# Question #67a25

Mar 20, 2017

$\frac{2.5 \cdot {10}^{-} 11 \text{mol}}{L}$

#### Explanation:

When Pb(IO)_3 dissociates in water it forms

$P {b}^{+} + I {O}_{3}^{-}$

If x = molar solubility then

$\text{Therefore Ksp} = \left[P {b}^{+}\right] {\left[I {O}_{3}^{-}\right]}^{2}$

$\text{Or Ksp} \left(2.5 \cdot {10}^{-} 13\right) = \left[x\right] {\left[2 x\right]}^{2}$

As it is in a $N a I {O}_{3}$ of 0.100M and you want the molar solubility of $P b {\left(I O\right)}_{3}$ in NaIO3 and not in water. .

$2.5 \cdot {10}^{-} 13 = \left[x\right] {\left[x + 0.1 M\right]}^{2}$

But x is so so so so small that we can ignore it.
Therefore x + 0.1 = 0.1

Solve for x

$2.5 \cdot {10}^{-} 13 = \left[x\right] {\left[0.1 M\right]}^{2}$

$2.5 \cdot {10}^{-} 13 = 0.01 x$

$x = \frac{2.5 \cdot {10}^{-} 13}{0.01}$

$x = 2.5 \cdot {10}^{-} 11$

= $\frac{2.5 \cdot {10}^{-} 11 \text{mol}}{L}$