1. The document provides solutions to mathematical problems involving calculations of wave number, wavelength, frequency, and energy of radiation emitted when an electron in a hydrogen atom moves between energy levels. 2. It also includes calculations of solubility, solubility product, and whether precipitation will occur for various salts (AgCl, Fe(OH)3, CaF2, BaSO4) at different temperatures and concentrations. 3. The relationships between solubility, solubility product, and the ionic product constant K are explained in determining precipitation. Equations for calculating solubility from mass and concentration are also provided.

1. 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:

2. 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 :
)()()( aqClaqAgsAgCl 


3. 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 :
)()()( aqClaqAgsAgCl 

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

4. 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

5. 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

6. 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

7. 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