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)`)

Explanation

Let solubility of AB is s so that s mol `L^(-1)` of `A^+` and s mol `L^(-1)` of `B^-` are formed . Further , let x mol of `B^-` react with `H^+` ions to form x mol of HB. Then (s-x) mol of `B^-` will be present at equilibrium
`{:(AB(s) hArr, A^(+)+,B^(-)),(s,s,(s-x)):}`
`{:(B^(-)+, H^(+) hArr, HB),(s-x, 1xx10^(-3),x):}` [Since pH=3 , `[H^+]=1xx10^(-3)`]
Now, `K_(sp)=s(s-x)=2xx10^(-10)`....(i)
`K_a=([B^-][H^+])/"[HB]"=1xx10^(-8)`
`((s-x)xx10^(-3))/x=1xx10^(-8)`
or , `(s-x)/x = 1xx10^(-5)` ...(ii)
Dividing eq. (i) by eq.(ii)
`sx=2xx10^(-5)`
Now, `s(s-x)=2xx10^(-10)`
`s^2-2xx10^(-5)=2xx10^(-10)`
`s^2=2xx10^(-10)+2xx10^(-5) approx 2 xx 10^(-5)`
`s=sqrt(2xx10^(-5))=sqrt20xx10^(-6)`
`=4.47xx10^(-3)`
`therefore` Y=4.47