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Solubility Product (Ksp)


What is dissolving? How is equilibrium reached?

How does an Ionic Compound dissolve?

The simplest way to describe this is the cation and the anions dissociate (break apart) into individual ions. Note****polyatomic ions stay together.

How do we write a chemical expression of dissolving?

When writing an ionic equation, any subscripts of anions and cations now become coefficients on the product side (right of the arrow). You must include the charges and the phase is now aqueous.


PbCrO4(s) <=> Pb2+(aq) + CrO42-(aq)

AgCl(s) = Ag+(aq) + Cl-(aq)

CaCO3(s) = Ca2+(aq) + CO32-(aq)

BaSO4 (s) <-->Ba2+ (aq) + SO42- (aq)


Practice writing equations-Highlight to reveal answer


<=> Products
AgBr(s) <=>Ag+(aq) + Br-(aq)
Na3PO4 (s) <=> 3Na+(aq) + PO4 3- (aq)


<=>2 Li+(aq) + CO32-(aq)


Writing the Ksp expression

Used when an ionic compound dissolves into its ions.

  For the reaction:

aA(s) <--> bB(aq) + cC(aq)

K =




Since the concentration of the solid is a constant, it is effectively incorporated into the equilibrium constant:

Ksp = [B]b[C]c


PbCrO4(s) <-->Pb2+(aq) + CrO42-(aq) Ksp=[Pb2+][CrO42-]

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


Na2CO3(s) <--> 2Na+(aq) + CO32-(aq)

Ksp=[Na+ ]2 [CO32- ]

BaSO4 (s) <-->Ba2+ (aq) + SO42- (aq)

Ksp=[Ba2+  ] [SO42-  ]
Li2CO3(s) <--> 2 Li+(aq) + CO32-(aq) Ksp=[Li+ ] [CO32- ]
Al2CO3(s)<-->2Al3+(aq) +3CO32- (aq) Ksp=[Al3+ ]2 [CO32- ]3
Li3PO4(s)<-->3Li+(aq) + PO43-(aq) Ksp=[Li+]3 [PO43- ]

Determining Ksp from Solubility

Example 1

Determine the Ksp of silver bromide, given that its molar solubility is 5.71 x 10¯7 moles per liter.

When AgBr dissolves, it dissociates like this:

AgBr (s)  <===>  Ag+ (aq)  Br¯ (aq)
5.71 x 10¯7M 5.71 x 10¯7M

The Ksp expression is:

Ksp = [Ag+] [Br¯]

Putting the values into the Ksp expression, we obtain:

Ksp = (5.71 x 10¯7) (5.71 x 10¯7) = 3.26 x 10¯13


Calculating solubility from Ksp

NORMAL Determine the molar solubility of silver bromide, given that its Ksp= 3.26 x 10¯13.

When AgBr dissolves, it dissociates like this:

AgBr (s) 


Ag+ (aq) 

Br¯ (aq)




Putting the values into the Ksp expression, we obtain:

Ksp = [Ag+] [Br¯]

Ksp = (X) (X) = 3.26 x 10¯13

X2=3.26 x 10¯13

X=5.71 x 10¯7M


HARDER Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8)

First, write the BALANCED REACTION:


In the above equation, however, we have two unknowns, [Ca2+] and [F-]2.  So, we have to write one in terms of the other using mole ratios.  According to the balanced equation, for every one mole of Ca2+ formed, 2 moles of F- are formed.  To simplify things a little, let's assign the variable X for the solubility of the Ca2+:

If we SUBSTITUTE these values into the equilibrium expression, we now only have one variable to worry about, X:

We can now SOLVE for X:

We assigned X as the solubility of the Ca2+ which is equal to the solubility of the salt, CaF2.  However, our units right now are in molarity (mol/L), so we have to convert to grams:


1. The solubility of PbCrO4 is 1.34 x 10-7 mol/L at 25ēC. Calculate the value of Ksp.

PbCrO4(s) <=> Pb2+(aq) + CrO42-(aq)
Ksp = [Pb2+][CrO42-]
Ksp = (1.34 x 10-7) (1.34 x 10-7)
Ksp = 1.80 x 10-14

2. The Ksp for Agcl is 1.7 x 10-10 at 25ēC. What is the solubility of AgCl?

Solution: Let's represent the concentration of Ag+ as "x". Since there will be an equal concentration of Cl-, this can be represented as "x" also. then,
AgCl(s) <=> Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-]
Ksp = [x] [x]
Ksp = x2
1.7 x 10-10 = x2
x = square root of 1.7 x 10-10
x = 1.3 x 10-5 mol/L


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