^{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)}

CaCO_{3}(s) = Ca^{2+}(aq)
+ CO_{3}^{2-}(aq)

BaSO_{4} (s) <-->Ba^{2+}
(aq) + SO_{4}^{2-}
(aq)

Practice writing equations-Highlight to reveal
answer

Reactant

<=> Products

AgBr_{(s)}

<=>Ag^{+}_{(aq)}
+ Br^{-}_{(aq)}

Na_{3}PO_{4 (s)}

<=> 3Na^{+}_{(aq) }
+ PO_{4}^{3-}_{(aq)}

Li_{2}CO_{3(s)}

<=>2 Li^{+}_{(aq)}
+ CO_{3}^{2-}_{(aq)}

Writing the K_{sp} expression

Used when an ionic compound dissolves into its ions.

For the reaction:

aA_{(s) }<--> bB_{(aq)}
+ cC_{(aq)}

K =

[B]^{b}[C]^{c}

[A]^{a}

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

Determine the K_{sp} 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¯^{7}M

5.71 x 10¯^{7}M

The K_{sp} expression is:

K_{sp} = [Ag^{+}] [Br¯]

Putting the values into the K_{sp} expression, we obtain:

K_{sp} = (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)

#-X

X

X

Putting the values into the K_{sp} expression, we obtain:

K_{sp} = [Ag^{+}] [Br¯]

K_{sp} = (X) (X) = 3.26 x 10¯^{13}

X^{2}=3.26 x 10¯^{13}

X=5.71 x 10¯^{7}M

HARDER Calculate the solubility of CaF_{2} in g/L
(K_{sp} = 4.0 x 10^{-8})

First, write the BALANCED REACTION:

Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM
EXPRESSION:

In the above equation, however, we have two unknowns, [Ca^{2+}] 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 Ca^{2+} formed, 2 moles of F^{-} are formed.
To simplify things a little, let's assign the variable X for the solubility
of the Ca^{2+}:

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 Ca^{2+} which is equal to
the solubility of the salt, CaF_{2}. However, our units right
now are in molarity (mol/L), so we have to convert to grams:

1. The solubility of PbCrO_{4} is 1.34 x 10^{-7} mol/L at
25ēC. Calculate the value of K_{sp}.

Solution:

PbCrO_{4}(s) <=> Pb^{2+}(aq) + CrO_{4}^{2-}(aq)

K_{sp} = [Pb^{2+}][CrO_{4}^{2-}]

K_{sp} = (1.34 x 10^{-7}) (1.34 x 10^{-7})

K_{sp} = 1.80 x 10^{-14}

2. The K_{sp} 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,