Use the value ksp=1.4x10-8 for PbI2 to solve the following problems?

A: What is the concentration of iodide ions in a saturated solution of PbI2? B: What is the solubility of PbI2 in a 0.010M solution of NaI?

Nov 3, 2016

Part- A

$P b {I}_{2} \left(s o \ln\right)$ ionises in its solution as follows

PbI_2("soln")rightleftharpoonsPb^"2+" +2I^-

If the concentration of $P b {I}_{2}$ in its saturated solution is xM then concentration of $P {b}^{\text{2+}}$ will be xM and concemtration of ${I}^{-}$ will be 2xM.

So K_"sp"=[Pb^"2+"][I^"-"]^2

$\implies 1.4 \times {10}^{-} 8 = x \cdot {\left(2 x\right)}^{2}$

$\implies {x}^{3} = \frac{1.4}{4} \times {10}^{-} 8 = 3.5 \times {10}^{-} 9$

$\implies x = 1.518 \times {10}^{-} 3$

So the concentrattion of ${I}^{-}$ ion in solution is $2 x = 3.036 \times {10}^{-} 3 M$

Part- B

Let the solubility of $P b {I}_{2}$ in 0.01 M NaI solution be s M.

Then in this case

$\left[P {b}^{\text{2+}}\right] = s M$

And

$\left[{I}^{\text{-}}\right] = \left(2 s + 0.01\right) M$

So K_"sp"=[Pb^"2+"][I^"-"]^2

$\implies 1.4 \times {10}^{-} 8 = s \times {\left(2 s + 0.01\right)}^{2}$

$\implies 1.4 \times {10}^{-} 8 = s \times \left(4 {s}^{2} + 4 s \cdot 0.01 + {\left(0.01\right)}^{2}\right)$

neglecting ${s}^{3} \mathmr{and} {s}^{2}$ terms we get

$\implies 1.4 \times {10}^{-} 8 = s \times {\left(0.01\right)}^{2}$

$\implies s = 1.4 \times {10}^{-} 4 M$

So solubilty of $P b {I}_{2}$ in this case is $= 1.4 \times {10}^{-} 4 M$