Chapter15-Equilibria in electrolytes

Ex1-pg486

In [1]:
import math
##Intitalisation of variables
c1= 0.1 ##M
c2= 0.2 ##M
c3= 0.4 ##M
n1= 1.
n2= 2.
##CALCULATIONS
u1= 0.5*(c1*n1**2+c1*n1**2)
u2= 0.5*(c3*n1**2+c2*n2**2)
u3= 0.5*((c3+c1)*n1**2+c1*n1**2+c2*n2**2)
##RESULTS
print'%s %.2f %s'% ('Ionic strength =   ',u1,'')
print'%s %.2f %s'% ('\n Ionic strength =   ',u2,'')
print'%s %.2f %s'% ('\n Ionic strength =   ',u3,'')

Ionic strength =    0.10 

 Ionic strength =    0.60 

 Ionic strength =    0.70

Ex2-pg488

In [2]:
import math
##Intitalisation of variables
c1= 0.01 ##M
c2= 0.001 ##M
n= 2. ##moles
k= -0.509
n1= 1. ##moles
##CALCULATIONS
f1= 10**(k*math.sqrt(c1))
f2= 10**(k*n*math.sqrt((c2*(n+n1))))
##RESULTS
print'%s %.2f %s'% ('activity coefficient =   ',f1,'')
print'%s %.2f %s'% ('\n activity coefficient =   ',f2,'')

activity coefficient =    0.89 

 activity coefficient =    0.88

Ex3-pg491

In [3]:
import math
##Intitalisation of variables
a= 6.4*10**-6 ##g ion per lit
a1= 0.05
n= 2.
##CALCULATIONS
Ksp= a**2*a1
s= (Ksp/n**2)**(1./3.)
##RESULTS
print'%s %.2e %s'% ('Ks =  ',Ksp,'')
print'%s %.2e %s'% ('\n solubility of Ag2CrO4 = ',s,' mole per litre')

Ks =   2.05e-12 

 solubility of Ag2CrO4 =  8.00e-05  mole per litre

Ex4-pg494

In [4]:
import math
##Intitalisation of variables
s1= -0.0059 ##mole per litre
x1= 0.0118 ##mole per lit
x2= 0.0269 ##mole per litre
##CALCULATIONS
S= s1+math.sqrt(0.25*x1**2+x2**2)
##RESULTS
print'%s %.4f %s'% ('Solubility = ',S,' mole per litre')

Solubility =  0.0216  mole per litre

Ex5-pg506

In [5]:
import math
##Intitalisation of variables
Ka= 1.752*10**-5
c= 0.1 ##M
##CALCULATIONS
ch= math.sqrt(Ka*c)
ch1= -0.5*Ka+math.sqrt(Ka*c)
r= ch1/c
##RESULTS
print'%s %.2e %s'% ('CH+ = ',ch,' g ion per litre')
print'%s %.3e %s'% ('\n CH+ = ',ch1,' g ion per litre')
print'%s %.2e %s'% ('\n degree of dissociation = ',r,'')

CH+ =  1.32e-03  g ion per litre

 CH+ =  1.315e-03  g ion per litre

 degree of dissociation =  1.31e-02

Ex6-pg508

In [6]:
import math
##Intitalisation of variables
c1= 0.1 ##M
cs= 0.05 ##M
Ka= 1.75*10**-5
##CALCULATIONS
ch= Ka*c1/cs
##RESULTS
print'%s %.1e %s'% ('CH+ = ',ch,' g ion per litre')

CH+ =  3.5e-05  g ion per litre

Ex7-pg511

In [7]:
import math
##Intitalisation of variables
ch= 5.46*10**-5 ##g ion per litre
ph= 8.752
##CALCULATIONS
pH= -math.log10(ch)
ch1= 10**(-ph)
##RESULTS
print'%s %.2f %s'% ('pH = ',pH,'') 
print'%s %.2e %s'% ('\n Hydrogen ion concentration = ',ch1,' g ion per litre') 

pH =  4.26 

 Hydrogen ion concentration =  1.77e-09  g ion per litre

Ex8-pg512

In [8]:
import math
##Intitalisation of variables
pt= 14.
ph= 4.75
##CALCULATIONS
coh= 10**(-(pt-ph))
##RESULTS
print'%s %.2e %s'% ('COH- = ',coh,' g ion per litre') 

COH- =  5.62e-10  g ion per litre