How do you use the $K_{sp}$ $K_"sp"$ value to calculate the molar solubility of the following compound in pure water?

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How do you use the $K_{sp}$ $K_"sp"$ value to calculate the molar solubility of the following compound in pure water?

${BaSO}_{4}$ $"BaSO"_4$

$K_{sp} = 1.07 \times 10^{- 10}$ $K_"sp" = 1.07xx10^(-10)$

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Answer 1 · 2016-06-08T00:06:06

Here's how you can do that.

Explanation:

What you need to do here is set up an ICE table based on the equilibrium reaction that describes the way barium sulfate, ${BaSO}_{4}$ $"BaSO"_4$ , dissolves in aqueous solution.

The salt is considered insoluble in water, so right from the start you know that the solution will contain very little amounts of dissolved ions, since most of the compound will remain undissociated as a solid.

So, barium sulfate will dissolve in very small quantities to produce

${BaSO}_{4 (a q)} ⇌ {Ba}_{(a q)}^{2 +} + {SO}_{4 (a q)}^{2 -}$ $"BaSO"_ (4(aq)) rightleftharpoons "Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-)$

Notice that for every mole of barium sulfate that dissolves in solution, you get one mole of barium cations, ${Ba}^{2 +}$ $"Ba"^(2+)$ , and one mole of sulfate anions, ${SO}_{4}^{2 -}$ $"SO"_4^(2-)$ .

If you take $s$ $s$ to be the molar solubility of the salt, i.e. the number of moles of barium sulfate that dissolve to release ions per liter of solution, you can use an ICE table to write

${BaSO}_{4 (a q)} ⇌ {Ba}_{(a q)}^{2 +} + {SO}_{4 (a q)}^{2 -}$ $"BaSO"_ (4(aq)) rightleftharpoons " ""Ba"_ ((aq))^(2+) " "+" " "SO"_ (4(aq))^(2-)$

$I a a a a - a a a a a a a a a a a 0 a a a a a a a a a a 0$ $color(purple)("I")color(white)(aaaacolor(black)(-)aaaaaaaaaaacolor(black)(0)aaaaaaaaaacolor(black)(0)$
$C a a a a - a a a a a a a a a (+ s) a a a a a a (+ s)$ $color(purple)("C")color(white)(aaaacolor(black)(-)aaaaaaaaacolor(black)((+s))aaaaaacolor(black)((+s))$
$E a a a a - a a a a a a a a a a a s a a a a a a a a a a s$ $color(purple)("E")color(white)(aaaacolor(black)(-)aaaaaaaaaaacolor(black)(s)aaaaaaaaaacolor(black)(s)$

By definition, the solubility product constant, $K_{s p}$ $K_(sp)$ , for this solubility equilibrium will be equal to

$K_{s p} = [{Ba}^{2 +}] \cdot [{SO}_{4}^{2 -}]$ $K_(sp) = ["Ba"^(2+)] * ["SO"_4^(2-)]$

Since the expression of $K_{s p}$ $K_(sp)$ uses equilibrium concentrations, you can say that

$K_{s p} = s \cdot s = s^{2}$ $K_(sp) = s * s = s^2$

In your case, you have

$1.07 \cdot 10^{- 10} = s^{2}$ $1.07 * 10^(-10) = s^2$

Solve for $s$ $s$ to find

$s = \sqrt{1.07 \cdot 10^{- 10}} = 1.03 \cdot 10^{- 5}$ $s = sqrt(1.07 * 10^(-10)) = 1.03 * 10^(-5)$

Since $s$ $s$ represents molar solubility, you can say that you have

$s = color(green)(|bar(ul(color(white)(a/a)color(black)(1.03 * 10^(-5)"M")color(white)(a/a)|)))$

This means that when you add barium sulfate to water, you can only hope to dissolve $1.03 * 10^(-5)$ moles for every liter of solution.

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How do you use the $K_{sp}$ $K_"sp"$ value to calculate the molar solubility of the following compound in pure water?

${BaSO}_{4}$ $"BaSO"_4$

$K_{sp} = 1.07 \times 10^{- 10}$ $K_"sp" = 1.07xx10^(-10)$

1 Answer

Explanation:

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How do you use the K_"sp"KspK_"sp" value to calculate the molar solubility of the following compound in pure water?

"BaSO"_4BaSO4"BaSO"_4 K_"sp" = 1.07xx10^(-10)Ksp=1.07×10−10K_"sp" = 1.07xx10^(-10)

1 Answer

Explanation:

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Impact of this question

How do you use the $K_{sp}$ $K_"sp"$ value to calculate the molar solubility of the following compound in pure water?

${BaSO}_{4}$ $"BaSO"_4$

$K_{sp} = 1.07 \times 10^{- 10}$ $K_"sp" = 1.07xx10^(-10)$