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Q.

A solution is 0.01M Ba(NO₃)₂ and 0.10M Sr(NO₃)₂. If solid Na₂CrO₄ is added to the solution, what [Ba²⁺] is when SrCrO₄ begins to precipitate?[K_sp (BaCrO₄ = $\large 1.2\, \times \,{10^{ - 10}}$ ; K_sp(SrCrO₄) = $\large 3.5\, \times \,{10^{ - 5}}$ ]

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a

$\large 7.4 \times \,{10^{ - 7}}$

b

$2.0 \times \,{10^{ - 7}}$

c

$6.1 \times \,{10^{ - 7}}$

d

$3.4 \times \,{10^{ - 7}}$

answer is D.

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Detailed Solution

${\left( {{K_{sp}}} \right)_{BaCr{O_4}}} < {\left( {{K_{sp}}} \right)_{SrCr{O_4}}}$

For precipitation of BaCrO₄, the concentration of $CrO_4^{ - 2}$ required

$= \frac{{{K_{sp}}}}{{\left[ {B{a^{ + 2}}} \right]}}$

$= \frac{{1.2 \times {{10}^{ - 10}}}}{{{{10}^{ - 2}}}}$

$={1.2 \times {{10}^{ - 8}}}$

For the precipitation of SrCrO₄, the concentration of $CrO_4^{ - 2}$ required

$= \frac{{{K_{sp}}}}{{\left[ {S{r^{ + 2}}} \right]}} = \frac{{3.5 \times {{10}^{ - 5}}}}{{{{10}^{ - 1}}}}$

$={3.5 \times {{10}^{ - 4}}}$

When the precipitation of SrCrO₄ begin, the equilibrium of undissolved BaCrO₄ and dissolved BaCrO₄ also exists.

$\therefore$ K_IP = K_sp

$\Rightarrow \left[ {B{a^{ + 2}}} \right]\left[ {CrO_4^{ - 2}} \right] = 1.2 \times {10^{ - 10}}$

$\Rightarrow \left[ {B{a^{ + 2}}} \right] = \frac{{1.2 \times {{10}^{ - 10}}}}{{3.5 \times {{10}^{ - 4}}}} = 3.43 \times {10^{ - 7}}$

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