How many grams of `KBr` can be added to `1L` of `0.12 M` solution of `AgNO_(3)` just to start the precipitation of `AgBr. (Mw "of" KBr = 120, K_(sp) of AgBr = 10^(-13))`

Options

A. `10^(-10)g`
B. `10^(-9)g`
C. `0.5 xx 10^(-10)g`
D. `0.5 xx 10^(-9)g`

Answer

A

Explanation

Let `x g` of `KBr`.
Moles of `KBr = (x)/(20), M = (x)/(120 xx1L)`
`:. [Ag^(o+)] = [AgNO_(3)] = 0.12M`
`[Br^(Theta)] = [KBr] = (x)/(120)`
When the precipitation just starts, `Q_(sp) = K_(sp)`
`[Ag^(o+)] [Br^(Theta)] = 10^(-13)`
`0.12 xx (x)/(120) = 10^(-13)`
`:. x = 10^(-10)g`