ag ] at this point, we simply solve for [Ag+] while substituting the other two values into this expression.

10 Ag  Br  K  sp 213  Ag 2.00 10 M 5.35 10  Ag  2.68 1011M 

(b) The AgCl will begin to ppt just beyond the conditions + - + where [Ag ][Cl ]=Ksp . So, if we want to know the [Ag ] at this point, we do like we did before by solving for [Ag+] while substituting the other two values into this expression.

Ag  Cl  K  sp 210  Ag 2.00 10 M 1.77 10  Ag  8.85 109M 

(c) Since the AgCl begins to ppt when the solution is saturated with AgBr, we can determine the [Br-] by examining things relative to the saturated AgBr solution. So,

Ag  Br  K  sp 913 8.85 10MBr 5.35 10  Br  6.05 105M 

11 (d) We started with (0.100 L)(2.00x10-2M) = 2.00x10-3 mol Br - . We know that (0.100 L)(6.05x10-5M) = 6.05x10-6 mol Br – can stay in solution when the Cl- starts to precipitate. Therefore 2.00x10-3 mol Br- - 6.05x10-6 mol Br- = 1.99x10-3 mol Br- are removed from the solution by precipitating with Ag+.

1.99x10-3 mol Br- 1 mol A g B r 187.8 g A g B r = 0.374g AgBr 1 mol Br - 1 mol AgBr

will precipitate from the solution before the AgCl begins to precipitate.

Solubility of Common Ions

Salts that share a common ion – like NaCl and KCl that both contain Cl- ions – do affect each other’s solubility when dissolved together.

Common ion situations can be put to good use. Sometimes they are utilized to force something out of solution that might otherwise be difficult to remove.

Common ion situations will also be discussed soon when we cover BUFFER solutions in the - topic.

12 Example Problem 1

(a) What is the molar solubility of Fe(OH)3 in a solution that contains 6.5 x10-5 M Fe+3 ? - (b) What is the [OH ] that results when Fe(OH)3 is dissolved as indicated in part (a)?

-39 Data: Ksp of Fe(OH)3 is 1.6 x10

Step 1 Write a balanced equation for the dissociation of the Fe(OH)3 in water and make an ICE box.

3  FeOH( )3  Fe  3 OH I Solid 6.5 x 10-5M 0 C -X + X + 3X E 6.5 x 10-5 +X 3X The common ion here is the Fe+3 and the “initial” solution already contains some Fe+3 dissolved.

Step 2 Write an equilibrium expression, substitute & solve.

3 3 KFeOHsp    3 1.6 1039 6.5 10 5 X  3X

But wait! This is a pretty hard equation to solve. This is beyond a simple quadratic problem.

Is there an easier way?

Yes. 13

Let’s use an approximation.

Since we know X is small, we will do what we have done with other equilibrium problems. Using this approach our equation becomes:

3 KFeOH 3    sp   3 1.6 1039 6.5 10 5  3X

39 27X 3  1.6 10 6.5 105

X  9.7 1013

14 Step 3 The molar solubility of Fe(OH)3 in a solution that contains 6.5 x10-5 M Fe+3 is what we are wanting to find. Now that we know the value of X, we know the amount of Fe+3 that is being added to this solution by the Fe(OH)3. We therefore know that the molar solubility of Fe(OH)3 is X.

If X = solubility of Fe(OH)3 , then;

The molar solubility of Fe(OH)3 in a solution containing 6.5 x10-5 M Fe+3 is 9.7 x10-13 M.

Now, question (b)

Step 4 Since we know the value of X, we can go back to the ICE box and do some .

3  FeOH( )3  Fe  3 OH I Solid 6.5 x 10-5M 0 C + X + 3X E 6.5 x 10-5 + X 3X Substitute (6.5 x 10-5) + (9.7 x 10-13) (3) (9.7 x 10-13) Solve 6.5 x 10-5 2.9 x 10-12

- This tells us that the [OH ] after the Fe(OH)3 is dissolved is 2.9 x10-12M.