What is the concentration of Ba2+ when BaF2(ksp=1.0×10−6) begins to precipitate from a solution that is 0.30M F−?

[Ba2+] [F−]2=Ksp [Ba2+] [F−]2=1.0×10−6 [Ba2+] =1.0×10−6(0.3)2 =1.0×10−6(3×10−1)2 =1×10−69×10−2 =0.11×10−4 =1.1×10−5

What is the concentration of Ba2+ when BaF2(ksp=1.0×10−6) begins to precipitate from a solution that is 0.30M F−?

[Ba2+] [F−]2=Ksp [Ba2+] [F−]2=1.0×10−6 [Ba2+] =1.0×10−6(0.3)2 =1.0×10−6(3×10−1)2 =1×10−69×10−2 =0.11×10−4 =1.1×10−5

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What is the concentration of $B a^{2 +}$ when $B a F_{2} (k s p = 1.0 \times 10^{- 6})$ begins to precipitate from a solution that is $0.30 M F^{-}$ ?

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$9.0 \times 10^{- 7}$

$3.0 \times 10^{- 7}$

$3.3 \times 10^{- 5}$

$1.1 \times 10^{- 5}$

answer is C.

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Detailed Solution

$[B a^{2 +}] {[F^{-}]}^{2} = K_{s p} [B a^{2 +}] {[F^{-}]}^{2} = 1.0 \times 10^{- 6} [B a^{2 +}] = \frac{1.0 \times 10^{- 6}}{{(0.3)}^{2}} = \frac{1.0 \times 10^{- 6}}{{(3 \times 10^{- 1})}^{2}} = \frac{1 \times 10^{- 6}}{9 \times 10^{- 2}} = 0.11 \times 10^{- 4} = 1.1 \times 10^{- 5}$