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PrecipitationEvery problem weve looked at so far has been one where we start with a solid, put it in some water and ask

(in one way or another) how much dissolves. The questions in this section are from the opposite perspective.

In this section we are going to check to see if a precipitate will form. To be clear, a precipitate (ppt) is an in-soluble solid formed when two soluble solutions are mixed. How does this fit into our Kspscheme? Remem-

ber that for an ionic substance, Kspmust be put into this format:

MX (s) M+ (aq) + X-(aq)

Where is the solid at in this equation? Clearly it is always on the left-hand side in this Kspsetup. This means

that for a precipitate (ppt) to form, the reaction has to shift to the left side. How does it do that? Remember inthe previous section our discussion of Q the reaction quotient? We determined earlier if:

Q = K System is at equilibrium

Q < K System must shift to the right

Q > K System must shift to the left

So what we are looking at here is performing a test-case for Q, and seeing what the value is relative to the Ksp

equation. If Q > K, the system will shift left and therefore form a precipitate. Lets look at some simple ex-

amples:

Simple Q CalculationsExample 1: Will a precipitate of BaCO3(Ksp= 2.6x10

-9) form if a solution is made so that [Ba+2] = 2x10-5M

and [CO3-2] = 2x10-4M?

Set up your Kspequation like you have this entire time:

BaCO3 (s) Ba+2 (aq) + CO3

-2(aq) Ksp= [Ba+2][CO3

-2] = 2.6x10-9

Now do a test-case for Q Q = [Ba+2][CO3-2]

Q = (2x10-5)(2x10-4)Q = 4x10-9

Now compare Q to Kspand we find that

Q (4x10-9) > Ksp(2.6x10-9) so your final answer is:

YES, a precipitate forms

If Q > Kspthen a precipitate (ppt) of the solid forms

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Example 3: Will a precipitate of Ag2CrO4(Ksp= 9x10-12) form if [Ag+1] = 2x10-4and [CrO4

-2] = 3x10-5?

Ag2CrO4(s) 2 Ag+1 (aq) + CrO4

-2(aq) Ksp= [Ag+1]2[CrO4

-2] = 9x10-12

Initial 2x10-4 3x10-5 Q = (2x10-4)2(3x10-5) = 1.2x10-12

Since Q < Ksp, No a ppt does not form

Example 4: Will a precipitate of Ca3(PO4)2(Ksp= 2x10-29) form if [Ca+2] = 3x10-5and [PO4

-3] = 4x10-6?

Ca3(PO4)2(s) 3 Ca+2 (aq) + PO4

-3(aq) Ksp= [Ca+2]3[PO4

-3]2= 2x10-29

Initial 3x10-5 4x10-6 Q = (3x10-5)3(4x10-6)2= 4.3x10-25

Since Q > Ksp, Yes a ppt forms

Example 5: What concentration of [I-1] will cause a precipitate of PbI2in a 4x10-5M solution of Pb(NO3)2?

KspPbI2= 9.8x10-9-

PbI2 (s) Pb+2(aq) + 2 I-1(aq) Ksp= [Pb

+2][I-1]2= 9.8x10-9

9.8x10-9= (4x10-5) ([I-1])2

[I-1] = 1.57x10-2M

Any [I-1] concentration greater than 1.57x10-

2will cause Q > Kspand cause a ppt.

Example 2: Will a precipitate of MgF2(Ksp= 5.2x10-11) form if [Mg+2] = 4x10-4and [F-1] = 2x10-3?

MgF2 (s) Mg+2(aq) + 2 F-1(aq) Ksp= [Mg

+2][F-1]2= 5.2x10-11

Q = (4x10-4)(2x10-3)2= 1.6x10-9

Since Q > Ksp, Yes a ppt forms

Please note that even though there is a 2 here in the equation, the amount of the F-1is NOT initially doubled.

Why is that? The 2 in the above formula is only taken into account when dealing with stoichiometric com-parisons. These only take place in the changeportion of an ICE table. The amounts we are using are all initial

amounts:

MgF2(s) Mg+2(aq) + 2 F-1(aq) Ksp= [Mg

+2][F-1]2= 5.2x10-11Initial 4x10-4 2x10-3

Change +x +2x

We dont evengetto this part of the table in these problems. If we did, then the 2 would be taken into ac-count here at the change line. Precipitation problems dont even get past the initial line so that 2 is not im-

portant.

Dont change the concentration of a given species becausethere is a stoichiometric coefficient in the equation

Note again that we are notchanging the

concentration of [Ag+1] ions here just be-

cause there is a 2 in the equation

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Compounds with Multiple Ions

Example 6: Will a precipitate of AgBr (Ksp= 5.4x10-13) form if solutions are mixed so that [AgNO3] = 3x10

-5and [CaBr2] = 2x10

-4?

The important thing to note here is that we are examine the Kspequation of AgBr so we have to exam-ine each compound given to us to look for the part we are interested in to solve the problem. Examine:

[AgNO3] = 3x10-5 What part of this compound are we interested in? The Ag+1part of it. How

many Ag+1ions are there in this compound? There is only one so we are leftwith:

[Ag+1] = 3x10-5 M

But

[CaBr2] = 2x10-4 What part of this compound are we interested in? The Br-1part of it. How

many Br-1ions are there in this compound? There are two Br-1ions so:[Br-1] = 2(2x10-4)

[Br-1] = 4x10-4M

AgBr(s) Ag+1(aq) + Br-1(aq) Ksp= [Ag

+1][Br-1] = 5.4x10-13

Initial 3x10-5 4x10-4 Q = (3x10-5)(4x10-4) = 1.2x10-8

Since Q > Ksp, Yes a ppt forms

Example 7: Will a ppt of Cu(OH)2(Ksp= 4.8x10-20), form if solutions are mixed so that [CuSO4] = 5x10

-6and[Ba(OH)2] = 3x10

-8?

[CuSO4] = 5x10-6 [Cu

+2] = 5x10-

6 (Note there is 1 Cu+2for every CuSO4)

[Ba(OH)2] = 3x10-8

[OH-1] = 6x10-8 (Note there are 2 OH-1for every Ba(OH)2)

Cu(OH)2 (s) Cu+2(aq) + 2 OH-1(aq) Ksp= [Cu

+2][OH-1]2= 4.8x10-20

Initial 5x10-6 6x10-8 Q = (5x10-6)(6x10-8)2= 1.8x10-20

Since Q < Ksp= There is no ppt

Example 8: Will a ppt of MgF2(Ksp= 5.2x10-11), form if solutions are mixed so that [MgCl2] = 3x10

-4and

[AlF3] = 7x10-3?

[MgCl2] = 3x10-4 [Mg

+2] = 3x10-

4 (Note there is 1 Mg+2for every MgCl2)

[AlF3] = 7x10-3

[F-1] = 2.1x10-2 (Note there are 3 F-1for every AlF3)

MgF2 (s) Mg+2(aq) + 2 F-1(aq) Ksp= [Mg

+2][F-1]2= 5.2x10-11

Initial 3x10-4 2.1x10-2 Q = (3x10-4)(2.1x10-2)2= 1.3x10-7

Since Q > Ksp= Yes, there is a ppt

If a compound has more than one of an ionic species

you are interested in for the Kspequation,

be sure to take that number into account for the initial concentration of that species

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Combining Solutions

The last wrinkle you may see in a problem involving precipitation is when the problem asks you to combine

multiple solutions together. The reason this gets a little tricky is because when you combine solutions, the

concentrations of those solutions get diluted. For example, will a ppt of AgCl (Ksp= 1.8x10-10) form if solu-

tion A and B are mixed?

Solution A: 50 mL of 0.2 M NaCl

Solution B:

50 mL of 0.2 M AgNO3

Solution A would go from 50 mL of 0.2 M NaCl to 100 mL (the new volume of both solutions). The

volume of the solution doubles. What happens to the concentration when this occurs? Remember from

earlier units that diluting a solution simply uses the following formula:

M1V1= M2V2

In these problems, however, we will alwaysbe looking for the new molarity (M2). Thus, we could

write the above equation as:

V1in the equation is going to be the original volume of the solution. V2is the new volume when the

solutions are added together (it is the sum of the 2 volumes). You then have to do this for each of the 2

solutions. Continuing the problem above we get:

Solution A original: 50 mL of 0.2 M NaCl

Solution A combined with B:

Solution B: 50 mL of 0.2 M AgNO3

Solution B combine with A:

Now that the solutions have been diluted, you may do the problems exactly as before.

Will a ppt of AgCl (Ksp= 1.8x10-10) form if solution A and B are mixed?

[NaCl] = 0.1 M [Cl-1] = 0.1 (Note there is 1 Cl-1for every NaCl)

[AgNO3] = 0.1 M [Ag+1

] = 0.1 (Note there is 1 Ag+1for every AgNO3)

AgCl(s) Ag+1(aq) + Cl-1(aq) Ksp= [Ag

+1][Cl-1] = 1.8x10-10

Initial 0.1 0.1 Q = (0.1)(0.1) = 1x10-2

Since Q >> Ksp= A ppt forms

When 2 solutions are combined, dilute each by the following equation:

Where V1is the volume of the individual solution and V2is the combined volume.

V1

V2M2= M1

V1

V2

M2= M1

50

100

M2= 0.2 M2= 0.1 M NaCl

V1

V2M2= M1

50

100M2= 0.2 M2= 0.1 M AgNO3

V1

V2M2= M1

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Note that the 100 mL is the original volume of

the each solution and the 200 mL is the com-

bined volume of both solutions

Example 9: Will a ppt of BaCO3(Ksp= 8.1x10-9) form if 100 mL of 6x10-5M BaCl2is mixed with 100 mL of

4x10-5M Na2CO3?

The first thing to do is to dilute each solution:

[BaCl2]combined= 6x10-5M [BaCl2]combined= 3x10

-5M

[Na2CO3]combined= 4x10-5M [Na2CO3]combined= 2x10

-5M

Now that we have diluted the solutions, we do the problem exactly as we did before but with dilutedconcentrations from above:

[BaCl2] = 3x10-5 [Ba+2] = 3x10-5 (Note there is 1 Ba+2for every BaCl2)

[Na2CO3] = 2x10-5 [CO3

-2] = 2x10-5 (Note there is 1 CO3

-2for every Na2CO3)

BaCO3 (s) Ba+2(aq) + CO3

-2(aq) Ksp= [Ba+2][CO3

-2] = 8.1x10-9

Initial 3x10-5 2x10-5 Q = (3x10-5)(2x10-5) = 6x10-10

Since Q < Ksp= No ppt forms

Example 10: Will a ppt of PbCl2(K

sp= 1.7x10-5) form if 50 mL of 5x10-3M BaCl

2is added to 120 mL of

4x10-3M Pb(NO3)2

[BaCl2]combined= 5x10-3M [BaCl2]combined= 1.47x10

-3M

[Pb(NO3)2]combined= 4x10-3M [Pb(NO3)2]combined= 2.82x10

-3M

[BaCl2] = 1.47x10-3 [Cl-1] = 2.94x10-3 (Note there are 2 Cl-1for every BaCl2)

[Pb(NO3)2] = 2.82x10-3 [Pb+2] = 2.82x10-3 (Note there is 1 Pb+2for every Pb(NO3)2)

PbCl2 (s) Pb+2(aq) + 2 Cl-1(aq)

Ksp= [Pb

+2][Cl-1]2= 1.7x10-5

Initial 2.82x10-3 2.94x10-3 Q = (2.82x10-3)(2.94x10-3)2= 2.4x10-8

Since Q < Ksp= No ppt forms

100 mL

200 mL

100 mL

200 mL

50 mL

170 mL

120 mL

170 mL