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Calcualate `pH,[H^(+)],[OH^(-)],[CH_(3)COOH],[SH^(-)],[H_(2)S],[S^(2-)]` in a solution obtained by mixing equal volume of `0.2M H_(2)S &0.02M` acetic acid .Given that `K_(a)(CH_(3)COOH)=2xx10^(-5),K_(a_(1))(H_(2)S)=10^(-7),K_(a_(2))(H_(2)S)=10^(-14)` Take `log21=-1.32,(1)/(sqrt(21))=0.218` |
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Answer» Now `[H_(2)S]=0.1M,[CH_(3)COOH]=0.01M` after mixing For `pH` calculation ,considering only first `[H^(+)]` of `H_(2)S` the system becomes similar to a mixture of two weak monoprotic acids. `2^(nd)H^(+)` coming form `H_(2)S` would be negligible because of very low value of `K_(a_(2))` & also because of common ion effect exerted by `H^(+)` from `CH_(3)COOH` `[H^(+)]=sqrt((10^(-1)xx10^(-7))+(10^(-2)xx2xx10^(-5)))=sqrt((0.1+2)xx10^(-7))=sqrt(21)xx10^(-4)M` `pH=4-(1)/(2)log21=3.34` `[OH^(-)]=(K_(W))/([H^(+)])=2.18xx10^(-11)M` For acetic acid `K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])=(sqrt(21)xx10^(-4)xx[CH_(3)COO^(-)])/(0.01)` `rArr [CH_(3)COOH^(-)]=4.36xx10^(-4)M` `rArr [CH_(3)COOH]=0.1M` For `H_(2)S,K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])=(sqrt(21)xx10^(-4)xx[s^(2-)])/(0.1)rArr [HS^(-)]=2.18xx10^(-5)M` For `HS^(-)K_(a_(2))=([H^(+)][S^(2-)])/([HS^(-)])=(sqrt(21)xx10^(-4)xx[S^(2-)])/(2.18xx10^(-5))rArr [S^(2-)]=4.76xx10^(-16)M` |
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