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 : The mass of formed is 164.4 grams.

Explanation :

The balanced chemical reaction will be,

First we have to calculate the moles of .

From the balanced chemical reaction we conclude that,

Moles of = Moles of = 0.375 mole

The moles of precipitate formed = 0.375 mole

Now we have to calculate the moles of that dissolve in .

The dissociation of lead bromate is written as:

The expression for solubility constant for this reaction will be,

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

The moles of that dissolve in = 0.020 mole

The moles of precipitate formed = 0.375 - 0.020 = 0.355 mole

Now we have to calculate the mass of formed.

Therefore, the mass of formed is 164.4 grams.

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