question_answer 89)

The ionisation constant of nitrous acid is\[4.5\times {{10}^{-4}}\].Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Answer:

Sodium nitrite is the salt of weak acid and strong base. \[pH=\frac{1}{2}[p{{K}_{w}}+p{{K}_{a}}+logC]\,\,\,\,\,\,\,\,\,\,\,......(i)\] \[p{{K}_{w}}=14\] \[p{{K}_{a}}=-{{\log }_{10}}{{K}_{a}}\] \[=-\log [4.5\times {{10}^{-4}}]\] \[=3.346\] \[\log C=\log 0.04=-1.3979\] Putting these values in Eqn. (i), we get \[pH=\frac{1}{2}[14+3.346-1.3979]\] \[=7.974\] Degree of hydrolysis, \[h=\sqrt{\frac{{{K}_{w}}}{C{{K}_{a}}}}\] \[=\sqrt{\frac{{{10}^{-14}}}{0.04\times 4.5\times {{10}^{-4}}}}\] \[=2.357\times {{10}^{-5}}\]