1.

What is the minimum mass of NaBr which should be added in 200 ml of 0.0004 M-AgNO_(3) solution just to start the precipitation of AgBr ? The value of K_(sp) of AgBr = 4 × 10^(-13) (B Solution : r = 80)

Answer»

`1.0 xx 10^(-9)g`
`2 xx 10^(-10) g`
`2.06 xx 10^(-8)g`
`1.03 xx 10^(-7)g`

Solution :`AgBr(s) For precipitation, `Q > K_(sp)`
or, `0.0004 xx [ Br^(-)] > 4 xx 10^(-13)`
`:. [ Br^(-)] > 10^(-9) M`
HENCE, MINIMUM mass of NaBr needed
`=((200xx10^(-9))/(1000)) xx 103 = 2.06 xx 10^(-8) gm`.


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