Question #67a25

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Answer 1 · 2017-03-20T04:44:06

$\frac{2.5 \cdot 10^{- 11} mol}{L}$ $(2.5*10^-11"mol")/L$

When Pb(IO)_3 dissociates in water it forms

$P b^{+} + I O_{3}^{-}$ $Pb^+ + IO_3^-$

If x = molar solubility then

$Therefore Ksp = [P b^{+}] {[I O_{3}^{-}]}^{2}$ $"Therefore Ksp" = [Pb^+][IO_3^-]^2$

$Or Ksp (2.5 \cdot 10^{- 13}) = [x] {[2 x]}^{2}$ $"Or Ksp"(2.5*10^-13) = [x][2x]^2$

As it is in a $N a I O_{3}$ $NaIO_3$ of 0.100M and you want the molar solubility of $P b {(I O)}_{3}$ $Pb(IO)_3$ in NaIO3 and not in water. .

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

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

Solve for x

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

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

$x = \frac{2.5 \cdot 10^{- 13}}{0.01}$ $x = (2.5*10^-13)/0.01$

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

= $\frac{2.5 \cdot 10^{- 11} mol}{L}$ $(2.5*10^-11"mol")/L$