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Solid Ba(NO3)2 is gradually dissolved in...

Solid `Ba(NO_3)_2` is gradually dissolved in a `1.0 xx 10^(-4)M, Na_2CO_3` solution. At what concentration of `Ba^(2+)` will a precipitate begin to form ? `(K_(sp)" for BaCO"_3= 5.1 xx 10^(-9))`

A

`4.1 xx 10^(-5)M`

B

`5.1 xx 10^(-5)M`

C

`8.1 xx 10^(-8)M`

D

`8.1 xx 10^(-7)M`

Text Solution

Verified by Experts

The correct Answer is:
B
`K_(sp)" for "BaCO_3=[Ba^(2+)] [CO_(3)^(2-)]`
given, `[CO_(3)^(2-)]=1.0 xx 10^(-4) M ("from Na"_(2)CO_3)`
`K_(sp)=5.1 xx 10^(-9)`
`5.1 xx 10^(-9)=[Ba^(2+)] xx [10^(-4)] rArr [Ba^(2+)]=5.1 xx 10^(-5)M`
Thus, when `[Ba^(2+)] =5.1 xx 10^(-5)M, BaCO_3` precipitate will begin to form.
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