# Question 0a80d

May 1, 2017
• 46: ${K}_{s p} = 1.1 \cdot {10}^{-} 38$
• 47: $x = 1.3 \cdot {10}^{-} 4 M$
• 48: $x = 0.0012 M$

#### Explanation:

• 46:

The first step to solving molar solubility problems is always to write down the balanced chemical equation of what's going on.

This would be:
$B {a}_{3} {\left(P {O}_{4}\right)}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s 3 B {a}^{2 +} \left(a q\right) + 2 P {O}_{4}^{3 -} \left(a q\right)$

Therefore the expression for ${K}_{s p}$ would be:

${K}_{s p} = {\left[B {a}^{2 +}\right]}^{3} {\left[P {O}_{4}^{3 -}\right]}^{2}$

Setting up an ICE table would give you:

${K}_{s p} = {\left(3 x\right)}^{3} {\left(2 x\right)}^{2} = 108 {x}^{5}$

Note: Watch your parentheses when simplifying these expressions

Now, the jump to take with all such problems is to realize that $x$ is your molar solubility. In the ICE table, it represents how much of your solid dissolved, and therefore is the molar solubility.

You're given a molar solubility, so all you need to do is plug and chug:

${K}_{s p} = 108 {\left(1.4 \cdot {10}^{-} 8\right)}^{5} = 1.1 \cdot {10}^{-} 38$

• 47:

Again, same process of setting up the expression:

$A {g}_{2} C r {O}_{4} \left(s\right) r i g h t \le f t h a r p \infty n s 2 A {g}^{+} \left(a q\right) + C r {O}_{4}^{2 -} \left(a q\right)$

${K}_{s p} = {\left[A {g}^{+}\right]}^{2} \left[C r {O}_{4}^{2 -}\right]$

By ICE Table:
${K}_{s p} = {\left(2 x\right)}^{2} \left(x\right) = 4 {x}^{3}$

The ${K}_{s p}$ value for this solid is $9.0 \cdot {10}^{-} 12.$ Hence:

$4 {x}^{3} = 9.0 \cdot {10}^{-} 12$
$x = 1.3 \cdot {10}^{-} 4 M$

• 48:

This problem is exactly the same process as the previous one, except with a different reaction:

$P b {I}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s 2 {I}^{-} \left(a q\right) + P {b}^{2 +} \left(a q\right)$

${K}_{s p} = \left[P {b}^{2 +}\right] {\left[{I}^{-}\right]}^{2}$

By ICE Table:
${K}_{s p} = \left(x\right) {\left(2 x\right)}^{2} = 4 {x}^{3}$

The ${K}_{s p}$ value for this solid is 7.1×10^(–9). Hence:

4x^3 = 7.1×10^(–9)
$x = 0.0012 M$

To better understand what is exactly going on here, I'd recommend you watch these two videos.

Hope that helps :)