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What is the pH of 0.001 M aniline solution? The ionisation constant of aniline is 4.27 times 10^-10. (i) Calculate degree of ionization of aniline in the solution.Also calculate the ionisation constant of the conjugate acid of anile. |
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Answer» SOLUTION :`(i) C_6H_5NH_2+H_2O leftrightarrowC_6H_5NH_3^(+)+OH` `K_a=([C_6H_5NH_3^(+)][OH^-])/([C_6H_5NH_2])=([OH^-]^2)/([C_6H_5NH_2])` `[OH^-]=sqrt(K_a[C_6H_5NH_2])=sqrt((4.27 times10^-10)(10^-3))=6.534 times10^-7M` `pOH=-log(6.534 times10^-7)=7-0.8152=6.18` `therefore =14-6.18=7.82` (ii) Also `C_6H_5NH_2+H_2O leftrightarrow C_6H_5NH_3^(+) +OH^-` INITIAL C At. eqm. C-Ca Ca Ca `K_b=(Ca.Ca)/(C(1-a))=(Ca^2)/(1-a)=Ca^2` `therefore a=sqrt(K_b//C)=sqrt(4.27 times10^-10//10^-3)=6.53 times 10^-4` (iii)`pK_a+pK_b=14` (for a pair of CONJUGATE ACID and base ) `pK_b=-log(4.27times10^-10)=10-0.62=9.38` `thereforepK_a=14-9.38=4.62` i.e., `-logK_a=4.62orlog K_a=-4.62=5.38` `K_a=`Antrilog `5.38=2.399 times 10^-5 APPROX 2.4 times 10^-5` |
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