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The solubility of a salt of weak acid (A...

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)`)

Text Solution

Verified by Experts

The correct Answer is:
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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Knowledge Check

  • If the solubility of MX_(2) ( a sparingly soluble salt is 2.5 xx 10^(-4) mol^(-1) . The value of K_(sp) of the salt would be

    A
    `6.25 xx 10^(-11)`
    B
    `12.5 xx 10^(-8)`
    C
    `6.25 xx 10^(-8)`
    D
    `3.125 xx 10^(-11)`

  • The solubility of a springly soluble salt AB_(2) in water is 1.0xx10^(-5) mol L^(-1) . Its solubility product is:

    A
    `10^(-15)`
    B
    `10^(-10)`
    C
    `4xx10^(-15)`
    D
    `4xx10^(-10)`

  • The solubility product of BaCl_(2) is 3.2xx10^(-9) . What will be solubility in mol L^(-1)

    A
    `4xx10^(-3)`
    B
    `3.2xx10^(-9)`
    C
    `1xx10^(-3)`
    D
    `1xx10^(-9)`
    • PRADEEP-EQUILIBRIUM-Competition Focus (VII. Numerical Value Type Questions) (In Decimal Notation)
    1. The solubility of a salt of weak acid (AB) at pH 3 is Y xx 10^(-3) "...

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