JEE MAIN - Chemistry (2018 (Offline) - No. 20)
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 $$\times$$ 10–10. What is the original concentration of Ba2+?
1.0 $$\times$$ 10–10 M
5 $$\times$$ 10–9 M
2 $$\times$$ 10–9 M
1.1 $$\times$$ 10–9 M
Explanation
Let initially concentration of Ba+2 = x m.
After adding 50 ml Na2SO4 in Ba+2 solution final volume becomes 500 ml.
$$\therefore\,\,\,$$ Initial volume of Ba+2 solution
= (500 $$-$$ 50) ml = 450 ml
As at the begining of precipitation, ionic product = solubility product.
$$ \Rightarrow \,\,\,$$ [Ba2+] [SO$${_4^{ - 2}}$$] = Ksp of BaSO4
$$ \Rightarrow \,\,\,$$ [Ba2+] $$\left( {{{50 \times 1} \over {500}}} \right)$$ = 1 $$ \times $$ 10$$-$$10
$$ \Rightarrow \,\,\,$$ [Ba2+] = 10$$-$$9 M.
So, the concentration of Ba+2 in final solution is 10$$-$$9 M.
$$\therefore\,\,\,$$ concentration of Ba+2 in original solution,
M1 V1 = M2 V2
$$ \Rightarrow \,\,\,$$ x $$ \times $$ 450 = 10$$-$$9 $$ \times $$ 500
$$ \Rightarrow \,\,\,$$ x = 1.1 $$ \times $$ 10$$-$$9 M
After adding 50 ml Na2SO4 in Ba+2 solution final volume becomes 500 ml.
$$\therefore\,\,\,$$ Initial volume of Ba+2 solution
= (500 $$-$$ 50) ml = 450 ml
As at the begining of precipitation, ionic product = solubility product.
$$ \Rightarrow \,\,\,$$ [Ba2+] [SO$${_4^{ - 2}}$$] = Ksp of BaSO4
$$ \Rightarrow \,\,\,$$ [Ba2+] $$\left( {{{50 \times 1} \over {500}}} \right)$$ = 1 $$ \times $$ 10$$-$$10
$$ \Rightarrow \,\,\,$$ [Ba2+] = 10$$-$$9 M.
So, the concentration of Ba+2 in final solution is 10$$-$$9 M.
$$\therefore\,\,\,$$ concentration of Ba+2 in original solution,
M1 V1 = M2 V2
$$ \Rightarrow \,\,\,$$ x $$ \times $$ 450 = 10$$-$$9 $$ \times $$ 500
$$ \Rightarrow \,\,\,$$ x = 1.1 $$ \times $$ 10$$-$$9 M
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