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The conductivity (k) of a saturated solu...

The conductivity `(k)` of a saturated solution of `AgBr`(1M) at `298K` is `8.5xx10^(-7)Scm^(-1)`. If `lambda_(Ag^(+))^(@)` and `lambda_(Br^(-))^(@)` are 62 and `78 S cm^(2)mol^(-1)` , respectively, then calculate the solubility and `K_(sp)` of `AgBr`.

Text Solution

Verified by Experts

`Lambda_(m)^(@) = lambda_(m)^(@) (Ag^(+) ) + lambda_(m)^(@) (Br^(-))`
`62 + 78 = 140.0 S cm^(2) mol^(-1)`
Now for saturated solution
`Lambda_(m) ^(@) (AgBr) = (kappa xx 1000)/(s)`
where s is the solubility
`140.0 = (8.5 xx 10^(-7) xx 1000)/(s)`
`or " " s = (8.5 xx 10^(-7) xx 1000)/(140.0) = 6.07 xx 10^(-6) mol L^(-1)`
Now `AgBr hArr Ag^(+) + Br^(-)`
`K_(sp) = [Ag^(+)] [Br^(-)]`
`= (6.07 xx 10^(-6)) xx (6.07 xx 10^(-6))`
`= 36.8 xx 10^(-12) mol^(2) L^(-2)`
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Knowledge Check

  • The specific conductivity of a saturated solution of AgCl is 3.40xx10^(-6) ohm^(-1) cm^(-1) at 25^(@)C . If lambda_(Ag^(+)=62.3 ohm^(-1) cm^(2) "mol"^(-1) and lambda_(Cl^(-))=67.7 ohm^(-1) cm^(2) "mol"^(-1) , the solubility of AgC at 25^(@)C is:

    A
    `2.6xx10^(-5)M`
    B
    `4.5xx10^(-3)M`
    C
    `3.6xx10^(-5)M`
    D
    `3.6xx10^(-3)M`

  • The specific conducitvity of a saturated solution of AgCI is 3.40 xx 10^6 "ohm"^(-1) cm^(-1) at 25^@C . If lambda_(Ag+) = 62.3 "ohm"^(-1) cm^2 "mol"^(-1) and lambda_(CI-) = 67.7 "ohm"^(-1) cm^2 "mol"^(-1) , the solubility of AgCI at 25^@C is.

    A
    ` 2.6 xx 10^(-5) "mol" L-1`
    B
    ` 3.731 xx 10^(-3) "mol" L-1`
    C
    ` 3.731 xx10^(-5) "mol" L-1`
    D
    ` 2.6 xx 10^(-3) g L-1`

  • The conductivity of 0.20M solution of KCl at 298K is 0.0248Scm^-1 its molar conductivity is

    A
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    B
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    C
    `12.4Omega^-1mol^-1cm^2`
    D
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