1.

The specific conductance of a saturated solution of AgCl at 25^(@)C after subtracting the specific conductance of conductivity of water is 2.28 xx 10^(-6) mho cm^(-1). Find the solublity product of AgCl at 25^(@)C. (Lambda_(AgCl)^(@) = 138.3 "mho cm"^(2))

Answer»

Solution :For equilibrium,
`AgCl = Ag^(+) + Cl^(-)`
`K_(sp) [Ag^(+)][Cl^(-)]`
If the solubility of AgCl in water is, SAY, x moles/litre or x eq./L,
`K_(sp) = x.x = x^(2)`
`therefore` volume containing 1 eq. of AgCl `= (1000)/(x)`
`Lambda_(AgCl) = "sp. cond. xx V`
`= 2.28 xx 10^(-6) xx (100)/(x)`.
Since AgCl is sparingly soluble in water, `Lambda_(AgCl) = Lambda_(AgCl)^(@) = 138.3`
`therefore 2.28 xx 10^(-6) xx (1000)/(x) = 138.3`
or `x = 1.644 xx 10^(-5)` eq./litre or mole/litre
`K_(sp) = x^(2) = (1.644 xx 10^(-5))^(2)`
`= 2.70 xx 10^(-10) ("mole/litre")^(2)`


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