1.

The solubility product (K_(sp) , mol^(3) dm^(-9) ) of MX_(2) at 298 K based on the information available for the given concentration cell is (take 2.303 xx R xx 298 // F = 0.059 V)

Answer»

`1 xx 10^(-15)`
`4 xx 10^(-15)`
`1 xx 10^(-12)`
`4 xx 10^(-12)`

Solution :The GIVEN concentration cell is `M | M^(2+) ` (saturated= `C_(1) ) || M^(2+) (0.001 M = C_(2))| M` EMF of concentration cell is :
`E_(cell) = (2.303 RT)/(nF) "LOG" (C_(2))/(C_(1)) = (0.059)/(N) "log" (C_(2))/(C_(1))`
`therefore 0.059 = (0.059)/(2) "log" (0.001)/(C_(1))`
`or 1 = (1)/(2) "log" (10^(-3))/(C_(1))or log 10^(-3) - log C_(1) = 2`
`or -3-log C_(1) = 2 or log C_(1) = -5`
or `C_(1) = 10^(-5)` M
For the SALT `MX_(2) , underset(S)(MX^(2)) to underset(S)(M^(2+)) + 2 underset(2 S)(X^(-))`
`K_(sp) = (S) (2S)^(2) = 4 S^(3) = 4 xx (10^(-5))^(3) = 4 xx 10^(-15)`


Discussion

No Comment Found

Related InterviewSolutions