Question #0e38d

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Question #0e38d

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Answer 1 · 2015-05-15T01:49:53

The solubility product constant for this salt will be $1.22 \cdot 10^{- 3}$ $1.22 * 10^(-3)$ .

When dissolved in water, your salt will dissociate into cations and anions according to the following equilibrium

$A B_{3 (s)} ⇌ A_{(a q)}^{3 +} + 3 B_{(a q)}^{-}$ $AB_(3(s)) rightleftharpoons A_((aq))^(3+) + color(red)(3)B_((aq))^(-)$

Since you're dealing with a 1.00-L solution, the molar solubility of your salt will be

$C = \frac{n}{V} = \frac{0.0820 moles}{1.00 L} = 0.0820 mol/L$ $C = n/V = "0.0820 moles"/"1.00 L" = "0.0820 mol/L"$

Take a look at the mole ratio that exists between the species involved in the equilibrium. You have 1 mole of $A B_{3}$ $AB_3$ dissociating to produce 1 mole of $A^{3 +}$ $A^(3+)$ and $3$ $color(red)(3)$ moles of $B^{-}$ $B^(-)$ .

You can use an ICE table to determine the value of the $K_{s p}$ $K_(sp)$ .

$A B_{3 (s)} ⇌ A_{(a q)}^{3 +} + B_{(a q)}^{-}$ $" "AB_(3(s)) rightleftharpoons A_((aq))^(3+) + B_((aq))^(-)$
I........ $-$ $-$ .............0..............0
C...... $-$ $-$ ............(+x)...........(+ $3$ $color(red)(3)$ x)
E....... $-$ $-$ .............x...............3x

By definition, $K_{s p}$ $K_(sp)$ will be

$k_{s p} = [A^{3 +}] \cdot {[B^{-}]}^{3} = x \cdot {(3 x)}^{3} = 27 \cdot x^{4}$ $k_(sp) = [A^(3+)] * [B^(-)]^(color(red)(3)) = x * (3x)^3 = 27 * x^4$

But $x$ $x$ is actually the molar solubility of the salt, 0.0820 mol/L, which means that the value of the solubility product constant will be

$K_{s p} = 27 \cdot {(0.0820)}^{4} = 1.22 \cdot 10^{- 3}$ $K_(sp) = 27 * (0.0820)^4 = color(green)(1.22 * 10^(-3))$

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