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  • Can someone explain An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1×10−10. W

An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4
just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1×10−10. What is the original
concentration of Ba2+ ?

  • Option 1)

    1.0×10−10 M

  • Option 2)

    5×10−9 M

  • Option 3)

    2×10−9 M

  • Option 4)

    1.1×10−9 M

 

Answers (1)

As we learned

Solubility product constant -

e.g.

BaSO_{4}\rightleftharpoons Ba^{2+}(aq)+SO_{4}^{2-}(aq)


K=\frac{[Ba^{2+}]\:[SO_{4}^{2-}]}{[BaSO_{4}]}

- wherein

For a pure solid substance the concentration remain constant

\therefore \:K_{sp}=K[BaSO_{4}]
 

               c
 

    K_{sp}=solubility\:product

 

 

 

Let original concentration of Ba2+ = xM

Volume of this solution is 500-50=450ml

On adding Na2SO4 solution. Concentration of Ba+2 =\frac{450\times x}{500}= \frac{9x}{10}M

Concentration of SO_{4}^{-2} in final solution = \frac{1\times 50}{500}= \frac{1}{10}M

Precipitation just starts so Q_{sp}= K_{sp}

\Rightarrow \left [ Ba^{+2} \right ]\left [ SO_{4}^{-2} \right ]= 1\times 10^{-10}

\Rightarrow \frac{9x}{10}\times \frac{1}{10}= 1\times 10^{-10}

\therefore x= \frac{1}{9}\times 10^{-8}\cong 1.1\times 10^{-9}M  


Option 1)

1.0×10−10 M

Option 2)

5×10−9 M

Option 3)

2×10−9 M

Option 4)

1.1×10−9 M

Posted by

Vakul

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