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Sulphide ions in alkaline solution react...

Sulphide ions in alkaline solution react with solid sulphur to form polyvalent sulphide ions. The equilibrium constant for the formation of `S_(2)^(2-)` and `S_(3)^(2-)` from `S` and `S^(2-)` ions is `1.7` and `5.3` respectively. Calculate equilibrium constant for the formation of `S_(3)^(2-)` from `S_(2)^(2-)` and `S`.

Text Solution

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`{:("Given",,S(s),+,S^(2-),hArr,S_(2)^(2-)),(,,2S(s),+,S^(2-),hArr,S_(3)^(2-)):}`
`:. K_(c_(1))=([S_(2)^(2-)])/([S^(2-)])=1.7 ...(i)`
`K_(c_(2))=([S_(3)^(2-)])/([S^(2-)])=5.3 ...(ii)`
Now for `S_(2)^(2-)+S(s) hArr S_(3)^(-2)`
`K_(c )=([S_(3)^(2-)])/([S_(2)^(2-)])`
`:.` By dividing equation (ii) by (i),
`K_(c_(2))/K_(c_(1))=([S_(3)^(2-)])/([S_(2)^(-2)])rArr K_(c )=3.11`
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