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A solution contains 0.1 M H_2S and 0.3 M HCl. Calculate the conc. of S^(2-) and HS^(-) ions in solution. Given K_(a_(1)) and K_(a_(2))" for "H_2S" are "10^(-7) and 1.3 xx 10^(-13)" respectively". |
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Answer» Solution :`H_(2)S LEFTRIGHTARROW H^(+)+HS^(-)""K_(a_(1))=10^(-7)` `HS^(-) Leftrightarrow H^(+)+S^(2-) K_(a_(2))=1.3 xx 10^(-13)` Due to COMMON ion effect the dissociation of `H_2S` is suppressed and the `[H^+]` in solution is due to HCl `therefore K_(a_(1))=([H^(+)][HS^(-1)])/([H_(2)S])` `10^(-7)= ([0.3][HS^(-1)])/([0.1]) [? [H^(+)]" fromHCl "=0.3]` `therefore [HS^(-1)]=(10^(-7) xx 0.1)/(0.3)` `=3.3 xx 106^(-8)M` Further `K_(a_(2))=([H^(+)][S^(2-)])/([HS^(-)]) and K_(a_(1))=([H^(+)][HS^(-1)])/([H_(2)S]` `therefore K_(a_(1)) xx K_(a_(2))=([H^(+)][S^(2-)])/([H_(2)S])` `10^(-7) xx 1.3 xx 10^(-13)=([0.3]^(3)[S^(2-)])/([0.1])` `therefore [S^(2-)]=(1.3 xx 10^(-20)xx 0.1)/(0.09)` `=1.44 xx 10^(-20)M` |
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