Answer: The correct answer is option A.
![[PO_4^{3-}]](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%3C%5BNO_3%5E%7B-%7D%5D%3C%5BNa%5E%7B%2B%7D%5D)
Explanation:
![Na_3PO_4+3AgNO_3\rightarrow Ag_3PO_4+3NaNO_3](https://tex.z-dn.net/?f=Na_3PO_4%2B3AgNO_3%5Crightarrow%20Ag_3PO_4%2B3NaNO_3)
![Concentration = \frac{Moles}{\text{Volume of Solution(L)}}](https://tex.z-dn.net/?f=Concentration%20%3D%20%5Cfrac%7BMoles%7D%7B%5Ctext%7BVolume%20of%20Solution%28L%29%7D%7D)
100 mL of 1.0 M ![Na_3PO_4](https://tex.z-dn.net/?f=Na_3PO_4)
Volume of
= 100 mL = 0.1 L
Moles of
= n
![n= 1.0 M\times 0.1 L=0.1 mol](https://tex.z-dn.net/?f=n%3D%201.0%20M%5Ctimes%200.1%20L%3D0.1%20mol)
1 mole of
gives 3 moles of sodium ions and 1 mole of phosphate ions.
Moles of sodium ions = ![0.1 \times 3 mol =0.3 mol](https://tex.z-dn.net/?f=0.1%20%5Ctimes%203%20mol%20%3D0.3%20mol)
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of sodium ions= ![\frac{0.3 mol}{0.2 L}=1.5 L](https://tex.z-dn.net/?f=%5Cfrac%7B0.3%20mol%7D%7B0.2%20L%7D%3D1.5%20L)
100 mL of 1.0 M ![AgNO_3](https://tex.z-dn.net/?f=AgNO_3)
Volume of
= 100 mL = 0.1 L
Moles of
= n'
![n'= 1.0 M\times 0.1 L=0.1 mol](https://tex.z-dn.net/?f=n%27%3D%201.0%20M%5Ctimes%200.1%20L%3D0.1%20mol)
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 ![Na_3PO_4](https://tex.z-dn.net/?f=Na_3PO_4)
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 ![Ag_3PO_4](https://tex.z-dn.net/?f=Ag_3PO_4)
Moles of phosphate ions left in solution= ![0.1 \times 0.0667 mol =0.0667 mol](https://tex.z-dn.net/?f=0.1%20%5Ctimes%200.0667%20%20mol%20%3D0.0667%20mol)
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of phosphate ions= ![\frac{0.0667 mol}{0.2 L}=0.3335 L](https://tex.z-dn.net/?f=%5Cfrac%7B0.0667%20mol%7D%7B0.2%20L%7D%3D0.3335%20L)
Moles of nitrate ions = ![0.1 \times 1 mol =0.1 mol](https://tex.z-dn.net/?f=0.1%20%5Ctimes%201%20mol%20%3D0.1%20mol)
Volume of solution after mixing = 200 mL = 0.2 L
Concentration of nitrate ions= ![\frac{0.1 mol}{0.2 L}=0.5 L](https://tex.z-dn.net/?f=%5Cfrac%7B0.1%20mol%7D%7B0.2%20L%7D%3D0.5%20L)
But is an excessive reagent its concentration will be less
![[PO_4^{3-}]](https://tex.z-dn.net/?f=%5BPO_4%5E%7B3-%7D%5D%3C%5BNO_3%5E%7B-%7D%5D%3C%5BNa%5E%7B%2B%7D%5D)