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A solution contains 0.1M H2S and 0.3M HC...

A solution contains 0.1M `H_2S` and 0.3M HCI. If `K_a` and `K_(a2)` for `H_2 S` are `1 xx 10^(-7)` and `1.3 xx 10^(-13)` respectively, calculate concentration of `HS^-` and `S^-` in the mixture.

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First ionisation, step 1 `,H_2 S hArr H^(+) +HS^(-) " "" K_a = 1 xx 10^(-7)`
Second ionisation, step 2,`HS^(-) hArr H^(+)+s^(2-) " " K_(a_2) = 3 xx 10^(-13)`
Presence of HCI supresses the dissociation of `H_2 S` due to common ion effect of `H^(+)`.
For step `1,1 xx 10^(-7) = ([H^+][HS^(-)])/( [H_2 S])=(0.3[HS^(-)])/( 0.1)`
Cocentration of `HS^(-) `,`[HS^(-)])= 3.3 xx 10^(-8) mol L^(-1)`
or step` 2,1,.3 xx 10^(-13) = ([H^+] [S^(2-)])/([HS^(-)]) = ( 0.3 [S^(2-)])/(3.3 xx 10^(-8))`
concentration of `s^(2-) , [S^(2-)] = 1.43 xx 10^(-20) mol L^(-1)`
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