Question #fb741

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Question #fb741

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Answer 1 · 2016-03-15T16:25:59

The first part of the question allows you to determine $K_{s p}$ $K_(sp)$ , and the second allows you to use $K_{s p}$ $K_(sp)$ to find the solubility of $C a C l_{2}$ $CaCl_2$ in the presence of a common ion ( $C l^{-}$ $Cl^-$ ).

Explanation:

The balanced reaction for ionization of $C a C l_{2}$ $CaCl_2$ is

$C a C l_{2} (s) \leftrightarrow C a^{2 +} (a q) + 2 C l^{-} (a q)$ $CaCl_2(s) harr Ca^(2+)(aq) + 2 Cl^(-)(aq)$
$K_{s p} = [C a^{2 +}] {[C l^{-}]}^{2}$ $K_(sp) = [Ca^(2+)][Cl^-]^2$

If $[C l^{-}] = 2 \times 10^{- 6}$ $[Cl^-]=2 times 10^-6$ M then $[C a^{2 +}] = 1 \times 10^{- 6}$ $[Ca^(2+)] = 1 times 10^-6$ M and
$K_{s p} = 4 \times 10^{- 18}$ $K_(sp) = 4 times 10^-18$

For the second part, the $B a C l_{2}$ $BaCl_2$ produces $[C l^{-}] = 10^{- 2}$ $[Cl^-]=10^(-2)$ M
Therefore, $[C a^{2 +}] = \frac{K_{s p}}{{[C l^{-}]}^{2}} = 4 \times 10^{- 14}$ $[Ca^(2+)] = K_(sp)/([Cl^-]^2) = 4 times 10^(-14)$ M

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Question #fb741

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