How do we produce a stoichiometrically balanced equation for the formation of $B a S O_{4}$ $BaSO_4$ ?

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Answer 1 · 2017-01-29T08:59:08

You balance it stoichiometrically.

And stoichiometry INSISTS that charge be balanced, AND mass be balanced. Experiment tells us that barium sulfate is insoluble in aqueous solution.

And thus:

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

The water solvate on the barium chloride is along for the ride in aqueous solution.

We could write the net ionic equation as follows:

$B a^{2 +} + S O_{4}^{2 -} \to B a S O_{4} (s) ⏐ ⏐ ↓$ $Ba^(2+) + SO_4^(2-) rarr BaSO_4(s)darr$

$K_{sp,}$ $K_"sp,"$ $B a S O_{4}$ $BaSO_4$ $=$ $=$ $1.0842 \times 10^{- 10}$ $1.0842 xx10^(−10)$ .

If you need further help working out the gram solubility from $K_{sp}$ $K_"sp"$ say the word..........

How do we produce a stoichiometrically balanced equation for the formation of BaSO4BaSO_4?