NAME:
SCH 4U1 SOLUBILITY EQUILIBRIA
/11 KNOW
/33 INQ
/44 TOTAL
PART ONE: KNOWLEDGE (11 KNOW)
Multiple Choice (3 marks KNOW)
Identify the letter of the choice that best completes the statement or answers the question.
_E_
1.
When solid lead(II) phosphate is in equilibrium with its ions, the ratio of lead(II) ions to
phosphate ions is which of the following?
a.
1:1
d.
2:3
b.
1:2
e.
3:2
c.
2:1
_B_
2.
In a saturated solution of silver phosphate, the concentration of silver ions is 4.5 x 10-4
mol/L. The Ksp of silver phosphate would be which of the following?
a.
6.8 x 10-8
d.
1.0 x10-11
-14
b.
1.4 x 10
e.
none of the above
c.
1.5 x 10-15
_B_
3.
In a saturated solution of PbBr2, the concentration of Br1- is 8.2 x 10-2 mol/L. The [Pb2+]
in mol/L is which of the following?
a.
8.2 x 10-2
d.
0.24
b.
4.1 x 10-2
e.
none of the above
c.
0.16
Modified True/False (8 marks KNOW)
Indicate whether the sentence or statement is true or false. If false, change the identified word or phrase
to make the sentence or statement true.
_F_
1.
A precipitate forms when the trial ion product for the precipitate is less than the Ksp.
____more than____
_F_
2.
A solution of calcium chloride is mixed with a solution of sodium carbonate and a
precipitate forms. The precipitate is sodium chloride. __calcium carbonate______
_F
3.
Ksp values are temperature independent. __temperature dependent_
_F_
4.
The equilibrium equation for lead(II) chloride is Pb2Cl(s) <==> 2Pb2+(aq) + Cl1-(aq).
__ PbCl2(s) <==> Pb2+(aq) + 2 Cl1-(aq)____
_T_
5.
The Ksp's of CuCl and TlCl are 1.910-7 and 1.8  10-4 respectively. This means TlCl is
more soluble than CuCl. _________________________
_F_
6.
In a saturated solution of copper (II) hydroxide, the quantity of copper (II) ion = the
quantity of hydroxide ion. ___quantity of copper=1/2 quantity of hydroxide__
Problems ( marks INQ)
1.
If the solubility of Pb(IO3)2 is 4.0 x 10-5 mol/L, what is its Ksp? (3 marks INQ)
Pb(IO3)2 (s) ⇌ Pb2+(aq) + 2 IO3-(aq)
Ksp= [Pb2+][IO3-]2
x= 4.0 x 10-5
Ksp= [Pb2+][IO3-]2
= (x)(2x)2
= 4(4.0 x 10-5)3
Ksp=2.6x10-13
2.
If the concentration of Zn2+ is found to be 1.7x10-5 mol/L in a saturated solution of this salt,
what is the Ksp of Zn(OH)2? (3 marks)
Zn(OH)2 (s) ⇌ Zn2+(aq) + 2 OH-(aq)
x
2x
Ksp= [Zn2+][OH-]2
Ksp= (x)(2x)2
Ksp= 4(1.7 x 10-5)3
Ksp=2.0x10-14
3.
What mass of PbCl2 would be found in 0.48 L of a saturated solution of PbCl2 if the Ksp of
PbCl2 is 1.6 x 10-5? (4 marks)
PbCl2 (s) ⇌ Pb2+(aq) + 2 Cl-(aq)
Ksp= 1.6 x 10-5
x
2x
Ksp= [Pb2+][Cl-]2
1.6x10-5=(x)(2x)2
1.6x10-5=4x3
x=1.6x10-2
moles of PbCl2=(0.48 L)( 1.6x10-2 mol/L)
= 7.7 x10-3 mol
m=nMM
m=(7.7 x10-3 mol)(278.1 g/mol)
m=2.1 g
4.
If 845 mL of a 2.5x10-5 mol/L solution of Ni(NO3)2 was mixed with 195 mL of a 4.86x10-2
mol/L solution of KOH, would a precipitate form?. The Ksp of Ni(OH)2(s) is 6x10-16. (5
marks)
[Ni2+] = (0.845)(2.5x10-5)/1.04 L
[OH-] =(0.195)(4.6x10-2)/1.04 L
-5
= 2.0x10 M
= 8.6x10-3 M
2+
- 2
Q = [Ni ][OH ]
= (2.0x10-5 M)( 8.6x10-3 M)2
= 1.5x10-9
Q is greater than Ksp therefore a precipitate will form.
5.
Calculate the concentration of both ions and the molar solubility for lead (II) iodide
(Ksp = 7.1x10-9) dissolved in 0.15 M sodium iodide. (6 marks)
PbI2 (s) ⇌ Pb2+(aq) + 2 I-(aq)
MR 1
2
I 0
0.15
C +x
2x
E x
0.15 + 2x
Ksp= [Pb2+][I-]2
7.1x10-9=(x)(0.15+2x)2
7.1x10-9= x(0.15)2
x=3.2x10-7
Therefore [Pb2+] = 3.2x10-7 M [I-] = 0.15 M
Ksp= 7.1 x 10-9
0.15/Ksp>1000
Approx works
6.
A 150.0 mL sample of 0.25 M magnesium nitrate solution reacts with 225.0 mL of 0.25M
sodium fluoride. Given the Ksp for magnesium fluoride is 7.9x10-8, calculate
A) The concentrations of all ions at equilibrium (8 marks)
B) The mass of the precipitate formed in the solution at the end of the reaction (4 marks)
Mg(NO3)2(aq) + 2 NaF (aq)  MgF2(s) + 2 NaNO3(aq)
c
0.25
0.25
V
0.150
0.225
n
0.038
0.056
LF 0.038/1= 0.038
0.056/2 =0.028
Limiting
Excess Mg(NO3)2 = 0.038-0.028 (amount used)
= 0.010 mol
Initial [Mg2+] = 0.010 mol/0.375 L = 0.027 M
MR
I
C
E
MgF2 (s) ⇌ Mg2+(aq) + 2 F-(aq)
1
1
2
--0.027
0
--+x
+2x
--0.027+x
2x
Ksp= 7.9 x 10-8
0.15/Ksp>1000
Approx works
Ksp= [Mg2+][F-]2
7.9x10-8=(0.027+x)(2x)2
7.9x10-88= 0.027(4)x2
x=8.6x10-4
a) Therefore [Mg2+] = 0.028 M [F-] = 1.7 x10-3 M [NO3-]= 2(0.15 L)(0.25 M)/0.375 L = 0.20 M
[Na+]= (0.25 M)(0.225 L)/0.375 L = 0.15 M
b) n of MgF2 made = 0.028 mol
n of ppt = 0.028 – 3.2x10-4
= 0.028 mol
m = n MM
= (0.028 mol)(62.31 g/mol)
= 1.7 g
n dissolved = (8.6x10-4 M)(0.375 L) = 3.2x10-4 mol