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Chemistry 1st
paper
C2: Qualitative chemistry
Mathematical problems
problems: electron of hydrogen atom moved to 2nd energy level from 4th(3rd line of Balmer
series). calculate-
(i) wave number (ii) wave length (iii)frequency (iv) energy of radiation
soln: we know,
)
11
(
1
2
2
2
1 nn
RH 

)
4
1
2
1
(10097.1 22
7

16
3
10097.1 7

wave number,

1
= 2056875 m-1
wave length,  = 4.86×10-7
m
Frequency calculation:


c

7
8
1086.4
103




= 6.17×1014
Hz
energy of radiation:
hE 
1434
1017.6106026.6  
J19
1008.4 

 solubility calculation:

1. at 350
c temperature solubility product of AgCl is 2.458×10-10
. calculate the solubility.
soln: dissociation of AgCl
)()()( aqClaqAgsAgCl 

we know,
]][[ 
 ClAgKsp
2
S
SS


so, KspS 
15
10
1056.1
10458.2




molL
2. at 250
c temperature solubility product of Fe(OH)3 is 2.69×10-38
. calculate the solubility of
Fe3+
.
soln: dissociation of Fe(OH)3 :
)(3)()()( 3
3 aqOHaqFesOHFe 

we know,
33
]][[ 
 OHFeKsp
4
3
27
)3(
S
SS


so, 4
27
Ksp
S 
110
34
1077.1
27
1069.2





molL
3. In different solution: (common ion effect)
at 350
c temperature solubilty product of AgCl is 2.458×10-10
. calculate the solubility in 0.01M NaCl
solution.
soln; dissociation of AgCl :


we know ,
]][[ 
 ClAgKsp
SS
SS
01.0
)01.0(
2


2
S can be ignored,
so that,
18
10
10458.2
01.0
10458.2
01.0
01.0







molL
Ksp
S
SKsp
 solubility product calculation:
1. at 350
c solubility of AgCl 1.22 ×10-5
molL-1
. Calculate the solubility product.
soln: dissociation of AgCl :

we know,
]][[ 
 ClAgKsp
2210
25
2
1048.1
)1022.1(






Lmol
S
SS
2. at 250
c temperature solubility of CaF2 is 1.34 ×10-6
molL-1
. Calculate the solubility product.
soln: dissociation of CaF2 :
)(2)()( 2
2 aqFaqCasCaF 

we know,
22
]][[ 
 FCaKsp

3318
36
3
2
106.9
)1034.1(4
4
)2(






Lmol
S
SS
 relation between solubility product and ionic product:
Ksp = Ki solution is saturated, no ppt occurred
Ksp > Ki solution is unsaturated, no ppt occurred
Ksp < Ki solution is over saturated, ppt occurred
problem:
1. at 250
c temperature solubility product of CaF2 1.69 ×10-11
Q. will ppt form in the mixed solution of i and ii?
soln: dissociation of CaF2 :
)(2)()( 2
2 aqFaqCasCaF 

14
4
2
1066.1
60
105.240
][






molL
Ca
14
3
104
60
102.120
][






molL
F
here,
11
244
22
1065.2
)104(1066.1
]][[





 FCaKi
since Ki>Ksp so ppt will form in the mixed solution.
2.5×10-4
M
40ml
CaCl2
1.2×10-3
M
20ml
NaF

2. 40ml 1.3×10-06
M Ba(NO3)2 and 20ml 1.0×10-02
M Na2SO4 solution mixed together. will ppt formin their
mixed solution? solubilty product of BaSO4 6.3×10-14
.
soln: here,
9
37
2
4
2
13
2
2
4
17
6
2
1088.2
1033.31066.8
]][[
1033.3
60
100.120
][
1066.8
60
103.140
][


















SOBaK
now
moll
SO
moll
Ba
i
since Ki>Ksp so ppt will form in the mixed solution.
 mathematical equation:
100


mM
m
S
here,
S = solubilty
m = mass of solute
M = mass of solution
M-m = mass of solvent
1. at 250
c temperature 30g of NaCl soluted in 75g of solution.calculate the solublity.
soln: we know,
66.66
100
3075
30
100







mM
m
S

2. at 250
c temperature solubility of KNO3 is 40. In 200g solution how much amount of
solid present?
soln: we know,
100
200
40
100






m
m
mM
m
S
gm
m
m
mm
14.57
140
8000
8000140
408000100




3. at 250
c and 350
c temperature solubility of KNO3 are respectively 40 and 60.
temperature of 200g solution changed 250
c to 350
c. How much gm of solute can be
added more?
soln: at 250
c temperature,
we know,
100
200
40
100






m
m
mM
m
S
gm
m
m
mm
14.57
140
8000
8000140
408000100




mass of solvent
gm86.142
)14.57200( 
again at 350
c temperature,
100
86.142
60
100




m
mM
m
S

71.85
6.8571100


m
m
mass of extra solute = )14.5771.85( 
=28.57g
4. at 250
c and 350
c temperature solubility of KNO3 are respectively 40 and 60.
temperature of 200g solution changed 350
c to 250
c. How much gm of solute will be precipited?
soln: at 350
c temperature,
we know,
100
200
60
100






m
m
mM
m
S
gm
m
m
mm
75
160
12000
12000160
6012000100




mass of solvent,
gm125
)75200( 
again at 250
c temperature,
100
125
60
100




m
mM
m
S
gm
m
50
5000100


amount of ppt = )5075( 
=25 g

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