Chapter15 Ionic Equilibria

Example 15.1,Page no.62

In [3]:
import math
#initialisation of variables
c=8*10**-5 # molar
n=2.0
#CALCULATIONS
Ksp=c**3*n**2
#RESULTS
Ksp=Ksp*10**15
x=Ksp
print 'solubility product =',Ksp,'* 10**-15'
print 'solubility=',x,'* 10**6'

solubility product = 2048.0 * 10**-15
solubility= 2048.0 * 10**6

Example 15.2,Page no.62

In [2]:
import math
#initialisation of variables
Ksp=2*10**-12 
M=8.84*10** -5 # molar 
n=2.0
#CALCULATIONS 
r=(Ksp/(n**2*M**3))**(1/3) 
#RESULTS 
print 'mean ionic activity cooeficient =',r

mean ionic activity cooeficient = 1.0

Example 15.3,Page no.63

In [3]:
import math
#initialisation of variables
from math import sqrt
n=2.0 
C=0.01 #M
#CALCULATIONS 
r=10**(-0.509*n*sqrt(C)) 
r=round(r,3)
#RESULTS 
print 'mean ionic activity coeficient =',r

mean ionic activity coeficient = 0.791

Example 15.4,Page no.63

In [17]:
import math
#initialisation of variables
M=18 #gms
k=5.5*10** -8 #ohmˆ−1 cmˆ−1
lc=349.8 #cmˆ2 equivˆ−1 ohmˆ−1
la=198 #cmˆ2 equivˆ−1 ohmˆ−1
#CALCULATIONS
A=M*k
A0= lc+la 
a=A/A0
a1= 1000*a/M 
Kw=a1*a1
#RESULTS
a1=a1* 10**5
a1=round(a1,2)
Kw=Kw* 10**10
Kw=round(Kw,4)
print 'degree of ionisation =',a1,'* 10**-5'
print 'ion product of water =',Kw,'* 10**-10'

degree of ionisation = 0.01 * 10**-5
ion product of water = 0.0001 * 10**-10

Example 15.5,Page no.64

In [5]:
import math
#initialisation of variables
Ka=1.772*10** -4 
#CALCULATIONS 
pK=-math.log10(Ka) 
#RESULTS
pK=round(pK,3)
print 'pKa =',pK

pKa = 3.752

Example 15.6,Page no.64

In [15]:
import math
#initialisation of variables
K=1.75*10** -5 
c=0.01 #M 
#CALCULATIONS 
r=10**( -0.509*sqrt(c)) 
Ka=K/r**2 
#RESULTS 
print 'ionisation constant =',format(Ka, '.8f')

ionisation constant = 0.00002212

Example 15.7,Page no.64

In [7]:
import math
#initialisation of variables
from math import sqrt
ka=1.75*10** -5 
ca=0.1 #mole lit 
#CALCULATIONS
pH=-math.log10(sqrt(ka*ca)) 
#RESULTS 
pH=round(pH,3)
print 'pH =',pH

pH = 2.878

Example 15.8,Page no.65

In [8]:
import math
#initialisation of variables
kw=10**-14
ka=2.69*10** -5
c=0.1 #N
#CALCULATIONS
pH=-math.log10(sqrt(kw*ka/c))
#RESULTS
pH=round(pH,3)
print 'pH =',pH

pH = 8.785

Example 15.9,Page no.65

In [9]:
import math
#initialisation of variables
pH=4.57
M=0.03 #mole litre ˆ−1
M1=0.1 #mole litre ˆ−1
#CALCULATIONS
pH1=pH+math.log10(M1/M)
#RESULTS
pH1=round(pH1,3)
print 'pH=',pH1

pH= 5.093

Example 15.10,Page no.65

In [10]:
import math
#initialisation of variables
pH=9.26
M=0.02 #N
M1=0.01 #/N
#CALCULATIONS
pH1=pH+math.log(M1/M)
#RESULTS
pH1=round(pH1,3)
print 'pH=',pH1

pH= 8.567

Example 15.11,Page no.66

In [16]:
import math
#initialisation of variables
pKa=6.84
n=0.04 #mole
n1=0.02 #mole
n2=0.001 #mole
pH3=7.0
#CALCULATIONS
pH=pKa+math.log10(n/n1) 
pH1=pKa+math.log10((n-n2)/(n1+n2)) 
dpH=pH-pH1 
pH2=-math.log10(n2)
dpH1=pH3-pH2
#RESULTS
pH1=round(pH1,2)
dpH=round(dpH,2)
print 'pH =',pH1
print 'dpH =',dpH
print 'dpH =',dpH1

pH = 7.11
dpH = 0.03
dpH = 4.0