An aqueous solution contains 0.10 M H_2Sand 0.20 M HCl. If the equilibrium constants for the formation of HS^- from H_2S is 1.0 xx 10^(-7) and that of S^(2-)from HS^-ions is 1.2 xx 10^(-13)then the concentration of S^(2-) ions in aqueous solution is :

An aqueous solution contains 0.10 M H_2Sand 0.20 M HCl. If the equilib

1.	An aqueous solution contains 0.10 M H_2Sand 0.20 M HCl. If the equilibrium constants for the formation of HS^- from H_2S is 1.0 xx 10^(-7) and that of S^(2-)from HS^-ions is 1.2 xx 10^(-13)then the concentration of S^(2-) ions in aqueous solution is :
Answer» `5 xx 10^(-8)` `3 xx 10^(-20)` `6 xx 10^(-21)` `5 xx 10^(-19)` Solution : ` therefore KA(iii) = ka(i) xx ka(ii)` ` = (1 xx 10^(-7) (1.2 xx 10^(-13))` ` = 1.2 xx 10^(-20)` (iii) of ka (iii) = ([H^+][S^(2-)])/([H_2S])` Where , `[H_2S] = 0.1 M, [H^+]= 0.2M = [HCl]` ` therefore 1.2 xx 10^(-20) = (0.2)^2 (S^(2-) ) // 0.1` `therefore [S^(2-)] = (0.1 xx 1.2 xx 10^(-20))/((0.2)^2)` ` = 0.3 xx 10^(-20) = 3 xx 10^(-21)`