Answer: The correct answer is option A.
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Explanation:


100 mL of 1.0 M 
Volume of
= 100 mL = 0.1 L
Moles of
= n

1 mole of
gives 3 moles of sodium ions and 1 mole of phosphate ions.
Moles of sodium ions = 
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of sodium ions= 
100 mL of 1.0 M 
Volume of
= 100 mL = 0.1 L
Moles of
= n'

1 mole of
gives 1 mole of silver ions and 1 mole of nitrate ions.
According to reaction 1 mole of
reacts with 3 moles of
.
Then 0.1 mole of
will react with:
of 
Moles of sodium phosphate left unreacted = 0.1 mol - 0.0333 mol = 0.0667 mol
1 mole of
gives 3 moles of sodium ions and 1 mole of phosphate ions.
As we can see that silver nitrate is in limiting amount
According to reaction 3 mole of
gives with 1 mole of
.
So, when 0.1 mol of
reacts it gives:
of 
Moles of phosphate ions left in solution= 
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of phosphate ions= 
Moles of nitrate ions = 
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of nitrate ions= 
But is an excessive reagent its concentration will be less
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