1.

If the solubility products of AgCl and AgBr are 1.2 xx 10^(-10) and 3.5 xx 10^(-13) respectively, then the relation between the solubilities (denoted by the symbol'S') of these

Answer»

S of AgBr is less than that of AgCl
S of AgBr is GREATER than that of AgCl
S of AgBr is EQUAL to that of AgCl
S of AgBr is `10^(6)` TIMES greater than that of AgCl

Solution :`AgCl K_(sp) = 1.2 xx 10^(-10)`
`S = sqrt((1.2 xx 10^(-10))), S = 1.09 xx 10^(-5)`
`AgBr K_(sp) = 3.5 xx 10^(-13)`
`S = sqrt(3.5 xx 10^(-13)) = 5.91 xx 10^(-6)`
So that S of AgBr is less than that of AgCl.


Discussion

No Comment Found

Related InterviewSolutions