Home
Class 11
CHEMISTRY
When 1.88 g of AgBr(s) is added to a 10^...

When `1.88 g` of AgBr(s) is added to a `10^(-3)` M aqueous solution of KBr, the concentrationof Ag is `5xx10^(-10)`M. If the same amout of AgBr(s) is added to a `10^(-2)` M aqueous solution of `AgNO_(3)`, the concentration of `Br^(-)` is

A

`9.4 xx 10^(-9)` M

B

` 5 xx 10^(-10)` M

C

`1 xx 10^(-11)` M

D

` 5 xx 10^(-11)` M

Text Solution

Verified by Experts

The correct Answer is:
D
`"AgBr"_((s))hArr Ag_((aq))^(+) +Br_((aq))^(-)`
`K_(sp(AgBr)) =[Ag^(+)] [Br^(-)] `
` = ( 5 xx 10^(-10)) (10^(-3))`
`= 5 xx10^(-13)`
Now, ` 5 xx 10^(-13) = (10^(-2)) [Br^(-)] `
`[Br^(-)] = 5 xx 10^(-11) M `.
Promotional Banner

Similar Questions

Explore conceptually related problems

The pH of 10M HCl aqueous solution is

The OH^- concentration of a solution is 1.0 xx 10^(-10) M. The solution is

In aqueous solution concentration of H^(+) ions is 10^(-4) mol/1. If equal volume of water is added to the solution, then the concentration of OH^(-) ion in mol/ dm^(3) is

In a saturated aqueous solution of AgBr, concentration of Ag^(+) ion is 1xx10^(-6) mol L^(-1) it K_(sp) for AgBr is 4xx10^(-13) , then concentration of Br^(-) in solution is

The concentration of OH^(-) in a solution is 1.0xx10^(-10)M . The solution is

The solubility product of Mg(OH)_(2) is 1.0 xx 10^(-12) . Concentrated aqueous NaOH solution is added to a 0.01 M aqueous solution of MgCl_(2) . The pH at which precipitation occur is -