At a certain temperature, the solubility of barium chromate ( $B a C r O_{4}$ $BaCrO_4$ ) is $1.8 x 10^{- 5}$ $1.8 x 10^-5$ mol/L. What is the Ksp value at this temperature?

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At a certain temperature, the solubility of barium chromate ( $B a C r O_{4}$ $BaCrO_4$ ) is $1.8 x 10^{- 5}$ $1.8 x 10^-5$ mol/L. What is the Ksp value at this temperature?

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Answer 1 · 2016-10-27T18:22:51

$K_{s p}$ $K_(sp)$ $=$ $=$ $[B a^{2 +}] [C r O_{4}^{2 -}]$ $[Ba^(2+)][CrO_4^(2-)]$ $=$ $=$ ${(1.8 \times 10^{- 5})}^{2}$ $(1.8xx10^-5)^2$

Explanation:

We write the solubility expression in this way:

$B a C r O_{4} (s) ⇌ B a^{2 +} + C r O_{4}^{2 -}$ $BaCrO_4(s) rightleftharpoons Ba^(2+) + CrO_4^(2-)$

Now $K_{s p}$ $K_(sp)$ $=$ $=$ $[B a^{2 +}] [C r O_{4}^{2 -}]$ $[Ba^(2+)][CrO_4^(2-)]$ $=$ $=$ ${(1.8 \times 10^{- 5})}^{2}$ $(1.8xx10^-5)^2$ $=$ $=$ $3.24 \times 10^{- 10} .$ $3.24xx10^-10.$

Why do we square this value? We were quoted the solubility of barium chromate. Because the solubility represents the concentrations of chromate ion and barium ion.

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At a certain temperature, the solubility of barium chromate ( $B a C r O_{4}$ $BaCrO_4$ ) is $1.8 x 10^{- 5}$ $1.8 x 10^-5$ mol/L. What is the Ksp value at this temperature?

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Explanation:

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At a certain temperature, the solubility of barium chromate (BaCrO4BaCrO_4) is 1.8x10−51.8 x 10^-5 mol/L. What is the Ksp value at this temperature?

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Explanation:

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At a certain temperature, the solubility of barium chromate ( $B a C r O_{4}$ $BaCrO_4$ ) is $1.8 x 10^{- 5}$ $1.8 x 10^-5$ mol/L. What is the Ksp value at this temperature?