# ksp-info by dandanhuanghuang

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```									                           Ksp: The Solubility-Product Constant

This principle was first put forth by Walther Nernst in 1899. It has to do with solid substances
usually considered insoluble in water. In each case, we will consider a saturated solution of the
insoluble substance that is in contact with some undissolved solid. Important points to consider
are:

1) Some of the solid does dissolve. Not very much, but enough.
2) The substance dissociates upon dissolving.
3) There exists an equilibrium between the undissolved solid and the solvated
ions.

Since equilibrium principles can be used, that is where we start. Our first example is silver
chloride, AgCl. When it dissolves, it dissociates like this:
AgCl (s) <===> Ag+ (aq) + Cl¯ (aq)

An equilibrium expression can be written:
Kc = ( [Ag+] [Cl¯] ) / [AgCl]

Now, we come to an important point. [AgCl] is the molar concentration of solid AgCl. This
value is a constant, since it can be directly related to the density, which is constant.
What we do is move it to the other side and incorporate it with the equilibrium constant.
Kc [AgCl] = [Ag+] [Cl¯]

Since Kc [AgCl] is a constant, we replace it with a single symbol. Like this:
Ksp = [Ag+] [Cl¯]

(Just a side point - as you go on in chemistry, you'll get introduced to the concept of activity. The
activity of a solid is defined as equal to the value of one. Since the activity of AgCl(s) = 1, it just
drops out. However, like I said, activity is for the future. Not right now). It turns out that the Ksp
value can be either directly measured or calculated from other experimental data. Knowing the
Ksp, we can calculate the solubility of the substance in a very straightforward fashion.
Here are several other examples of dissociation equations and their Ksp expressions:

Sn(OH)2 (s) <===> Sn2+ (aq) + 2 OH¯ (aq) Ksp = [Sn2+] [OH¯]2
Ag2CrO4 (s) <===> 2 Ag+ (aq) + CrO42¯ (aq) Ksp = [Ag+]2 [CrO42¯]
Fe(OH)3 (s) <===> Fe3+ (aq) + 3 OH¯ (aq) Ksp = [Fe3+] [OH¯]3
In order to write Ksp expressions properly, you must know how ionic substances dissociate in
water. That means, you have to know your chemical nomenclature, polyatomic ions, and the
charges associated with ion. Also, and this is important, so pardon the shouting:

EACH CONCENTRATION IN THE Ksp EXPRESSION IS RAISED TO
THE POWER OF ITS COEFFICIENT IN THE BALANCED EQUATION.

Here are some practice problems for writing Ksp expressions. Write the chemical equation
showing how the substance dissociates and write the Ksp expression:

1) AlPO4

2) BaSO4

3) CdS

4) Cu3(PO4)2

5) CuSCN

6) Hg2Br2

7) AgCN

8) Zn3(AsO4)2

9) Mn(IO3)2

10) PbBr2

11) SrCO3

12) Bi2S3
Equilibrium Chapter Review

Terms & Concepts
Chemical equilibrium            Keq                              Reversible reaction
Forward reaction
Reverse reaction                Equilibrium expression           LeChatelier’s Principle          Ksp

Problems
Write an equilibrium expression for each of the following reactions.

1) 3 O2 <===> 2 O3

2) N2 + 3 H2 <===> 2 NH3

3) H2 + I2 <===> 2 HI

4) PCl5 <===> PCl3 + Cl2

5) SO2 + (1/2) O2 <===> SO3

6) H2 + I2 <==> 2 HI

a) Calculate the equilibrium constant (Keq) for the following reaction: H2 + I2 <===> 2 HI , when the
equilibrium concentrations at 25 °C were found to be:

[H2] = 0.0505 M
[I2] = 0.0498 M
[HI] = 0.389 M

b) The same reaction as above was studied at a slightly different temperature and the following
equilibrium concentrations were determined: Solve for the Keq.

[H2] = 0.00560 M
[I2] = 0.000590 M
[HI] = 0.0127 M

7. Hydrogen sulfide gas decomposes and establishes equilibrium at 1400ºC
2H2S(g) <==> 2H2(g) + S2(g)

A liter of this mixture contains 0.18 mol H2S, 0.014 mol H2, and 0.035 mol S2. Calculate Keq for this
reaction.
8. Calculate Keq for the above reaction if [H2S]= 0.25 M, [H2]=0.88M, and [S2]=0.44 M

9. The equilibrium constant, Keq, for the following reaction is 5.6. If the concentration of N2O4
is 0.66M, what is the concentration of NO2?        2NO2(g) <==>N2O4(g)                Keq = 5.6

Ksp
10. Write the Ksp for each of the following.

a. BaSO4    b.BaF2        c. AgCl          d. CaF2

11. Solve for the Ksp using the equations listed above:

a.     [BaSO4] = 0.34M          [Ba+2] =0.25M              [SO4-2] = 0.12M

b. [BaF2] = 0.01M              [Ba+2] =0 .02M             [F1-] = 0.001M

c.    [AgCl] = 0.14M           [Ag1+] = 0.15M             [Cl1-] = .10M

d. [CaF2] = 0.001M             [Ca+2] = 0.02M             [F1-] = 0.04M

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