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Calcualate pH,[H^(+)],[OH^(-)],[CH(3)COO...

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`

Text Solution

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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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