Question

When 100 mL of 1.0 M Na3PO4 is mixed with 100 mL of 1.0 M AgNO3, a yellow precipitate forms and [Ag ] becomes negligibly small. Which of the following is a correct listing of the ions remaining in solution in order of increasing concentration

(A) PO43- < NO3- < Na+
(B) PO43- < Na+ < NO3-
(C) NO3- < PO43- < Na+
(D) Na+ < NO3- < PO43-
(E) Na+ < PO43- < NO3-

Answer 1

Answer: The correct answer is option A.

`[PO_4^(3-)]<[NO_3^(-)]<[Na^(+)]`

Step-by-step explanation:

`Na_3PO_4+3AgNO_3\rightarrow Ag_3PO_4+3NaNO_3`

`Concentration = \frac{Moles}{\text{Volume of Solution(L)}}`

100 mL of 1.0 M
`Na_3PO_4`

Volume of
`Na_3PO_4` = 100 mL = 0.1 L

Moles of
`Na_3PO_4` = n

`n= 1.0 M* 0.1 L=0.1 mol`

1 mole of
`Na_3PO_4` gives 3 moles of sodium ions and 1 mole of phosphate ions.

Moles of sodium ions =
`0.1 * 3 mol =0.3 mol`

Volume of solution after mixing = 200 mL = 0.2 L

Concentration of sodium ions=
`(0.3 mol)/(0.2 L)=1.5 L`

100 mL of 1.0 M
`AgNO_3`

Volume of
`AgNO_3` = 100 mL = 0.1 L

Moles of
`AgNO_3` = n'

`n'= 1.0 M* 0.1 L=0.1 mol`

1 mole of
`AgNO_3` gives 1 mole of silver ions and 1 mole of nitrate ions.

According to reaction 1 mole of
`Na_3PO_4` reacts with 3 moles of
`AgNO_3`.

Then 0.1 mole of
`AgNO_3` will react with:

`(1)/(3)* 0.1=0.0333mol` of
`Na_3PO_4`

Moles of sodium phosphate left unreacted = 0.1 mol - 0.0333 mol = 0.0667 mol

1 mole of
`Na_3PO_4` 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
`AgNO_3` gives with 1 mole of
`Ag_3PO_4`.

So, when 0.1 mol of
`AgNO_3` reacts it gives:

`(1)/(3)* 0.1 = 0.3 mol` of
`Ag_3PO_4`

Moles of phosphate ions left in solution=
`0.1 * 0.0667  mol =0.0667 mol`

Volume of solution after mixing = 200 mL = 0.2 L

Concentration of phosphate ions=
`(0.0667 mol)/(0.2 L)=0.3335 L`

Moles of nitrate ions =
`0.1 * 1 mol =0.1 mol`

Volume of solution after mixing = 200 mL = 0.2 L

Concentration of nitrate ions=
`(0.1 mol)/(0.2 L)=0.5 L`

But is an excessive reagent its concentration will be less

`[PO_4^(3-)]<[NO_3^(-)]<[Na^(+)]`