Solid `Ba(NO_(3))` is gradually dissolved in a `1.0xx10^(-4) M Na_(2)CO_(3)` solution. At what concentrations of `Ba^(2+)`, will a precipitate begin to form?
(`K_(SP)` for `BaCO_(3)=5.1xx10^(-9)`)

Options

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`

Answer

B

Explanation

`[CO_(3)^(2-)] = 1.0 xx 10^(-4) M`
`K_(sp) =5.1 xx 10^(-9)`
`[Ba^(2+)] = (K_(sp))/[[CO_(3)^(2-)]] =(5.1 xx 10^(-9))/(1.0 xx 10^(-4)) =5.1 xx 10^(-5) M`
At `5.1 xx 10^(-5) M [Ba^(2+)]`, Precipitate will began to form.