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Solid Ba(NO_{3})_{2} is gradually dissolved in a 1.0\times 10^{-4}\; M\; Na_{2}CO_{3} solutions. At which concentration of Ba^{2+}, precipitate of BaCO_{3} begins to form ?(K_{sp}\; for \; BaCO_{3}=5.1\times 10^{-9}) 

Option: 1

5.1\times 10^{-5} M
 


Option: 2

7.1\times 10^{-8} M


Option: 3

4.1\times 10^{-5} M

 


Option: 4

8.1\times 10^{-7} M


Answers (1)

best_answer

As we have learnt,

 

Relation between Solubility(s) and Solubility Product (Ksp)
$\mathrm{AxBy} \leftrightarrow \mathrm{xA}^{+y}+\mathrm{yB}^{-\mathrm{x}}$
    a               0             0
 a - s            xs            ys

$\mathrm{Thus,\: Ksp}=\mathrm{x}^{x} \mathrm{y}^{y}(\mathrm{s})^{\mathrm{x}+\mathrm{y}}$
If ? is given, then:
$\mathrm{Ksp}=\mathrm{x}^{\mathrm{x}} \mathrm{y}^{\mathrm{y}}(\alpha \mathrm{s})^{\mathrm{x}+\mathrm{y}}$

-

Na_2CO_3\rightarrow 2Na^+ + CO^{2-}

BaCO_3\rightarrow BA^{2+}+CO_{3}^{2-}

K_{sp} = [ Ba^{2+}] [ CO_{3}^{2-}]

5.1\times 10^{-9} = [Ba ^{2+}][1 \times 10^{-4}]

[Ba^{2+}]= \frac{5.1 \times 10^{-9}}{1\times 10^{-4}}= 5.1 \times 10 ^{-5}M

Therefore, Option(1) is correct

Posted by

Gunjita

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