Question #005a0

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Question #005a0

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Answer 1 · 2017-05-15T23:22:59

$B a C l_{2}$ $BaCl_2$ and $N a_{2} S O_{4}$ $Na_2SO_4$

Explanation:

In aqueous solution, which is assumed for an exchange reaction like this one, the reaction is:

$B a S O_{4} (s) + 2 N a C l (a q) \to B a C l_{2} (a q) + N a_{2} S O_{4} (a q)$ $BaSO_4 (s) + 2NaCl (aq) rarr BaCl_2 (aq) + Na_2SO_4 (aq)$ .

The products are those species on the right side of the reaction arrow.

Answer 2 · 2017-05-15T23:24:17

No Reaction

Explanation:

$B a r i u m . Sulfate . K_{s p} = 1.5 x 10^{- 9}$ $Barium. "Sulfate". K_(sp) = 1.5 x 10^(-9)$ => $Solubility = \sqrt{K_{s p}}$ $"Solubility" = sqrt(K_(sp)$ = $\sqrt{1.5 x 10^{- 9}}$ $sqrt(1.5 x 10^(-9)$ = $3.9 x 10^{- 5}$ $3.9 x 10^-5$ M. This means that there is very little chance $N a^{+}$ $Na^+$ will react with $S O_{4}^{2 -} i o n$ $SO_4^(2-) ion$ and very little chance that the GpIIA $B a^{+ 2}$ $Ba^(+2)$ will react with the $C l^{-}$ $Cl^-$ . ion. Simply, there is no driving force for formation of product with Barium Sulfate + Sodium Chloride as reactants in the problem.

Theoretically, $B a S O_{4} + 2 N a C l \Rightarrow N a_{2} S O_{4} + B a C l_{2}$ $BaSO_4 + 2NaCl => Na_2SO_4 + BaCl_2$ but neither sodium sulfate or barium chloride would form an insoluble precipitate (required) to be the driving force. Therefore, no reaction occurs.

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Question #005a0

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