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2 years ago

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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-

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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