Question

Calculate the mass of precipitate that will form if 0.75 L of 0.50 M lead nitrate is mixed with 0.850 L of concentrated sodium bromate. Ksp for lead bromate = 7.9 x 10-6.

Answer 1

Answer : The mass of
`Pb(BrO_3)_2` formed is 164.4 grams.

Explanation :

The balanced chemical reaction will be,

`PbNO_3(aq)+2NaBrO_3(aq)\rightarrow Pb(BrO_3)_2(s)+2NaNO_3(aq)`

First we have to calculate the moles of
`PbNO_3`.

`\text{Moles of }PbNO_3=\text{Molarity}* \text{Volume in L}=0.50M* 0.75L=0.375moles`

From the balanced chemical reaction we conclude that,

Moles of
`PbNO_3` = Moles of
`Pb(BrO_3)_2` = 0.375 mole

The moles of precipitate formed
`Pb(BrO_3)_2` = 0.375 mole

Now we have to calculate the moles of
`Pb(BrO_3)_2` that dissolve in
`PbNO_3`.

The dissociation of lead bromate is written as:

`Pb(BrO_3)_2\rightleftharpoons Pb^(2+)+2BrO_3^-`

The expression for solubility constant for this reaction will be,

`K_(sp)=[Pb^(2+)][BrO_3^-]^2`

`K_(sp)=(s)* (2s)^2`

`K_(sp)=4s^3`

`7.9* 10^(-6)=4s^3`

`s=0.0125M`

Total volume of the solution = 0.75 + 0.850 = 1.60 L

`\text{Moles of }Pb(BrO_3)_2=\text{Molarity}* \text{Volume in L}=0.0125M* 1.60L=0.020moles`

The moles of
`Pb(BrO_3)_2` that dissolve in
`PbNO_3` = 0.020 mole

The moles of precipitate formed
`Pb(BrO_3)_2` = 0.375 - 0.020 = 0.355 mole

Now we have to calculate the mass of
`Pb(BrO_3)_2` formed.

`\text{ Mass of }Pb(BrO_3)_2=\text{ Moles of }Pb(BrO_3)_2* \text{ Molar mass of }Pb(BrO_3)_2`

`\text{ Mass of }Pb(BrO_3)_2=(0.355moles)* (463g/mole)=164.4g`

Therefore, the mass of
`Pb(BrO_3)_2` formed is 164.4 grams.