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Solid Ba(NO(3))(2) is gradually dissolve...

Solid `Ba(NO_(3))_(2)` is gradually dissolved in a `10^(-4)M Na_(2)CO_(3)` solution. At what concentration of `Ba^(2+)` will a precipitate begin to form ? (`K_(sp)` for `BaCO_(3)=5.1xx10^(-9)`)

A

`4.1xx10^(-5)M`

B

`8.1xx10^(-8)M`

C

`5.1xx10^(-5)M`

D

`8.1xx10^(-7)M`

Text Solution

Verified by Experts

The correct Answer is:
C
`K_(sp)=[Ba^(2+)][CO_(3)^(2-)]`
`5.1xx10^(-9)=[Ba^(2+)]xx10^(-4)`
`[Ba^(2+)]=5.1xx10^(-5)M`.
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Knowledge Check

  • Solid Ba(NO_3)_2 is gradually dissolved in 1xx10^(-4) M Na_2CO_3 solution. At what concentration of Ba^(2+) will a precipitate begin to form ?

    A
    `8.1 xx10^(-8)` M
    B
    `4.0 xx 10^(-10)` M
    C
    `4.1 xx 10^(-5)` M
    D
    `5.1 xx 10^(-5)` M

  • Solubility product of a salt AB is 1xx10^(-8) in a solution in which concentration of A is 10^(-3) M. the salt will precipitate when the concentration of B becomes more than

    A
    `10^(-4)M`
    B
    `10^(-7)M`
    C
    `10^(-6)M`
    D
    `10^(-5)M`

  • Solubility product of a salt AB is 1xx10^(-8)M^(2) in a solution in which the concentration of A^(+) ions is 10^(-3)M . The salt will precipitate when the concentration of B^(-) ions is kept

    A
    between `10^(-8)M` to `10^(-7)M`
    B
    between `10^(-7)M` to `10^(-8)M`
    C
    `gt 10^(-5)M`
    D
    `lt 10^(-8)M`

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