InterviewSolution
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InterviewSolution
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The solubility of a salt of weak acid (AB) at pH 3 is `Y xx 10^(-3)` mol `L^(-1)`. The value of Y is ............ (Given that the value of solubility product of aB `(K_(sp))=2xx10^(-10)` and the value of ionization constant of HB `(K_(a))=1xx10^(-8)`) |
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Answer» Correct Answer - 4.47 `AB (s) hArr A^(+) (aq) + B^(-) (aq) ` ...(i) `B^(-)(aq)+ H^(+) (aq) hArrHB (aq) ` ...(ii) Adding eqns (i) and (ii), we get `{:(,AB(s),+,H^(+)(aq),hArr,A^(+)(aq),+,HB(aq)),("Initial conc",x,,10^(-3),,0,,0),("Final conc.",x-S,,10^(-3),,S,,S),(,,,,,,,):}` (S = solubility when pH = 3) From eqn. (i), `K_(sp)=[A^(+)][B^(-)]` From eqn. (ii), `K_(a)=([HB])/([B^(-)][H^(+)])` `:. (K_(sp))/(K_(a))=([A^(+)][HB])/([H^(+)])` `(2xx10^(-10))/(10^(-8))=(SxxS)/(10^(-3))` or `S^(2) = 2 xx10^(-5) = 20 xx 10^(-6) or S = 4.47 xx 10^(-3)` Hence , y = 4.47. |
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