Solubility product of silver bromide is `5.0xx10^(-13)`. The quantity of potassium bromide (molar mass taken as `120g mol^(-1)`) to be added to `1 L` of `0.05 M` solution of silver nitrate to start the precipitation of `AgBr` isA. `6.2 xx 10^(-5)g`B. `5.0 xx 10^(-8) g`C. `1.2 xx 10^(-10)g`D. `1.2 xx 10^(-9)g`

Solubility product of silver bromide is `5.0xx10^(-13)`. The quantity

1.	Solubility product of silver bromide is `5.0xx10^(-13)`. The quantity of potassium bromide (molar mass taken as `120g mol^(-1)`) to be added to `1 L` of `0.05 M` solution of silver nitrate to start the precipitation of `AgBr` isA. `6.2 xx 10^(-5)g`B. `5.0 xx 10^(-8) g`C. `1.2 xx 10^(-10)g`D. `1.2 xx 10^(-9)g`
Answer» Correct Answer - D `\|Ag^(+)\|=0.05 M ,K_(sp)(AgBr)=[Ag^(+)][Br^(-)]` `:. Br^(-) =(K_(sp))/(\|Ag^(+)\|)=(5.0 xx 10^(-13))/(0.05)=10^(-11)M` `i.e., ` amount of KBr to be added to `=10^(-11)` mole `=10^(-11)xx120=1.2 xx 10^(-9) ` g

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