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the addition of 0.3800 l of 1.150 m kcl to a solution containing ag and pb2 ions is just enough to precipitate all of the ions as agcl and pbcl2. the total mass of the resulting precipitate is 61.90 g. find the masses of pbcl2 and agcl in the precipitate.



Answer :

The mass of PbCl2 and AgCl is respectively 23.82 grams and 38.08 grams.

Solution -

Molarity = moles of solute / liters of solution

Moles of Cl- from KCl = molarity × volume in liter = 1.150 * 0.380 = 0.437

Let mass of AgCl = x grams

Moles of AgCl = x/143.32

So moles of Cl- from AgCl = x/143.32

Moles of PbCl2 = (61.90 - x) / 278.106

So moles of Cl- from PbCl2 = 2 × (61.90-x) / 278.106 = (61.90-x) / 139.053

Now,

(x/143.32) + (61.90-x) / 139.053 = 0.437

139.053x + 8871.508 - 143.32x = 8709.0062

- 4.267x = -162.5018

x = 38.08

 

Thus,

Mass of AgCl = x = 38.08 grams

Mass of PbCl2 = 61.9 - 38.08 = 23.82 grams

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