AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

The ionization constant of conjugate acid is calculated as:
Ka(NH4+).Kb(NH3)=10−14
Ka(NH4+)=

10−14

Kb(NH3 )

=

10−14

2.5×10−5

The dissociation reaction of ammonia is as follows:
The degree of ionization is:
α=√

Kb

C

=√

2.5×10−5

0.01

=0.05
Therefore concentraion of hydroxide ion becomes:
[OH− ]=Cα
=0.01×0.05
=0.0005
The pH of the ammonia solution is calculated as:
pH=14−pOH
=14−(−log[OH−])
=14+log(0.005)
=14+(−3.3)
=10.7
Therefore, pH is 10.7 and ionization constant is equal to 4×10−10