If equal volumes of BaCl $_{2}$ (aq) and NaF(aq) solutions are mixed together, which will give a precipitate ?
[K $_{s p}$ of BaF $_{2} = 1.7 \times 10^{- 7}$ ]?

0.004 M BaCl

_{2}

and 0.002 M NaF

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0.010M BaCl

_{2}

and 0.015M NaF

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0.015 M BaCl

_{2}

and 0.002 M NaF

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0.020 M BaCl

_{2}

and 0.002 M NaF

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Solution

The correct option is B 0.010M BaCl $_{2}$ and 0.015M NaF
When equal volumes of 0.010 ${B a C l}_{2}$ and 0.015 $N a F$ solutions are mixed, the ionic product exceeds solubility product. Hence, precipitation will occur.

${B a C l}_{2} + N a F ⟶ {B a F}_{2} + N a C l$

$K_{a} = [{B a}^{2 -}] [F^{-}]^{2}$
$= (0.010) (0.015)^{2} = 2.25 \times 10^{- 6}$

$∴ K_{a} = 2.25 \times 10^{- 6}$

$K_{a} > K_{s p}$ where, $K_{s p} = 1.7 \times 10^{- 7}$ ...... (given)

Hence, the correct option is $B$

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If equal volumes of BaCl2(aq) and NaF(aq) solutions are mixed together, which will give a precipitate ?[Ksp of BaF2=1.7×10−7]?

If equal volumes of BaCl $_{2}$ (aq) and NaF(aq) solutions are mixed together, which will give a precipitate ?
[K $_{s p}$ of BaF $_{2} = 1.7 \times 10^{- 7}$ ]?