[Win,odd Dhamnekar]

 How to answer the following problem?  I shall try to post the answer given by author in my next  post, if required

[Borek]

you are here long enough to know it is always required

[Win,odd Dhamnekar]

My answer to question a.
[itex]\frac{5.6921 g ore \times 0.216 g Ag}{1.00 g ore}= 1.23 g Ag[/itex]
[itex] \frac{123 g Ag}{107.87 g/gmol}=[/itex] 1.14e-2 mol Ag
[itex]\frac{1.14e-2 mol Ag \times 1 Ag_3PO_4}{ 3 Ag}=[/itex] 3.78e-3 mol
0.00378 mol Ag₃PO₄ × 418.6 g mol Ag₃PO₄ =1.59 g Ag₃PO₄

My answer to question b.
K_sp= 8.89e-17=[Ag⁺]³ [PO₄^3-]=[Ag₊]³[0.12]= [Ag⁺]= 2.20e-6 M
Silver lost in 2.5 l solution would be
2.20e-6 M × 2.5 l= 5.50e-6 mol Ag
5.50e-6 mol Ag × 107.87 g/gmol Ag = 5.94e-4 g Ag lost [1]
Filteration and washing of Ag₃PO₄, No common ion, 0.5 l of wash water[ 3(PO₄^3-)]=Ag⁺
K_sp=8.89e-17 =[Ag⁺]³[PO₄^3-]=[3[PO₄^3-]]³[PO₄^3-]
[PO₄^3-]=4.26e-5 M, so Ag⁺= 3(4.26e-5 M)=1.28e-4 M

Silver lost in 0.5 l of solution would be:
1.28e-4 mol/l × 0.5 l= 6.39e-5 mol Ag

6.39e-5 mol Ag × 107.87 g/gmol Ag = 6.89e-3 g  Ag lost in washing Ag₃PO₄ [2]
Total silver lost= [1]+[2]=7.49e-3 g

My answer to question c  is 6.58 × 10³ g Ag lost.

My answer to question d[itex]\frac{6.58e3 g Ag}{28.35g/ounce}× $5.00/ounce=$1160[/itex] Author's answer is $1700

What are your answers to questions e and f?

Author's answers to question a, b, c are given below:

 



 

[mjc123]

Once again, we are dealing with different values of K_sp. Are you given a value in the question/textbook, or expected to look it up?

First step of b: how do you get [Ag⁺] = 2.2e-6 M? I get 9e-6 M.

e and f: as Borek said, you need to show an attempt of your own.

[Win,odd Dhamnekar]

Because of one wrong result given by my calculator, I change my answer  to silver lost during precipitation. It would be 2.44e-3 M. So total silver lost  would be 9.33e-3 g.

My answer to question c is 8.198 × 10³ g would be lost.
 My answer to question d is $1446.

[sjb]

Seems sensible - you might want to check your definition of an ounce (but this might be sneaky)