# Question 005a0

May 15, 2017

$B a C {l}_{2}$ and $N {a}_{2} S {O}_{4}$

#### Explanation:

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

$B a S {O}_{4} \left(s\right) + 2 N a C l \left(a q\right) \rightarrow B a C {l}_{2} \left(a q\right) + N {a}_{2} S {O}_{4} \left(a q\right)$.

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

May 15, 2017

No Reaction

#### Explanation:

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

Theoretically, $B a S {O}_{4} + 2 N a C l \implies N {a}_{2} S {O}_{4} + B a C {l}_{2}$ but neither sodium sulfate or barium chloride would form an insoluble precipitate (required) to be the driving force. Therefore, no reaction occurs.