# Question #71344

Apr 1, 2015

The solubility $= 9.22 \times {10}^{- 10} \text{mol/l}$

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

${K}_{s p} = 8.5 \times {10}^{- 19} m o {l}^{2} . {l}^{- 2}$

${K}_{s p} = \left[P {b}_{\left(a q\right)}^{2 +}\right] \left[{C}_{2} {O}_{4 \left(a q\right)}^{2 -}\right]$

Since the concentration of lead(II) and ethandioate are equal we can write:

${K}_{s p} = {\left[P {b}_{\left(a q\right)}^{2 +}\right]}^{2} = 8.5 \times {10}^{- 19} m o {l}^{2} . {l}^{- 2}$

So $\left[P {b}_{\left(a q\right)}^{2 +}\right] = \sqrt{8.5 \times {10}^{- 19}} = 9.22 \times {10}^{- 10} \text{mol/l}$

This is the concentration of lead(II) ions in the saturated solution.

Since $P b {C}_{2} {O}_{4}$ is 1 molar with respect to Pb(II) then this is also the solubility of $P b {C}_{2} {O}_{4}$ in $\text{mol/l}$. I assume this is what you mean by "dissolution value".

You must always quote the temperature for which Ksp refers. You should be given this with the question.

Apr 1, 2015

Dissolution is the process by which a solute forms a solution in a solvent. The dissolution value may be the ion concentration after dissolution. This would be $9.2 \cdot {10}^{- 10} M$

The Ksp value in this process is the relationship

${K}_{s p} = \left[P {b}^{2 +}\right] \left[{C}_{2} {O}_{4}^{2 -}\right] = 8.5 \cdot {10}^{- 19}$

For each formula unit of lead II oxalate that dissolves you get one lead II ion and one oxalate ion. Let x = these concentrations.

$x = \left[P {b}^{2 +}\right] = \left[{C}_{2} {O}_{4}^{2 -}\right]$
${x}^{2} = 8.5 \cdot {10}^{- 19}$

so, after dissolution the ion concentrations are

$x = \left[P {b}^{2 +}\right] = \left[{C}_{2} {O}_{4}^{2 -}\right] = 9.2 \cdot {10}^{- 10} M$