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Question: How many milliliters of water are needed to dissolve \( 10mg \) of \( Pb{{F}_{2}} \) ?...

How many milliliters of water are needed to dissolve 10mg10mg of PbF2Pb{{F}_{2}} ?

Explanation

Solution

Hint : The solubility product of an electrolyte is equal to the product of ionic concentrations of the electrolyte and for lead fluoride the solubility product is as follows; Ksp=[Pb2+][F−]2{{K}_{sp}}=[P{{b}^{2+}}]{{[{{F}^{-}}]}^{2}} where Ksp={{K}_{sp}}= Solubility product, [Pb2+]=[P{{b}^{2+}}]= Concentration of the Lead and [F−]2={{[{{F}^{-}}]}^{2}}= Concentration of the fluoride.

Complete Step By Step Answer:
In the question it is asked to find the minimum volume of water required to dissolve 10mg10mg of lead fluoride. The given solubility product of lead fluoride is 4×10−84\times {{10}^{-8}} . The given mass of the lead fluoride is 10mg10mg . When calcium is added to water it dissolves and the ionic parts are in equilibrium with the calcium sulphate and it is represented as follows: PbF2(s)⇌Pb(aq)2++2F(aq)−Pb{{F}_{2(s)}}\rightleftharpoons Pb_{(aq)}^{2+}+2F_{(aq)}^{-} .
By definition, the solubility product constant for lead (II) fluoride is Ksp=[Pb2+][F−]2{{K}_{sp}}=[P{{b}^{2+}}]{{[{{F}^{-}}]}^{2}}
If we take s to be the molar solubility of the salt in water, which is the number of moles of lead(II) fluoride that dissociate to produce the lead(II) cations along with fluoride anions, we can say that we have; Ksp=s×(2s)2=4s3{{K}_{sp}}=s\times {{(2s)}^{2}}=4{{s}^{3}}
By rearranging the for finding the value of s, s=Ksp43=4×10−843=2.154×10−3Ms=\sqrt[3]{\dfrac{{{K}_{sp}}}{4}}=\sqrt[3]{\dfrac{4\times {{10}^{-8}}}{4}}=2.154\times {{10}^{-3}}M
This means that you can only dissolve 2.154×10−32.154\times {{10}^{-3}} moles of lead (II) fluoride for every liter of solution. For all intended purposes, volume of the solution could be approximated with volume of water, the solvent. To convert this into grams per liter, by using molar mass of lead (II) fluoride,
[2.154×10−3]×[moles(PbF2)1L(H2O)]×[245.2g1mole(PbF2)]=0.5282g⋅L−1\left[ 2.154\times {{10}^{-3}} \right]\times \left[ \dfrac{moles(Pb{{F}_{2}})}{1L({{H}_{2}}O)} \right]\times \left[ \dfrac{245.2g}{1mole(Pb{{F}_{2}})} \right]=0.5282g\cdot {{L}^{-1}}
Thus, the solubility of the salt is equivalent to, 0.5282g⋅L−1=0.5282mg⋅L−10.5282g\cdot {{L}^{-1}}=0.5282mg\cdot {{L}^{-1}}
Therefore, you will be able to dissolve 10mg10mg of lead (II) fluoride in
10mg×1mL(H2O)0.5282mg=19mL(H2O)10mg\times \dfrac{1mL({{H}_{2}}O)}{0.5282mg}=19mL({{H}_{2}}O)
Therefore, 1919 milliliters of water is needed to dissolve 10mg10mg of PbF2Pb{{F}_{2}} .

Note :
Initially we have to calculate in one liter of water how much lead (II) fluoride is going to get soluble. Then only we will get how much water is needed to dissolve 10g10g of lead (II) fluoride. Make sure to use the method given above to recalculate using the value you have for the Ksp{{K}_{sp}} of the salt.