Question #854b9

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Question #854b9

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Answer 1 · 2017-07-05T00:26:09

$K_{s p} = 3.97 \cdot 10^{- 3}$ $K_(sp) = 3.97 * 10^(-3)$

Explanation:

Notice that the problem provides you with the molar solubility of the salt, i.e. the number of moles that can be dissolved per liter of solution in order to have a saturated solution of $AB$ $"AB"$ .

In your case, you know that $0.0630$ $0.0630$ moles of $AB$ $"AB"$ can be dissolved in $1.00 L$ $"1.00 L"$ of water to form $1.00 L$ $"1.00 L"$ of saturated solution at $25^{\circ} C$ $25^@"C"$ .

You can thus say that the molar solubility of the salt, $s$ $s$ , is equal to

$s = {0.0630 mol L}^{- 1}$ $s = "0.0630 mol L"^(-1)$

Now, you know that when $AB$ $"AB"$ is dissolved in water, the following dissociation equilibrium is established

${AB}_{(s)} ⇌ A_{(A q)}^{+} + B_{(a q)}^{-}$ $"AB"_ ((s)) rightleftharpoons "A"_ ((Aq))^(+) + "B"_ ((aq))^(-)$

Notice that every mole of $AB$ $"AB"$ that dissociates produces $1$ $1$ mole of $A$ $"A"$ and $1$ $1$ mole of $B$ $"B"$ , so right from the start, you can say that the saturated solution will contain

$[A] = [B]$ $["A"] = ["B"]$

Since you already know the molar solubility of the salt at $25^{\circ} C$ $25^@"C"$ , i.e. the maximum number of moles of $AB$ $"AB"$ that dissociate to produce solvated ions, you can say that

$[A] = [B] = s$ $["A"] = ["B"] = s$

which means that you have

$[A] = [B] = {0.0630 mol L}^{- 1}$ $["A"] = ["B"] = "0.0630 mol L"^(-1)$

By definition, the solubility product constant, $K_{s p}$ $K_(sp)$ , is equal to

$K_{s p} = [A] \cdot [B]$ $K_(sp) = ["A"] * ["B"]$

This means that you have--I won't add the units here

$K_(sp) = 0.0630 * 0.0630 = color(darkgreen)(ul(color(black)(3.97 * 10^(-3))))$

The answer is rounded to three sig figs.

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Question #854b9

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