# Chapter 9 Ionic Equilibria and Buffer Action¶

## Example 9.1 , Page no:182¶

In [1]:
import math
from __future__ import division

#initialisation of variables
c= 0.1 #M
p= 1.34 #per cent
T= 25 #C

#CALCULATIONS
C1= c*p/100
C2= c*p/100
C3= c-C1
Ka= C1*C2/C3

#RESULTS
print"ionization constant =",'%.2E'%Ka;

ionization constant = 1.82E-05


## Example 9.2 , Page no:184¶

In [2]:
import math
from __future__ import division

#initialisation of variables
k= 1.8*10**-5
C= 0.2 #M
T= 25 #C

#CALCULATIONS
x= math.sqrt(C*k)
a= x/C
C1= a*C

#RESULTS
print"hydronium-ion concentration =",'%.2E'%C1,"mole per litre";

hydronium-ion concentration = 1.90E-03 mole per litre


## Example 9.3 , Page no:185¶

In [3]:
import math
from __future__ import division

#initialisation of variables
K= 1.8*10**-5
V= 500 #ml
c1= 0.3 #M
c2= 0.2 #M

#CALCULATIONS
x= V*c1/1000
y= V*c2/1000
C= K*y/x

#RESULTS
print"hydronium-ion concentration =",'%.2E'%C,"mole per litre";

hydronium-ion concentration = 1.20E-05 mole per litre


## Example 9.4 , Page no:186¶

In [4]:
import math
from __future__ import division

#initialisation of variables
K= 1.4*10**-5
T= 25 #C
V= 200 #ml
m= 3.7 #gms
m1= 4.8 #gms
M= 74 #gms
M1= 96 #gms

#CALCULATIONS
x= m*1000/(V*M)
y= m1*1000/(V*M1)
X= K*x/y

#RESULTS
print"hydronium-ion concentration =",'%.2E'%X,"mole per litre";

hydronium-ion concentration = 1.40E-05 mole per litre


## Example 9.5 , Page no:188¶

In [5]:
import math
from __future__ import division

#initialisation of variables
c= 0.050 #M
Ksp= 4.3*10**-7

#CALCULATIONS
C= math.sqrt(Ksp*c)

#RESULTS
print"concentration of hydronium-ion =",'%.1E'%C,"mole per litre";

concentration of hydronium-ion = 1.5E-04 mole per litre


## Example 9.6 , Page no:189¶

In [6]:
import math
from __future__ import division

#initialisation of variables
C= 0.050 #M
K= 2.4*10**-17
c= 0.1 #M

#CALCULATIONS
c1= K*C/c**2

#RESULTS
print"concentration of carbonate-ion =",'%.1E'%c1,"mole per litre";

concentration of carbonate-ion = 1.2E-16 mole per litre


## Example 9.7 , Page no:192¶

In [7]:
import math
from __future__ import division

#initialisation of variables
n= 1.31*10**-4 #mole
T= 25 #C

#CALCULATIONS
N= 2*n
Ksp= N**2*n

#RESULTS
print"Ksp =",'%.1E'%Ksp;

Ksp = 9.0E-12


## Example 9.8 , Page no:193¶

In [8]:
import math
from __future__ import division

#initialisation of variables
Ksp= 1.4*10**-11
V= 200 #ml
M= 24.3 #/gms

#CALCULATIONS
x= (Ksp/4)**(1/3)
m= x*M*V/1000

#RESULTS
print"grams of Mg+2 present =",'%.1E'%m,"gms per mol";

grams of Mg+2 present = 7.4E-04 gms per mol


## Example 9.9 , Page no:193¶

In [9]:
import math
from __future__ import division

#initialisation of variables
c= 0.010 #M
Ksp= 1.56*10**-10
M= 108 #gms
C= 10**-3 #M

#CALCULATIONS
K= Ksp/C
m= M*K
m1= M*c

#RESULTS
print"quantity =",'%.2E'%m,"gms";
print"quantity =",round(m1,2),"gms";

quantity = 1.68E-05 gms
quantity = 1.08 gms


## Example 9.10 , Page no:199¶

In [10]:
import math
from __future__ import division

#initialisation of variables
c= 0.1 #M
Kb= 1.8*10**-5
Kw= 10**-14

#CALCULATIONS
C= math.sqrt(c*Kw/Kb)

#RESULTS
print"concentration of hydronium ion =",'%.2E'%C,"mol per litre";

concentration of hydronium ion = 7.45E-06 mol per litre


## Example 9.11 , Page no:201¶

In [11]:
import math
from __future__ import division

#initialisation of variables
c= 0.050 #M
Kb= 1.8*10**-5
T= 25 #C
Kw= 10**-14

#CALCULATIONS
C= math.sqrt(Kw*c/Kb)

#RESULTS
print"concentration of hydronium ion =",'%.2E'%C,"mol per litre";

concentration of hydronium ion = 5.27E-06 mol per litre


## Example 9.12 , Page no:203¶

In [12]:
import math
from __future__ import division

#initialisation of variables
kw= 10**-14
Ka= 1.8*10**-5

#CALCULATIONS
Kb= Ka
B= math.sqrt(kw/(Ka*Kb))

#RESULTS
print"degree of hydrolysis =",'%.2E'%B;

degree of hydrolysis = 5.56E-03


## Example 9.13 , Page no:205¶

In [13]:
import math
from __future__ import division

#initialisation of variables
k1= 3.5*10**-7
k2= 4.4*10**-11

#CALCULATIONS
c= math.sqrt(k1*k2)

#RESULTS
print"concentration of solution =",'%.2E'%c,"mol per litre";

concentration of solution = 3.92E-09 mol per litre


## Example 9.14 , Page no:207¶

In [14]:
import math
from __future__ import division

#initialisation of variables
c= 1.92*10**-5 #mole per litre

#CALCULATIONS
pH= -math.log10(c)

#RESULTS
print"pH of solution =",round(pH,2);

pH of solution = 4.72


## Example 9.15 , Page no:208¶

In [15]:
import math
from __future__ import division

#initialisation of variables
pH= 7.36

#CALCULATIONS
C= 10**-pH

#RESULTS
print"concentration of solution =",'%.2E'%C,"mol per litre";

concentration of solution = 4.37E-08 mol per litre


## Example 9.16 , Page no:210¶

In [16]:
import math
from __future__ import division

#initialisation of variables
c= 1 #M
Kb= 5.3*10**-5
pKw= 14

#CALCULATIONS
pH= pKw+0.5*math.log10(Kb)+0.5*math.log10(c)

#RESULTS
print"pH of solution =",round(pH,2);

pH of solution = 11.86


## Example 9.17 , Page no:211¶

In [17]:
import math
from __future__ import division

#initialisation of variables
c= 0.1 #M
Ka= 6.3*10**-5
pKw= 14
#CALCULATIONS
pH= -0.5*math.log10(Ka)+0.5*pKw+0.5*math.log10(c)

#RESULTS
print"pH of a buffer solution =",round(pH,2);

pH of a buffer solution = 8.6


## Example 9.18 , Page no:214¶

In [18]:
import math
from __future__ import division

#initialisation of variables
Ka= 1.8*10**-5
a= 0.1 #molar

#CALCULATIONS
pH= -math.log10(Ka)

#RESULTS
print"pH of a buffer solution =",round(pH,2);

pH of a buffer solution = 4.74


## Example 9.19 , Page no:214¶

In [19]:
import math
from __future__ import division

#initialisation of variables
pH= 7.10
pH1= 7.21

#CALCULATIONS
r= 10**(pH-pH1)

#RESULTS
print"ratio of salt to acid =",round(r,2);

ratio of salt to acid = 0.78