Solubility Equilibrium
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Q: 1 2 3
4 5
6 7 8 9
10 11

1

Silver acetate is a sparingly soluble salt. How is its solubility
affected by the addition of the following substances to a saturated
solution containing some solid? Write "increased", "decreased",
or "no change". Give a justification for your answer in
each case.
(1.) Sodium acetate
(2.) Sodium chloride
(3.) Nitric acid
(4.) Potassium cyanide


2 
At 298 K the water solubility
of magnesium hydroxide is 1.26 x 10^{2} gram litre^{1}.
Calculate the solubility product constant of magnesium hydroxide
at this temperature.


3 
At a certain temperature the solubility of lead(II) sulfate in water is 1.25 x 10^{4} M. Calculate the solubility product constant of lead(II) sulfate at this temperature. 

4 
The concentration of Ag^{+} ion in a saturated solution of silver oxalate (Ag_{2}C_{2}O_{4}) at a certain temperature is 2.2 x 10^{4} M. Calculate the solubility product constant of silver oxalate at this temperature.


5 
Using K_{sp} values, calculate the molar solubility
of both salts in each of the following pairs of insoluble salts.
Also state which is the more soluble salt of each pair.
(1.) Fe(OH)_{2}, Fe(OH)_{3}
(2.) Ag_{2}SO_{4}, SrCrO_{4}


6 
Find the solubility at 298 K of PbSO_{4} in:
(1.) pure water
(2.) 0.10 M lead(II) nitrate
(3.) 0.10 M sodium sulfate


7 
At 298 K the solubility of barium sulfate in water is 1 x 10^{5} M. What is its solubility in 0.1 M potassium sulfate at 298 K?


8 
For each of the following, decide whether or not a precipitate could be expected to form at 298 K:
(1.) Silver nitrate (0.0050 g) is added to 0.0010 M NaCl (2.0 litre)
(2.) Calcium nitrate (2.0 x 10^{3} g) and sodium carbonate (6.0 x 10^{3} g) are dissolved in water (500 ml).
(3.) 0.0010 M magnesium nitrate (50 ml) is added to 0.0010 M sodium hydroxide (200 ml).


9 
A metal from Periodic
Group 2 is converted to the hydroxide by reaction with water. The
saturated solution of this hydroxide has a pH of 9.60. What is the
concentration of metal ions in this solution?


10 
The [Ag^{+}]
of a solution is 4.0 x 10^{3} M. Calculate the [Cl^{}]
that must be exceeded before silver chloride can precipitate at 298
K. 

11 
(1.) A concentrated water solution of sodium carbonate is added slowly with stirring to a solution that contains 0.10 M calcium ion and 0.10 M magnesium ion. What is the first solid to precipitate?
(2.) Neglecting dilution, what is the concentration of carbonate ion in the reaction mixture at the instant when a precipitate first appears?
(3.) Neglecting dilution, what is the concentration of calcium ion in solution at the instant when a precipitate of magnesium carbonate first appears?


Solubility
Equilibrium (Answers) 

1

The dissolution of silver
acetate in water can be described as an equilibrium:
CH_{3}COOAg(s)
CH_{3}COO^{}(aq) + Ag^{+}(aq)
Like other equilibria
it is subject to LeChatelier's principle, which qualitatively describes
the effect of concentration changes in the system. There are two
possible ways an added species can effect solubility. If one of
the species present in the dissolution equation is added it will
have a direct effect on the equilibrium, eg. part (1.).
Alternatively, if an added
species can react with one of the species present in the
dissolution equilibrium there may be an indirect effect on the dissolution
equilibrium (eg. parts 2, 3 and 4). In either case, the direction
the dissolution equlibrium is forced will determine if solubility
is increased or decreased.
(1.)
The addition of sodium acetate will result in a decrease
in solubility. It is evident from the dissolution equation that
increasing [CH_{3}COO^{}] will shift the equilibrium
to the left, hence favouring solid CH_{3}COOAg (ie an effective
decrease in solubility). This is known as the common ion effect.
(2.)
AgCl is less soluble than CH_{3}COOAg, so Cl^{}
withdraws Ag^{+} from solution more than CH_{3}COO^{} ^{
}
Ag^{+}(aq) +
Cl^{}(aq) AgCl(s)
Referring to the dissolution
equation, removing Ag^{+} ions will shift the equilibrium
to the right, hence increasing the solubility of CH_{3}COOAg.
(3.)
Nitric acid reacts with the acetate ion
to produce acetic acid:
H^{+} + CH_{3}COO^{}
® CH_{3}COOH
Acetic acid is a weak
acid, so does not dissociate to H^{+} and CH_{3}COO^{}
to a great extent. As a result, the addition of nitric acid effectively
reduces [CH_{3}COO^{}], hence shifting the equilibrium
to the right, increasing the solubility of CH_{3}COOAg.
(4.)
The cyanide ion reacts with Ag^{+} to form the dicyanoargentate(I)
ion:
Ag^{+} + 2CN^{}
[Ag(CN)_{2}]^{} ^{
}
This reaction lowers
[Ag^{+}], and hence forces the equilibrium to the right,
increasing the solubility of CH_{3}COOAg.


2 
Mg(OH)_{2}
dissolves in water to a small extent according to the following equation:
Mg(OH)_{2}(s)
Mg^{2+} + 2OH^{} ^{
}
K_{SP} Mg(OH)_{2}
= [Mg^{2+}] [OH^{}]^{2} ^{
}
Note that in the expression
for K_{SP}, [OH^{}] is raised to the power
of 2 since two OH^{} ions are produced as the equation
is written.
No. mol Mg(OH)_{2}
in 1.26 x 10^{2} g = 1.26 x 10^{2} g / 58.33 g
mol^{1} = 2.16 x 10^{4} mol.
[Mg^{2+}] = 2.16
x 10^{4} mol L^{1} ^{
}
[OH^{}] = 4.32
x 10^{4} mol L^{1} ^{
}
\
K_{SP}
= (2.16 x 10^{4} mol L^{1}) x (4.32 x 10^{4}
mol L^{1})^{2} = 4.03 x 10^{11} mol^{3}
L^{3} ^{
} 

3 
PbSO_{4} dissolves
to a small extent in water according to the following equation:
PbSO_{4}(s)
Pb^{2+} + SO_{4}^{2} ^{
}
K_{SP} PbSO_{4}
= [Pb^{2+}] [SO_{4}^{2}]
No. mol PbSO_{4}
in 1.25 x 10^{4} mol
[Pb^{2+}] = 1.25
x 10^{4} mol L^{1} ^{
}
[SO_{4}^{2}]
= 1.25 x 10^{4} mol L^{1} ^{
}
\
K_{SP}
= (1.25 x 10^{4} mol L^{1}) x (1.25 x 10^{4}
mol L^{1}) = 1.56 x 10^{8} mol^{2} L^{2} ^{
} 

4

Ag_{2}C_{2}O_{4}
dissolves in water to a small extent according to the following
equation:
Ag_{2}C_{2}O_{4}(s)
2Ag^{+} + C_{2}O_{4}^{2} ^{
}
K_{SP} =
[Ag^{+}]^{2} [C_{2}O_{4}^{2}]
From the above equation
it is evident that [C_{2}O_{4}^{2}] = 0.5
x [Ag^{+}]. Substituting into the equation for K_{SP}:
K_{SP} =
[Ag]^{2} x 0.5 x [Ag^{+}] = (2.2 x 10^{4}
mol L^{1})^{2} x 0.5 x (2.2 x 10^{4} mol
L^{1}) = 5.3 x 10^{12} mol^{3} L^{3} ^{
} 

5 
(1.)
Fe(OH)_{2}(s) Fe^{2+} + 2OH^{} ^{
}
From the data sheet,
K_{SP} Fe(OH)_{2} = 8 x 10^{16}
= [Fe^{2+}] [OH^{}]^{2} ^{
}
Let the molar solubility
of Fe(OH)_{2} be x.
Now [Fe^{2+}]
= x, [OH^{}] = 2x. These figures are arrived
at by considering that one mole Fe(OH)_{2} dissolves to
give one mole Fe^{2+} and two moles OH^{}.
\
K_{SP}
= (x)(2x)^{2} = 4x^{3} = 8
x 10^{16} mol^{3} L^{3} ^{
}
\
Molar solubility
of Fe(OH)_{2} = x = 6 x 10^{6} mol L^{1} ^{
}
Fe(OH)_{3}(s)
Fe^{3+} + 3OH^{} ^{
}
From the data sheet,
K_{SP} Fe(OH)_{3} = 4 x 10^{40}
= [Fe^{3+}] [OH^{}]^{3} ^{
}
Let the molar solubility
of Fe(OH)_{3} be x.
Now [Fe^{2+}]
= x, [OH^{}] = 3x
\
K_{SP}
= (x)(3x)^{3} = 27x^{4} = 4
x 10^{40} mol^{4} L^{4} ^{
}
\
Molar solubility
of Fe(OH)_{2} = x = 6 x 10^{11} mol L^{1} ^{
}
Therefore Fe(OH)_{2}
is the more soluble salt.
(2.)
Ag_{2}SO_{4}(s) 2Ag^{+} + SO_{4}^{2} ^{
}
From the data sheet:
K_{SP} Ag_{2}SO_{4} = 2 x 10^{5}
= [Ag^{+}]^{2} [SO_{4}^{2}]
Let the molar solubility
of Ag_{2}SO_{4} be x.
Now [Ag^{+}]
= 2x, [SO_{4}^{2}] = x
\
K_{SP}
= (2x)^{2}(x) = 4x^{3} = 2
x 10^{5} mol^{3} L^{3} ^{
}
\
Molar solubility
of Ag_{2}SO_{4} = x = 2 x 10^{2}
mol L^{1} ^{
}
SrCrO_{4}(s)
Sr^{2+} + CrO_{4}^{2} ^{
}
From the data sheet:
K_{SP} SrCrO_{4} = 2 x 10^{5} =
[Sr^{2+}] [CrO_{4}^{2}]
Let the molar solubility
of SrCrO_{4} be x.
Now [Sr^{2+}]
= x, [CrO_{4}^{2}] = x
\
K_{SP}
= x^{2} = 2 x 10^{5} mol^{2} L^{2} ^{
}
\
Molar solubility
of Ag_{2}SO_{4} = x = 4 x 10^{3}
mol L^{1} ^{
}
Ag_{2}SO_{4}
is the more soluble.


6 
(1.)
PbSO_{4}(s) Pb^{2+} + SO_{4}^{2} ^{
}
From the data sheet,
K_{SP} PbSO_{4} = 2 x 10^{8} = [Pb^{2+}]
[SO_{4}^{2}]
Let the molar solubility
of PbSO_{4} be x
Now [Pb^{2+}]
= x, [SO_{4}^{2}] = x
\
K_{SP}
= x^{2} = 2 x 10^{8} mol^{2} L^{2} ^{
}
\
Molar solubility
of PbSO_{4} = x = 1 x 10^{4} mol L^{1} ^{
}
(2.)
The addition of Pb(NO_{3})_{2}
increases [Pb^{2+}], and effects the equilibrium (the common
ion effect).
K_{SP} PbSO_{4}
= 2 x 10^{8} = [Pb^{2+}] [SO_{4}^{2}]
Upon addition of Pb(NO_{3})_{2},
[Pb^{2+}] equals 0.10 mol L^{1} (this is ignoring
the relatively tiny contribution from the dissolved lead sulfate).
Again, we let the molar
solubility of PbSO_{4} = x
Now [Pb^{2+}]
= 0.10 mol L^{1}, [SO_{4}^{2}] = x
K_{SP} =
(0.10 mol L^{1}) x (x) = 2 x 10^{8} mol^{2}
L^{2} ^{
}
\
Molar solubility
of PbSO_{4} = x = 2 x 10^{7} mol L^{1} ^{
}
Note that this figure
is 1 / 500 the value for pure water, due to the common ion effect.
(3.)
The addition of Na_{2}SO_{4}
increases [SO_{4}^{2}] and affects the equilibrium.
K_{SP} PbSO_{4}
= 2 x 10^{8} = [Pb^{2+}] [SO_{4}^{2}].
Upon addition of Na_{2}SO_{4}
, [SO_{4}^{2}] equals 0.10 mol L^{1} (this
is ignoring the relatively tiny contribution from the dissolved
lead sulfate).
Again, we let the molar
solubility of PbSO_{4} = x.
Now [Pb^{2+}]
= x, [SO_{4}^{2}] = 0.10 mol L^{1} ^{
}
K_{SP} =
(x) x (0.10 mol L^{1}) = 2 x 10^{8} mol^{2}
L^{2} ^{
}
\
Molar solubility
of PbSO_{4} = x = 2 x 10^{7} mol L^{1} ^{
} 

7 
K_{SP} BaSO_{4}
= 1 x 10^{10} mol^{2} L^{2} = [Ba^{2+}]
[SO_{4}^{2}]
Let the molar solubility
of BaSO_{4} be x.
[Ba^{2+}] = x
[SO_{4}^{2}]
= 0.10 M
K_{SP} =
(x) x (0.10 M)
\
Molar solubility
of BaSO_{4} = x = 1 x 10^{8} mol L^{1} ^{
} 

8 
(1.)
AgNO_{3} + NaCl AgCl +
NaNO_{3} _{
}
Total volume of solution
= 2.0 L
[Cl^{}] = 0.0010
mol L^{1} ^{
}
No. mol AgNO_{3}
= 0.0050 g / 169.91 g mol^{1} = 2.94 x 10^{5}
mol
\
[Ag^{+}]
= 1.47 x 10^{5} mol L^{1} ^{
}
K_{SP} AgCl
= [Ag^{+}] [Cl^{}] = 2 x 10^{10} mol^{2}
L^{2} from data sheet.
In the solution, [Ag^{+}]
[Cl^{}] = (1.47 x 10^{5} mol L^{1}) (0.0010
mol L^{1}) = 1.47 x 10^{8} mol^{2} L^{2}.
This value is greater
than the data sheet value of K_{SP}, so a precipitate
will form.
(2.)
Ca(NO_{3})_{2} + Na_{2}CO_{3}
CaCO_{3} + 2NaNO_{3} _{
}
Total volume of solution
= 500 mL
No. mol Ca(NO_{3})_{2}
= 2.0 x 10^{3} g / 164.12 g mol^{1} = 1.22 x 10^{5}
mol
\
[Ca^{2+}]
= 2.44 x 10^{5} mol L^{1} ^{
}
No. mol Na_{2}CO_{3}
= 6.0 x 10^{3} g / 105.99 g mol^{1} = 5.66 x 10^{5}
mol
\
[CO_{3}^{2}]
= 1.13 x 10^{4} mol L^{1} ^{
}
K_{SP} CaCO_{3}
= [Ca^{2+}] [CO_{3}^{2}] = 5 x 10^{9}
mol^{2} L^{2} from data sheet.
In the solution, [Ca^{2+}]
[CO_{3}^{2}] = (2.44 x 10^{5} mol L^{1})
(1.13 x 10^{4} mol L^{1}) = 2.76 x 10^{9}
mol^{2} L^{2}.
This value is less than
the data sheet value of K_{SP}, so a precipitate
will not form.
(3.)
Mg(NO_{3})_{2} + 2NaOH Mg(OH)_{2}
+ 2NaNO_{3} _{
}
Total volume of solution
= 250 mL, so the concentration of each species becomes:
[Mg^{2+}] = 2
x 10^{4} mol L^{1} ^{
}
[OH^{}] = 8
x 10^{4} mol L^{1} ^{
}
K_{SP} Mg(OH)_{2}
= [Mg^{2+}] [OH^{}]^{2} = 1 x 10^{11}
mol^{3} L^{3} from data sheet.
In the solution, [Mg^{2+}]
[OH^{}]^{2} = (2 x 10^{4} mol L^{1})
(8 x 10^{4} mol L^{1})^{2} = 1.28 x 10^{10}
mol^{2} L^{2}.
This value is more than
the data sheet value of K_{SP}, so a precipitate
will form.


9 
Since the metal is from
group 2, it must have a valency of 2. Therefore the hydroxide will
have the formula M(OH)_{2}, using M to denote the symbol
for the element.
pH = 9.60
\
pOH = 14.00  9.60
= 4.40
[OH^{}] = 10^{4.40}
= 3.98 x 10^{5} mol L^{1} ^{
}
From the empirical formula
we know there is one M^{2+} ion for every two OH^{}
ions,
\
[M^{2+}]
= [OH^{}] / 2 = 2.0 x 10^{5} mol L^{1} ^{
} 

10 
K_{SP}(AgCl)
= 2 x 10^{10} = [Ag^{+}] [Cl^{}]
[Ag^{+}] = 4.0
x 10^{3} mol L^{1} ^{
}
Minimum [Cl^{}]
required for precipitate to form = K_{SP} / [Ag^{+}]
= 2 x 10^{10} mol^{2} L^{2} / 4.0 x 10^{3}
mol L^{1} = 5 x 10^{8} mol L^{1} ^{
}
Therefore, when [Cl^{}]
exceeds 5 x 10^{8} mol L^{1} AgCl will precipitate.


11 
(1.)
K_{SP}(CaCO_{3}) = 5 x 10^{9} mol^{2}
L^{2} = [Ca^{2+}] [CO_{3}^{2}]
[Ca^{2+}] = 0.10
mol L^{1} ^{
}
Minimum [CO_{3}^{2}]
required for precipitate to form = K_{SP} / [Ca^{2+}]
= 5 x 10^{9} mol^{2} L^{2} / 0.10 mol
L^{1} = 5 x 10^{8} mol L^{1} ^{
}
When [CO_{3}^{2}]
exceeds 5 x 10^{8} mol L^{1} CaCO_{3}
will precipitate.
K_{SP} MgCO_{3}
= 1 x 10^{5} mol^{2} L^{2} = [Mg^{2+}]
[CO_{3}^{2}]
[Mg^{2+}] = 0.10
mol L^{1} ^{
}
Minimum [CO_{3}^{2}]
required for recipitate to form = K_{SP} / [Mg^{2+}]
= 1 x 10^{5} mol^{2} L^{2} / 0.10 mol
L^{1} = 1 x 10^{4} mol L^{1}.
When [CO_{3}^{2}]
exceeds 1 x 10^{4} mol L^{1} MgCO_{3}
will precipitate.
Therefore CaCO_{3}
will precipitate first.
Since both MgCO_{3}
and CaCO_{3} share the same basic MCO_{3} formula,
this qualitative result could have been determined by simply noting
that the solubility product of MgCO_{3} is larger than that
of CaCO_{3}.
(2.)
From the above working, [CO_{3}^{2}] = 5 x 10^{8}
mol L^{1} at the instant that precipitation begins.
(3.)
K_{SP} CaCO_{3} = [Ca^{2+}] [CO_{3}^{2}]
= 5 x 10^{9} mol^{2} L^{2} ^{
}
When MgCO_{3}
starts precipitating, [CO_{3}^{2}] = 1 x 10^{4}
mol L^{1} ^{
}
\
[Ca^{2+}]
= K_{SP} CaCO_{3} / [CO_{3}^{2}]
= 5 x 10^{9} mol^{2} L^{2} / 1 x 10^{4}
mol L^{1} = 5 x 10^{5} mol L^{1}.

