Respuesta :
Answer: The correct answer is option A.
[tex][PO_4^{3-}]<[NO_3^{-}]<[Na^{+}][/tex]
Explanation:
[tex]Na_3PO_4+3AgNO_3\rightarrow Ag_3PO_4+3NaNO_3[/tex]
[tex]Concentration = \frac{Moles}{\text{Volume of Solution(L)}}[/tex]
100 mL of 1.0 M [tex]Na_3PO_4[/tex]
Volume of [tex]Na_3PO_4[/tex] = 100 mL = 0.1 L
Moles of [tex]Na_3PO_4[/tex] = n
[tex]n= 1.0 M\times 0.1 L=0.1 mol[/tex]
1 mole of [tex]Na_3PO_4[/tex] gives 3 moles of sodium ions and 1 mole of phosphate ions.
Moles of sodium ions = [tex]0.1 \times 3 mol =0.3 mol[/tex]
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of sodium ions= [tex]\frac{0.3 mol}{0.2 L}=1.5 L[/tex]
100 mL of 1.0 M [tex]AgNO_3[/tex]
Volume of [tex]AgNO_3[/tex] = 100 mL = 0.1 L
Moles of [tex]AgNO_3[/tex] = n'
[tex]n'= 1.0 M\times 0.1 L=0.1 mol[/tex]
1 mole of [tex]AgNO_3[/tex] gives 1 mole of silver ions and 1 mole of nitrate ions.
According to reaction 1 mole of [tex]Na_3PO_4[/tex] reacts with 3 moles of [tex]AgNO_3[/tex].
Then 0.1 mole of [tex]AgNO_3[/tex] will react with:
[tex]\frac{1}{3}\times 0.1=0.0333mol[/tex] of [tex]Na_3PO_4[/tex]
Moles of sodium phosphate left unreacted = 0.1 mol - 0.0333 mol = 0.0667 mol
1 mole of [tex]Na_3PO_4[/tex] 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 [tex]AgNO_3[/tex] gives with 1 mole of [tex]Ag_3PO_4[/tex].
So, when 0.1 mol of [tex]AgNO_3[/tex] reacts it gives:
[tex]\frac{1}{3}\times 0.1 = 0.3 mol[/tex] of [tex]Ag_3PO_4[/tex]
Moles of phosphate ions left in solution= [tex]0.1 \times 0.0667 mol =0.0667 mol[/tex]
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of phosphate ions= [tex]\frac{0.0667 mol}{0.2 L}=0.3335 L[/tex]
Moles of nitrate ions = [tex]0.1 \times 1 mol =0.1 mol[/tex]
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of nitrate ions= [tex]\frac{0.1 mol}{0.2 L}=0.5 L[/tex]
But is an excessive reagent its concentration will be less
[tex][PO_4^{3-}]<[NO_3^{-}]<[Na^{+}][/tex]
