The solubility of a salt of weak acid (AB) at pH 3 is Y xx 10^(-3) mol L^(-1). Thevalue 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))

The solubility of a salt of weak acid (AB) at pH 3 is Y xx 10^(-3) mol

1.	The solubility of a salt of weak acid (AB) at pH 3 is Y xx 10^(-3) mol L^(-1). Thevalue 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))
Answer» Solution :`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^(-)]` Fromeqn. (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.