Respuesta :
Answer : The mass of [tex]Pb(BrO_3)_2[/tex] formed is 164.4 grams.
Explanation :
The balanced chemical reaction will be,
[tex]PbNO_3(aq)+2NaBrO_3(aq)\rightarrow Pb(BrO_3)_2(s)+2NaNO_3(aq)[/tex]
First we have to calculate the moles of [tex]PbNO_3[/tex].
[tex]\text{Moles of }PbNO_3=\text{Molarity}\times \text{Volume in L}=0.50M\times 0.75L=0.375moles[/tex]
From the balanced chemical reaction we conclude that,
Moles of [tex]PbNO_3[/tex] = Moles of [tex]Pb(BrO_3)_2[/tex] = 0.375 mole
The moles of precipitate formed [tex]Pb(BrO_3)_2[/tex] = 0.375 mole
Now we have to calculate the moles of [tex]Pb(BrO_3)_2[/tex] that dissolve in [tex]PbNO_3[/tex].
The dissociation of lead bromate is written as:
[tex]Pb(BrO_3)_2\rightleftharpoons Pb^{2+}+2BrO_3^-[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Pb^{2+}][BrO_3^-]^2[/tex]
[tex]K_{sp}=(s)\times (2s)^2[/tex]
[tex]K_{sp}=4s^3[/tex]
[tex]7.9\times 10^{-6}=4s^3[/tex]
[tex]s=0.0125M[/tex]
Total volume of the solution = 0.75 + 0.850 = 1.60 L
[tex]\text{Moles of }Pb(BrO_3)_2=\text{Molarity}\times \text{Volume in L}=0.0125M\times 1.60L=0.020moles[/tex]
The moles of [tex]Pb(BrO_3)_2[/tex] that dissolve in [tex]PbNO_3[/tex] = 0.020 mole
The moles of precipitate formed [tex]Pb(BrO_3)_2[/tex] = 0.375 - 0.020 = 0.355 mole
Now we have to calculate the mass of [tex]Pb(BrO_3)_2[/tex] formed.
[tex]\text{ Mass of }Pb(BrO_3)_2=\text{ Moles of }Pb(BrO_3)_2\times \text{ Molar mass of }Pb(BrO_3)_2[/tex]
[tex]\text{ Mass of }Pb(BrO_3)_2=(0.355moles)\times (463g/mole)=164.4g[/tex]
Therefore, the mass of [tex]Pb(BrO_3)_2[/tex] formed is 164.4 grams.
