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Grade: 12th pass
        Please answer the question shown in the image above .
7 months ago

Answers : (1)

Chhavi Jain
92 Points
							
Dear student ,
 
                           NH3 (aq)  +     H2(l)     --------->     NH4+ (aq)     +       OH- (aq)
           t=0           c mol L-1                                            –                          – 
 
           t=eqbm    c( 1 – α )        –                                  cα                        cα
 
( c ---> is the concentration in mol/ L;   α----> degree of dissociation  )
 
Now ,     K = [ NH4+]  [ OH-] /   [ NH3]
                
                    =      (cα * cα ) /   c( 1 – α )
                    =      cα2 /    ( 1 – α ) 
 
Assuming the value of ‘ α ‘ is very less than 1 we can take  (1 – α ) ≈  1 
   
So,     Kb = c α2
 
             1.8 *10-5 =  1.75 α2       
               
               α =  0.32 *  10-2    (our assumption is correct )
 now ,     [ OH- ] =  cα   =  1.75 * 0.32 *10-2   = 0.56 * 10-2 
 
So,  pOH  =  -log [OH-]  = – log ( 0.56 * 10-2 ) =  2 – log( 0.56)
 
pH + pOH =  14 
Therefore ,  pH = 14 – {2 – log(0.56)}  = 12+ (-0.251) = 11.749
 
Chemistry Consultant
Askiitians
 
7 months ago
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