Question #ac64e

Question

Question #ac64e

1 Answer

Impact of this question

1762 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-04T01:17:20

Here's what I got.

Explanation:

Interestingly enough, the name sodium phosphate is actually used to describe a family of salts that contain sodium cations and phosphate anions in various ratios.

My guess is that you are supposed to treat "sodium phosphate" as trisodium phosphate, ${Na}_{3} {PO}_{4}$ $"Na"_3"PO"_4$ .

The idea here is that a formula unit of sodium phosphate, ${Na}_{3} {PO}_{4}$ $"Na"_3"PO"_4$ , contains

three sodium cations, $3 \times {Na}^{+}$ $3 xx "Na"^(+)$

one phosphate anion, $1 \times {PO}_{4}^{3 -}$ $1 xx "PO"_4^(3-)$

This, of course, implies that $1$ $1$ mole of sodium phosphate will contain $3$ $3$ moles of sodium cations and $1$ $1$ mole of phosphate anions.

In other words, you have

${no. of moles of Na}^{+} = 3 \times {no. of moles of Na}_{3} {PO}_{4}$ $"no. of moles of Na"^(+) = 3 xx "no. of moles of Na"_3"PO"_4$

${no. of moles of PO}_{4}^{3 -} = {no. of moles of Na}_{3} {PO}_{4}$ $"no. of moles of PO"_4^(3-) = "no. of moles of Na"_3"PO"_4$

Since you know that your sample contains $0.500$ $0.500$ moles of sodium phosphate, you can say that it contains $1.50$ $1.50$ moles of sodium cations and $0.500$ $0.500$ moles of phosphate anions.

Science

Math

Humanities

... and beyond

Question #ac64e

1 Answer

Explanation:

Related questions

Impact of this question