Ncert Exercises & Solutions

7.52 What is the pH of 0.001M aniline solution ? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

K_{b} = 4.27 \times 10^{- 10}

c = 0.001 M pH

= ?

α = ?

K_{b} = {cα}^{2}

4.27 \times 10^{- 10} = 0.001 \times α^{2}

4270 \times 10^{- 10} = α^{2}

65.34 \times 10^{- 5} = α = 6.53 \times 10^{- 4}

Then, [anion] = cα = . 001 \times 65.34 \times 10^{- 5}

= . 065 \times 10^{- 5}

pOH = - \log (. 065 \times 10^{- 5})

= 6.187

pH = 7.813

Now,

K_{a} \times K_{b} = K_{w}

∴ 4.27 \times 10^{- 10} \times K_{a} = K_{w}

K_{a} = \frac{10^{- 14}}{4.27 \times 10^{- 10}}

= 2.34 \times 10^{- 5}

Thus, the ionization constant of the conjugate acid of aniline is 2.34 ×

10^{- 5}

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