Chapter 2: Solubility Part 3 Page 1 SABIS Grade 11 (Level M) Chemistry

2.3 The Equilibrium Law


Table 2.1 and the schematic presentations of Figures 2.8, 2.9, and 2.10 are useful for a quick and qualitative view of solubilities. But chemists are not satisfied with the statement that a substance has low solubility. We must know how much of the substance dissolves. We must be able to treat solubility in a quantitative fashion. Fortunately this can be done with the aid of the principles of equilibrium, developed in Chapter 1. 

 As mentioned earlier, the quantitative concentration relationship that exists at equilibrium is shown in the equilibrium law relation:
K=[E]e[F]f[A]a[B]b         (2)

Expression (2) applies to a solubility equilibrium, provided we write the chemical reaction to show the important molecular species present. In Section 1 we considered the solubility of iodine in alcohol. Since iodine dissolves to give a solution containing molecules of iodine, the concentration of iodine itself fixed the solubility. The situation is quite different for AgCl, because AgCl solid is made up of ions. When solid AgCl is placed in water, a little of ions dissolve and the rest remain in the solid state. At equilibrium the equation will be
AgCl(s) ⇌ Ag+(aq) + Cl–(aq)         (19)

Equilibrium will exist when the concentrations are in agreement with the expression
K = a constant = [Ag+] × [Cl–]        (20)

The “concentration” of the solid (silver chloride) does not appear in the equilibrium expression (20); it does not vary. To consider a more complicated example, consider the application of expression (2) to the solubility of lead chloride, PbCl2:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl–(aq)     (21)
At equilibrium,
K = a constant = [Pb2+] × [Cl–]2      (22)

Solubility equilibrium constants, such as (20) and (22), are given a special name—the solubility product. It is symbolized Ksp. A low value of Ksp means the concentrations of ions are low at equilibrium. Hence the solubility must be low. Table 2.2 lists solubility products for some common compounds.

Table 2.2 Some solubility products at room temperature.


Table 2.1 Solubility of common compounds in water.




Figure 2.8 Almost all alkali metals, hydrogen ions, and ammonium ions are soluble in water.



Figure 2.9 Positive ions (colored in pink) form compounds of low solubilities with various anions (colored in blue).



Figure 2.10 More examples positive ions (colored in pink) forming compounds of low solubilities with various anions (colored in blue).


To work out a general expression for the solubility in terms of Ksp. It is required to calculate the solubility of CuCl(s) in water in terms of the Ksp:
                                                  CuCl(s)       ⇌      Cu+(aq)     +      Cl–(aq)
Initial concentrations                      —                          0 M                   0 M
Equilibrium concentration                                             a                        b
What expression can we use to calculate the solubility?

a = s/2, b = s, and Ksp = s.s/2

a = b = 2s and Ksp = 4s2

a = s, b = 2s, and Ksp = 2s2

a = b = s/2, and Ksp = (s/2)2

a = b = s, and Ksp = s2


















 answer :
a = b = s, and Ksp = s2
Response:
That's the correct answer

2.3.1 Calculation Of The 

Solubility Of Copper (I) Chloride In Water


Calculate the solubility of a salt, given the Ksp
The solubility product is learned from measurements of the solubility. 
In turn, it can be used as a basis for calculations of solubility. 
Suppose we wish to know how much copper (I) chloride, CuCl, will dissolve in one litre 
of water. We begin by writing the balanced equation for the reaction:
CuCl(s) ⇌ Cu+(aq) + Cl–(aq)       (23)

From this equation, we can write the equilibrium expression:
Ksp = [Cu+][Cl–]          (24)

Now the numerical value of Ksp is found in Table 2.2:
Ksp = 3.2 × 10–7 = [Cu+][Cl–]         (25)

Expression (25) indicates that copper (I) chloride dissolves, according to reaction (23), 
until the molar concentrations of copper (I) ion and chloride ion rise enough to 
make their product equal to 3.2 × 10−7.
Suppose we designate the solubility of copper (I) chloride in water by a symbol, s. 
This symbol s equals the number of moles of solid copper (I) chloride that dissolve in 
one liter of water. Remembering equation (23), we see that s moles of 
solid copper (I) chloride will produce s moles of copper (I) ion, Cu+, 
and s moles of chloride ion, Cl–. Hence these concentrations must be equal, 
as shown below:
                                          CuCl(s)  ⇌  Cu+(aq) + Cl–(aq)
Initial concentrations            —                0 M          0 M
Part that dissolves                –s                +s            +s
Equilibrium concentration                          s              s
                                   
[Cu+] = [Cl–] = s moles/liter = moles CuCl dissolved           (26)

Substituting (26) into (25), we have:






How many milligrams of silver bromide dissolve in 20 liters of water? Give your answer to three significant figures.
[Ag = 108, Br = 80]
Ksp of AgBr = 5 ×10−13

M(AgBr)  =
 mg

Calculate the solubility, in moles per liter, of calcium sulfate in water, given Ksp = 2.4 × 10−4. Give your answer to three significant figures.
s =
 M


Calculate the solubility of SrCrO4 in water, given Ksp = 3.6 × 10−5.
s = a × 10b
Give your answer to two significant figures.
a =
 Calculate the solubility of SrCrO4 in water, given Ksp = 3.6 × 10−5.

s = a × 10b
Give your answer as an integer.
b =
0.0010 mol HCl and 1.0 × 10–6 mol of Pb(NO3)2 are mixed forming 2.0 liter solution . Will a precipitate form? For PbCl2 Ksp = 1.3 × 10–8.





Q = a×10b


answer :

a = = 1.3
b = = –13
Response:
That's the correct answer

Match the correct answer.
Equal volumes of 0.020 M CaCl2 and 0.00040 M Na2SO4 are mixed.
For CaSO4 Ksp = 2.4 × 10–4.
Q = a×10b

a =

b =



 answer :

a =
= 2.0
b =
= –6
Response:
That's the correct answer

To 200 cm3 of 0.10 M HCl is added 300 cm3 of 0.20 M Pb(NO3)2.
For PbCl2 Ksp = 1.3 × 10–8.
Q = a×10b

a =

b =

answer :

a =
= 1.9
b =
= −4
Response:
That's the correct answer