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 7.52     What is the pH of 0.001M aniline solution?  Calculate the degree of ionization of aniline in the solution. Also, calculate the ionization constant of the conjugate acid of aniline.(K= 4.27\times 10^{-10})

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We have,
C = 0.001 M
K_b = 4.27\times 10^{-10}
Degree of ionization of aniline (a) = ?
Ionization constant of the conjugate acid (K_a) = ?

We know that 
  K_b = C.a^2
  4.27\times 10^{-10}     =  (0.001)a^2
Thus  a={\sqrt{4270\times 10^{-10}}}
               = 65.34\times 10^{-4}

Then [Base] = C.a = (65.34\times 10^{-4})(0.001)
                               = 0.653\times 10^{-5}

Now, P^{OH} = -\log (0.65\times 10^{-5})=6.187
P^H = 14-P^{OH} 
           = 14 - 6.187
           = 7.813

It is known that,

K_a\times K_b = K_w
So, K_a = \frac{10^{-14}}{4.27\times10^{-10}}
               =2.34\times 10^{-5}   This is ionization constant.

Posted by

manish

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