Question

How do you calculate the concentration of iodate ions in a saturated solution of\[lead\left( {II} \right){\text{ }}iodate\;\]?

Answer 1

Hint:When we have to describe saturated solutions of ionic compounds of relatively low solubility, we use solubility product constants. When we state a solution as a saturated one we state it to be in dynamic equilibrium between dissolved, dissociated, ionic compound and undissolved solid. The equation for the same will be:
\[{M_x}{A_y}\left( s \right){\text{ }} - - > {\text{ }}x{\text{ }}{M^{y + }}\left( {aq} \right){\text{ }} + {\text{ }}y{\text{ }}{A^{x - }}\left( {aq} \right)\]
The general equilibrium constant for such processes can be written as:
\[{K_c}\; = {\text{ }}{\left[ {{M^{y + }}} \right]^x}{\left[ {{A^{x - }}} \right]^y}\]

Complete step-by-step answer:Here to calculate the concentration of iodate ions in a saturated solution of \[lead\left( {II} \right){\text{ }}iodate\;\] we need to understand that lead (II) iodate is known to be insoluble in water, thus there we can find a very low concentration of iodate anions, $I{O_3}^ - $ in a saturated solution of the same.
The dissociation equilibrium that describes the dissociation of lead(II) iodate looks like this

\[Pb{(I{O_3})_{2(s)}} \rightleftharpoons P{b^{2 + }}_{(aq)} + 2I{O^ - }_{3(aq)}\]

Here, every mole of \[lead\left( {II} \right){\text{ }}iodate\;\]which is known to dissociate produces one mole of lead cations and two moles of iodate anions in solution.
Furthermore, at equilibrium, this would means that a saturated solution of \[lead\left( {II} \right){\text{ }}iodate\;\]will have

\[[I{O_3}^ - ] = 2\,X\,[P{b^{2 + }}]\]
Now, the solubility product constant for this dissociation equilibrium:

\[{K_{sp}} = [P{b^{2 + }}]\,X\,{[I{O_3}^ - ]^2}\]
If s is taken to be the concentration of lead cations in the solution, i.e. the molar solubility of the salt, then it can be stated that

\[{K_{sp}} = s\,X\,{(2s)^2}\]

which is equivalent to
\[2.6\,X\,{10^{ - 13}} = 4{s^3}\]

Rearrange to solve for s
\[s = 3\sqrt {2.6\,X\,{{10}^{ - 13}}/4} = 4.02\,X\,{10^{ - 5}}\]

Thus a saturated solution of lead(II) iodate will have
\[[P{b^{ + 2}}] = 4.02\,X\,{10^{ - 5}}\;M\]
and
\[[I{O_3}^ - ] = 2\,X\,4.02\,X\,{10^{ - 5}}.\]

And thus,
\[M = 8.0\,X\,{10^{ - 5}}M\]

Note:When we talk about the equilibrium constant, we must know that it refers to the product of the concentration of ions which are there in a saturated solution of an ionic compound. the symbol for solubility product constant is\[{K_{sp}}\]. There is a need of the equation of the dissolving process to find this constant so the equilibrium expression can be written. Along with this, we need concentrations of each ion expressed in terms of molarity.