Ksp Notes | Mr. Grodski Chemistry

Ksp Notes – link to pdf

Please use the links to move to the skill you want to review:

(SKILL # 1) – determining solubility by comparison on Ksp values

(SKILL # 2a) – Calculating the molar solubilities of individual ions ( 1 : 1 mole ratio)

(SKILL # 1b) – determining solubility by comparison on Ksp values (that have different mole ratios):

(SKILL # 2b) – Calculating the molar solubilities of individual ions (with 1:2 mole ratio)

(SKILL # 3) – Calculating with the molar solubilities (g or L)

(SKILL # 4) – Calculating the Ksp from molar solubilities

(SKILL # 5) – Solving for whether a added solution will cause the solution to form precipitate.

(SKILL # 6) Determining the which salt precipitates first

(SKILL # 7) Prevent the Quadratic Catastrophe by approximating – Embedded in Skill 6 and Skill 8

(SKILL # 8) – Common Ion effect

Intro:

First and foremost K_sp is just a equilibrium constant K_eq of one specific type of reactions called precipitating reactions. Remember that ∆H has many names that mean the same thing; Heat of reaction (∆H_rxn ), Heat of combustion (∆H_comb), Heat of formation (∆Hfº), and Heat of Fusion (∆H _fusion) to name just a few.

Remember that we learned about Ka’s from out Acid stoichiometry unit. We used Ka’s to determine the strength of acids. Ka is the equilibrium for acids dissociating For example, the strong acid HCl has the following reaction:

HCl (aq) <—> H⁺ (aq) + Cl^– (aq)

K_eq = K_a = 1.3 x 10⁶

The equilibrium value (Keq) of this acid dissociation reaction is called a Ka when the reaction is an acid dissociating reaction like the HCl reaction above. The value of the Ka for HCl dissociating is extremely large because Strong Acids like HCl almost completely dissociate into the products. Plain and simple, the products are extremely favored with strong acids!! A weak acid like HF has a much lower Ka because they barely dissociate into their products and the reactants are favored.

So just like a Ka is the equilibrium constant for acids breaking apart (dissociating), Ksp is the equilibrium constant for Ionic Solids dissociating into ions when they dissolve. We use Ksp’s to measure how much INSOLUBLE salts (Ionic Solids) actually dissolve. In most lower level chemistry courses salts are labeled Soluble (they dissolve) or Insoluble (they do not dissolve). The real truth is that most insoluble salts actually do dissolve a little bit. We can quantitatively measure the small solubility of these seemingly insoluble salt by using the Keq of these reactions of salts dissolving into their ions in water.

We call these reactions precipitating reactions and they all are written the same with the solid salt written as a reactant and aqueous (dissolved) ions as the product. In this way the Ksp describes how much products are favored in these type of reactions. Because these reactions are always written in this format and the aqueous ions are always written as the product we can use the Ksp or the equilibrium constant as a direct measure of the solubility of these salts that are labeled insoluble.

For example in Table L of your Regents Reference Tables:

You can see that carbonates are always insoluble except with group 1 ions (Li, Na, K, Rb, Cs ) and NH₄⁺.

Calcium carbonate, based on this table, is insoluble which normally means that no calcium or carbonate ions dissolve but that is not accurate based on its Ksp.

CaCO₃ (s) <—> Ca⁺² + CO₃^-2

Ksp = [Ca⁺²][CO₃^-2]

Notice that there is no denominator AND THERE IS NEVER a denominator with precipitating reactions because with equilibrium expressions:

1. products over reactants
2. Coefficients become exponents
3. Solids and Liquids do not apply.

So since all precipitating reactions have solids salts as the reactants the Ksp will always have no denominator and the Ksp equation or equilibrium expression will just be the PRODUCT of the IONS:

CaCO₃ (s) <—> Ca⁺² + CO₃^-2Ksp = [Ca⁺²][CO₃^-2] = 1.4 x 10^-11

So the value of the product of the ions is just the Ksp also called the solubility product = product of the ions.

In other words, Ksp = product of the ions that dissolved! Ksp is a measure of how soluble these insoluble salts are.

The Ksp of the reaction above ( 1.4 x 10^-11) is very small and thus describes an equilibrium positions that favors the reactants BUT because this value is not zero there must be some products (ions ) are formed however small.

We label salts with low Ksp’s Insoluble because it is very hard to have Q’s (reaction quotients) that are smaller than the Ksp.

This means that most mixtures of Ca⁺²and CO₃^-2 will have molarities that result in Q > Ksp! And preciptation occurs.

At the left I bring back the U diagram which is now specific for the equilibrium constant for the precipitating reaction of CaCO₃.

Notice that the Ksp is very small for the salt which is considered insoluble in many cool chemistry circles.

Didn’t we use CaCO₃ in a gravimetric lab to determine whether a another salt was K₂CO₃ or Na₂CO₃ ?

The very small Ksp of CaCO₃reveals that MORE reactants are favored at equilibrium than products and it also reveals that equilibrium position exists in a position a bit forward of the 100% Reactant.

This means that we start with the solid and mix it with water some small amount of the ions will dissolve before equilibrium is reached!

Since all Insoluble Salts have non zero Ksp’s they are must have a small amount of solubility!

(SKILL # 1) – determining solubility by comparison on Ksp values

CaCO₃ (s) <—> Ca⁺² + CO₃^-2 Ksp = 1.4 x 10^-11

MgCO₃(s) <—> Mg⁺²(aq) + CO₃^-2 (aq) Ksp = 6.82 x 10^-6

We can compare the solubility of Ca⁺² vs Mg⁺² by looking directly at the Ksp’s. Clearly MgCO₃is more soluble than CaCO₃ because its Ksp is larger (products are MORE favored) than MgCO₃ as more ions are dissolved at equilibrium than CaCO₃! Any possible reason for this?

Careful here. We can only directly compare solubilities through solubilities that have the same Ksp formula. Notice that the two above precipitating reactions have ions that are in a 1 : 1 ratio. If the salts produce ions in different ratios we will have to calculate the molar concentrations of the individual ions to compare.

(SKILL # 2a) – Calculating the molar solubilities of individual ions ( 1 : 1 mole ratio)

Remember that Ksp is an equilibrium constant that has no units and is just the product of the molarities of the

ions at equilibrium (saturated solution) at their respective exponents (coefficients). Solubility is a measure of how many ions dissolve in certain volume of solution = molarity.

If we want the measure the actual solubility (molarity) of the Ca⁺² or Mg⁺² we will have to solve for them.

For instance the molar solubility of Ca⁺² is solved the following way (SKILL # 2)

CaCO₃ (s) <—> Ca⁺² + CO₃^-2 Ksp = 1.4 x 10^-11

Ksp = [Ca⁺²][CO₃^-2]

We need an Equilibrium Ice Table to solve this problem since we need the value of [Ca⁺²] at equilibrium.

We have used stoichiometry tables already that use moles but now we are using equilibrium tables and they require molarity [ ] or partial pressures ( ) from equilibrium expressions!

CaCO₃ (s) <—> Ca⁺² + CO₃^-2

Initial : I: 0 0 we like to start from the zero position

and determine the change from this position

Change: C: +x +x each ion is has 1 : 1 ratio;

there is 1 Ca⁺² for every 1 CO₃^-2

@ Equilibrium: E: x x change from zero is just the change

we need this value at equilibrium to place

in our equilibrium expressions

Equilibrium expression = Ksp = [Ca⁺²][CO₃^-2]

1.4 x 10^-11 = [x] [x] from the E : part of the table

1.4 x 10^-11 = [x]²

√1.4 x 10^-11 = √[x]²

3.74 x 10^-6M = x = [Ca⁺² ] = Molar solubility of calcium ions

3.74 x 10^-6M = x = [CO₃^-2] = Molar solubility of calcium ions

You should notice that x also = the molar solubility of the CO₃^-2 ion as well because the stoichiometric

ratio of the ions are 1 : 1 in the precipitation reaction.

(SKILL # 1b) – determining solubility by comparison on Ksp values (that have different mole ratios):

We cannot evaluate the molar solubility from Ksp’s of salts that have different Ksp formulas that result from different the product ions that have different stoichiometric solubilities. For example:

CaCO₃ (s) <—> Ca⁺² + CO₃^-2 Ksp = 1.4 x 10^-11

PbCl₂ (s) <—> Pb⁺² + 2Cl^-1 Ksp = 1.7 x 10^-5

One might think that Pb⁺²is more soluble at equilibrium and thus would have a greater molar solubility than Ca⁺² but we must solve for Pb⁺² first because of the different Ksp equations that result for precipitating reactions that have different number ions or coefficients.

(SKILL # 2b) – Calculating the molar solubilities of individual ions (with 1:2 mole ratio)

Lets use are equilibrium ICE table to solve for the molar concentration of Pb⁺² ions at equilibrium when PbCl₂ (s) is added to water at a constant temperature.

PbCl₂ (s) <—> Pb⁺² + 2Cl^-1

Initial : I: 0 0 we like to start from the zero position

and determine the change from this position

Change: C: +x +2x each ion is has 1 : 2 ratio;

there is 1 Pb⁺² for every 2 Cl^-1

@ Equilibrium: E: x 2x change from zero is just the change

we need this value at equilibrium to place

in our equilibrium expressions

*The +2x is only a stoichiometric value meaning that we double the x when we have a value that relates to it!

Equilibrium expression = Ksp = [Pb⁺²][Cl^-1]²

1.7 x 10^-5 = [x] [2x]² from the E : part of the ICE table

1.7 x 10^-5 = [4x]³

∛ 1.7 x 10^-5 = ∛ [x]³

1.62 x 10^-2 = [x] = [Pb⁺² ] = Molar solubility of lead ions

3.24 x 10^-2 = [2x] = [Cl⁺² ] = Molar solubility of chloride ions

*Notice a 1:1 ion salt like CaCO₃ always will have Ksp = [x][x] = [x]²

*Notice a 1:2 or 2:1 ion salt like PbCl₂ (s) will Always have a Ksp = [x][2x]² = [4x]³

There are other possibilities but these are the most common!

-SO if we want to continue the comparison of which salt is more soluble (free ions dissolved):

CaCO₃ (s) <—> Ca⁺² + CO₃^-2 Ksp = 1.4 x 10^-11

PbCl₂ (s) <—> Pb⁺² + 2Cl^-1 Ksp = 1.7 x 10^-5

From skill 2a above:

Equilibrium expression = Ksp = [Ca⁺²][CO₃^-2]

1.4 x 10^-11 = [x] [x]

√1.4 x 10^-11 = √[x]²

3.74 x 10^-6M = x = [Ca⁺² ] = Molar solubility of calcium ions or CO₃^-2 ions

PbCl₂ (s) <—> Pb⁺² + 2Cl^-1 Ksp = 1.7 x 10^-5

From skill 2b above:

∛ 1.7 x 10^-5 = ∛ [x]³

1.62 x 10^-2 = [x] = [Pb⁺² ] = Molar solubility of lead ions

3.24 x 10^-2 = [2x] = [Cl⁺² ] = Molar solubility of chloride ions

PbCl₂ (s) (1.62 x 10^-2 = [Pb⁺² ] ) then is more soluble than CaCO3 (s) , (3.74 x 10^-6M = x = [Ca⁺² ] )

The molar solubilities of the ions needed to be calculated to justify!

(SKILL # 3) – Calculating with the molar solubilities (g or L) TOP

Once you solve for the molar solubilities (skill # 2a or 2b above) you can calculate the all other values related to molarity which include grams or Liters.

In the example below I am showing you how to solve for grams:

How many grams of Ca⁺² will dissolve in 1.50 L of saturated solution of CaCO₃ (aq) (lime water).

You will need to solve for [Ca⁺²] in skill #2 which we have already done and then understand that this is a molarity.

Molarity = moles of ions dissolved

Liter of solution

We have already done step 1 above (skill #2a) for CaCO₃ and 3.74 x 10^-6M = x = [Ca⁺² ]

(3.74 x 10^-6 mol x (1.50L) = 5.61 x 10^-6~~mol~~ x 40g = 2.24 x 10^-4 grams

L ↑ 1 ~~mol~~

M x V = moles….silly!

* Ca has a gram atomic mass of 40. grams per mol

So ONLY 2.24 x 10 ^-4 grams of CaCO₃ can maximally dissolve in 1.5L in solution at constant temperature.

Does this makes sense for “Insoluble Compounds”?

Now what if you wanted to solve for the amount of volume of solution needed to dissolve an “insoluble” compound (with a small Ksp)? In this example, Lets say we want to dissolve 2.0 grams of CaCO₃ (s) ?

Given the same molar solubility of Ca⁺² ions:

– We have already done step 1 above (skill #2a) for CaCO₃ and 3.74 x 10^-6M = x = [Ca⁺²]

2.0 g of Ca⁺² x 1 mol = .050 ~~mol~~ of Ca⁺²x 1 L = 13,400 L

40. g 3.74 x 10^-6~~mol~~

Whoa!! It takes 13,400 Liters of water to dissolve 2 grams? Does that make sense?

Yes because the “insoluble compound” will need a tremendous amount of water to dissolve

the salt if only a small amount will dissolve according the VERY small Ksp!

(SKILL # 4) – Calculating the Ksp from molar solubilities. TOP

This is just going in the reverse direction of solving for molar solubilities like we did in Skill 2a and 2 b.

Instead of a given Ksp you will be given molar solubilities of one ion of the products AND you will have to use STOICHIOMETRY to solve for other ion.

*This the only place where you will use the 2X that appears in the PbCl₂ (s) problem.

*The +2x is only a stoichiometric value meaning that we double the x when we have a value that relates to it!

Equilibrium expression = Ksp = [1.62 x 10^-2][3.24 x 10^-2]²

Ksp = [ 1.62 x 10^-2][3.24 x 10^-2]² = 1.7 x 10^-5

(SKILL # 5) – Solving for whether an added solution will cause the solution to form precipitate. TOP

A) You will need to dilute the 2 solutions which will be given in molarities.

Use the Dilution equation: MV = MV

B) Write the net ion reaction

C) Calculate the reaction quotient Q which is the same as the Ksp formula BUT use molarities that are Added and thus are not at equilibrium.

D) Compare Q vs Ksp (Keq) to determine if the reverse reaction is more spontaneous = precipitation.

Is Q larger or smaller than Ksp?

– If Q > Ksp the reverse reaction will be favored – Solubility decreases as precipitate forms.

-If Q < Ksp the forward reaction will be favored – Solubility increases as more ions dissolve.

Example:

25 molar of a 1.3 x 10^-3molar solution of Pb(NO₃)₂ is added to 15 ml of a 4.4 x 10^-4 molar

solution of CaCl_2 .Will a precipitate of PbCl₂ form?

New Diluted [ ] Total Volume

A) Dilute: M x V = M V

(1.3 x 10^-3 M )(25 ml) = M x ( 40. ml)

New diluted molarity: [Pb(NO₃)₂] = 8.1 x 10^-4 M

New Diluted [ ] Total Volume

M x V = M V

(4.4 x 10^-3 M )(15 ml) = M x ( 40. ml)

New diluted molarity: [CaCl₂] = 1.65 x 10^-3 M

B) Write the net ion reaction: (remove spectators)

Pb⁺² (aq) + 2 Cl^– (aq) —-> PbCl₂ (s)

You should flip this reaction so that the Ksp reflects ions as products.

PbCl₂ (s) —-> Pb⁺² (aq) + 2 Cl^– (aq)

C) From this net ion reaction above determine the reaction quotient:

Q = [Pb⁺²][Cl^-1]²

Q = [8.1 x 10^-4][*1.65 x 10^-3]²

*Notice that I did not double the [Cl^-1]

because that IS the actual value of [Cl^-1]!

Q = 2.2 x 10^-9

D) Now you could be asked to calculate the Ksp (Skill 4) or it will be given at a particular temperature

which will stay the same. Ksp = 1.7 x 10^-5

Lets compare Q vs K(sp)!

Q = 2.2 x 10^-9 < Ksp = 1.7 x 10^-5

Thus the two solutions reaction will not make a precipitate when mixed and the forward reaction is favored.

PbCl₂ (s) —-> Pb⁺² (aq) + 2 Cl^– (aq)

More spontaneous –—————————>

(SKILL # 6) Determining the which salt precipitates first. TOP

AND (SKILL # 7) Prevent the Quadratic Catastrophe by approximating.

A beaker contains a 400. ml solution of .20 M of Cl^–and .10 M of I^–. If a solution of Pb(NO₃)₂ was slowly added to the beaker, which salt precipitate’s first?

The Ksp of PbI₂ is 1.4 x 10^-8 . The Ksp of PbCl₂ is 1.7 x 10^-5

This precipitation reaction (net ion): Pb⁺² (aq) + 2I^–(aq) —> PbI₂ (s) was what I did as demo in class:

We should realize the that precipitating reactions that will occur due to the soluble Pb(NO₃)₂that will deliver the soluble Pb⁺²ions are: (Based on soluble rules nitrates are always soluble!)

PbI₂ (s) <—> Pb⁺² + 2I^– Ksp = 1.4 x 10^-8

PbCl₂ (s) <—> Pb⁺² + 2Cl^-1 Ksp = 1.7 x 10^-5

If both ions, .20 M of Cl^– and .10 M of I^–, were dissolved with the same concentration then all we would have to do is look at the Ksp values to determine which salt would precipitate first because both salts have the same formula (same stoichiometric ratios of product ions that lead to the same Ksp formula). The salt with the smaller Ksp value must precipitate first as it the most insoluble (produces the lest amount of ions before it reaches equilibrium). Remember that Ksp = Keq and that when Keq < 1 reactants are more favored.

BUT this is NOT the case so we must calculate the equilibrium molar concentration of Pb⁺² for each of the precipitating reaction and compare [Pb⁺²] to see which one is the smaller. The smaller [Pb⁺²] will be the one that precipitates first because this concentration Pb⁺² is the value that must be exceeded first to cause the Q > Ksp to cause precipitation.

The lower concentration will be reached first as you pour Pb⁺² ions into the beaker.

So here we are calculating the molar concentration of I^– which is Skill 2b with a twist. The twist is that we are starting with .10 M before we add any Pb⁺² .

PbI₂ (s) <—> Pb⁺² + 2I^–

Initial : I: 0 .10 The solution already has .10 M

of I^– ions.

Change: C: +x +2x each ion is has 1 : 2 ratio;

there is 1 Pb⁺² for every 2 I^-1

@ Equilibrium: E: x .10 + 2x we are adding 2x as equilibrium

is reached

Equilibrium expression = Ksp = [Pb⁺²][I^-1]²

Ksp = [x][.10 + 2x]²

Ksp = 4x³ + .4x² + .01x (CUBIC Equation!)

The Quadratic Catastrophe!

So eliminate the need to solve for these three roots of the cubic equation by using the idea that because the Ksp is so small that the reaction will move only a very small amount forward to reach equilibrium. Thus x is very tiny as compared to the .10 M that was initially added. We eliminate the Quadratic Catastrophe by approximating the 2x away.

(this is skill 7):

Ksp = [x][.10 + 2x]²

Ksp = [x][.10]²

1.4 x 10^-8 = [x][.10]²

1.4 x 10^-6 = [x] = [Pb⁺²]

This all means the when molarity of Pb⁺² is greater than 1.4 x 10^-6PbI₂starts Precipitating!

Or Better Yet it means that the value of Pb⁺² if increased from 1.4 x 10^-6 will cause Q to become bigger

than K(sp) and precipitation BEGINs (reverse reaction becomes more spontaneous!)

Okay so lets calculate the [Pb⁺²] when PbCl₂ starts precipitating:

PbCl₂ (s) <—> Pb⁺² + 2Cl^-1 Ksp = 1.7 x 10^-5

Initial : I: 0 .20 The solution already has .20 M

of I^– ions.

Change: C: +x +2x each ion is has 1 : 2 ratio;

there is 1 Pb⁺² for every 2 Cl^-1

@ Equilibrium: E: x 20 + 2x we are adding 2x as equilibrium

is reached .

Equilibrium expression = Ksp = [Pb⁺²][Cl^-1]²

Approximate away the quadratic or cubic formula!

Ksp = [x][.20 + 2x]²

Ksp = [x][.20]²

1.7 x 10^-5 = [x][.20]²

4.25 x 10^-4 = [x] = [Pb⁺²]

This all means the when molarity of Pb⁺² is greater than4.25 x 10^-4M PbCl₂starts Precipitating!

Any value above 4.25 x 10^-4 M will cause the Q to be greater than K(sp) and reaction will shift to the formation of the precipitate due to Le Chatelier’ s Principle.

Compare the molar solubilities of the [Pb⁺²] from both calculations:

When we pour Pb(NO₃)₂ ( delivering Pb⁺² ions) into the beaker which Threshold [Pb⁺²] is reached first?

From PbI₂ From PbCl₂

[Pb⁺²] = 1.4 x 10^-6 [Pb⁺²] = 4.25 x 10^-4

PbI₂ precipitates first because the concentration of Pb⁺²ions is smaller and thus equilibrium is reached first.

SKILL # 8 – Common Ion effect TOP

Also (SKILL # 7) Prevent the Quadratic Catastrophe by approximating.

If an ion that is common to the Ksp expression is added to a solution that is saturated the molar concentration of the of the other ion is calculated for.

If there is a 1000.0 ml of saturated solution of PbCl₂ at 20˚C and .250 moles of NaCl is added to the solution calculate the molar concentration of the Pb⁺² ion.

There is way to do this which is harder way and an easier way. I will show you both!

Harder way:

Now since we have an equilibrium of PbCl₂ (s) :

PbCl₂ (s) <—> Pb⁺² + 2Cl^-1

Initial : I: 0 0 Start solid in pure water.

Change: C: +x +2x each ion is has 1 : 2 ratio;

there is 1 Pb⁺² for every 2 Cl^-1

@ Equilibrium: E: x 2x we are adding to zero

Equilibrium expression = Ksp = [Pb⁺²][Cl^-1]²

1.6 x 10^-5 = [x] [2x]² from the E : part of the table

1.6 x 10^-5 = [4x]³

∛ 1.6 x 10^-5 = ∛ [x]³

1.59 x 10^-2 = [x] = [Pb⁺² ] = Molar solubility of lead ions

3.17 x 10^-2 = [2x] = [Cl⁺² ] = Molar solubility of lead ions

Okay so we now we have the initial concentrations of the ions in equilibrium, lets put them in a new equilibrium table that reflects the common ion the was added in our problem. Before we do lets make sure we are putting molarity NOT moles in our equilibrium table!

The common ion comes from NaCl.

.250 mol of NaCl was added to 1000ml = 1 Liters. NaCl is completely soluble:

NaCl (s) —–> Na⁺ (aq) + Cl^– (aq)

Thus .250 mol / 1 L = .250 M of NaCl = [Cl^–] = .250M

Remember that there is 1 NaCl to 1 Cl^– ratio!

Okay back to out new equilibrium table that reflects the added .250 M for the Cl- that came from NaCl added.

PbCl₂ (s) <—> Pb⁺² + 2Cl^-1

Initial : I: 1.59 x 10^-2 3.17 x 10^-2 + .250 Start with molar solubilities.

Change: C: –x – 2x Q > Ksp here so the reverse

reaction wil occur!

@ Equilibrium: E: 1.59 x 10^-2 –x .250 + 3.17 x 10^-2 -2x

Equilibrium expression = Ksp = [Pb⁺²][Cl^-1]²

1.6 x 10^-5 = [1.59 x 10^-2 –x][.250 + 3.17 x 10^-2 -2x]²

^{This will be a catastrophe if we do not approximate away the cubic formula!}

1.6 x 10^-5 = [1.59 x 10^-2 –x][.250 + 3.17 x 10^-2 -2x]

Remember that 2x and the initial molar concentration are so small they are insignificant to the Much Larger .250!

1.6 x 10^-5 = [1.59 x 10^-2 –x][.250 ]²

1.6 x 10^-5 = 9.94 x 10^-4-.0625x

1.6 x 10^-5 – 9.94 x 10^-4 = -.065x

-9.78 x 10-4 = -.065x

-9.78 x 10-4 = x = .0156

-.0625

From equilibrium table above : Pb⁺²

@ Equilibrium: E: 1.59 x 10^-2 – x

1.59 x 10^-2 – .0156 = 2.56 x 10^-4 = [Pb⁺² ]

Easier way:

Or we could think that we already have [Cl^–] = .250M amount of Cl- ions present in a 1000 ml solution and we add enough solid PbCl₂ (s) to reach equilibrium with the ions present.

If we think real hard we can visualize that there are No Pb⁺²present so the reaction has to move forward till enough Pb⁺² ions dissolve to to reach the Ksp value with the [Cl^–] = .250M. That value or amount of Pb⁺²

was initially there that we are now NOT considering ( 1.59 x 10^-2 M) by approximation is because this amount of Pb⁺² will be be decreased so much by the addition of the VERY Large amount of the common ion (Cl-) that it DOES NOT MATTER how much Pb⁺² we initially have!

In this line of thought all we need to solve is: Ksp = [Pb⁺²][Cl^-1]²

1.6 x 10^-5= [↓x][↑.250]²

2.56 x 10^-4 = [x] = [Pb⁺² ]

*So in a common ion problem you can approximate away the initial concentration of the Pb⁺² as well:

1.6 x 10^-5 = [~~1.59 x 10^-2~~ –x][.250 + 3.17 x 10^-2 -2x]

and THUS all you have to do is set up your problem like this:

1.6 x 10^-5= [x][.250]²

Which way do you like?

The [Pb⁺² ] with .250 moles of NaCl added must be lower than the molar equilibrium WITHOUT a large amount of Cl- ions present as the common ion.

The molar solubility of Pb⁺² without .250 moles of NaCl added = 1.59 x 10^-2

The molar solubility of Pb⁺² with .250 moles of NaCl added = 2.56 x 10^-4

Why is it lower? Ksp is constant at the same temperature, thus a greater [Cl^– ] leads to a lower [Pb⁺² ] to maintain the same value of Ksp. Also remember that adding a common ion will cause shift in the reaction toward the formation of more precipitate and less ions (shift Left) and thus less product or ions will be present to reestablish equilibrium at the same temperature due to Le Chatelier’s Principle.

PbCl₂ (s) ⇠⇢ Pb⁺² + 2Cl^-1

Shift Left ↓

<————————–

___________________________________________________________________________________________