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anygoal [31]
2 years ago
11

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-
Chemistry
1 answer:
steposvetlana [31]2 years ago
7 0

Answer: The correct answer is option A.

[PO_4^{3-}]

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\times 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 \times 3 mol =0.3 mol

Volume of solution after mixing = 200 mL = 0.2 L

Concentration of sodium ions= \frac{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\times 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:

\frac{1}{3}\times 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:

\frac{1}{3}\times 0.1 = 0.3 mol of Ag_3PO_4

Moles of phosphate ions left in solution= 0.1 \times 0.0667  mol =0.0667 mol

Volume of solution after mixing = 200 mL = 0.2 L

Concentration of phosphate ions= \frac{0.0667 mol}{0.2 L}=0.3335 L

Moles of nitrate ions =  0.1 \times 1 mol =0.1 mol

Volume of solution after mixing = 200 mL = 0.2 L

Concentration of nitrate ions= \frac{0.1 mol}{0.2 L}=0.5 L

But is an excessive reagent its concentration will be less

[PO_4^{3-}]

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