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Rufina [12.5K]
3 years ago
15

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:
lana [24]3 years ago
5 0

Answer:

Na3PO4 + 3AgNO3 -------> Ag3PO4 + 3NaNO3

In which [Ag+] in negligibly small and the concentration of each reactant is 1.0 M

The answer is A)  PO43- < NO3- < Na+

Explanation:

Ag+ is removed from the solution just like PO43-, so there are just 2 possible answers at this point: a or b. Then we can notice that Na3PO4 releases 3 moles of Na+ and just 1 mole of NO3-

We have 100mL of each reactant with the same concentration for both (1.0 M) so:

(0.1)(1)(3)= 0.3 mol Na+

(0.1)(1)= 0.1 mol NO3-

so PO43- < NO3- < Na+

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