Question

Nitrate anion solution is prepared by dissolving 3g of potassium nitrate in 250ml of water. What is the concentration of nitrate anion in ppm, molarity, mass percentage?

Answer 1

(a) 7000 ppm; (b) 0.12 mol/L; (c) 0.7 % m/m.

Explanation:

(a) ppm

For `"KNO"_3`

`M_text(r) = "K"^"+" + "NO"_3^"-" = 39.10 + 62.00 = 101.10`

If we have 3 g `"KNO"_3,`

`"Mass of NO"_3^"-" = 3 color(red)(cancel(color(black)("g KNO"_3))) × ("62.00 g NO"_3^"-")/(101.10 color(red)(cancel(color(black)("g NaNO"_3)))) = "1.8 g NO"_3^"-"`

ppm is the number of "parts" of something in one million parts.

For example, we could express 1 ppm as a concentration of 1 g of solute in `10^6` g solution.

If 250 mL of the solution has a mass of 250 g then

`"ppm" = ("1.8 g NO"_3^"-")/(253 color(red)(cancel(color(black)("g solution"))))× (10^6 color(red)(cancel(color(black)("g solution"))))/(10^6color(white)(l) "g solution")= ("7000 g NO"_3^"-")/(10^6color(white)(l) "g solution") = "7000 ppm NO"_3^"-"`

(b) Molar concentration

`c = "moles"/"litres"`

`n = 1.8 color(red)(cancel(color(black)("g NO"_3^"-"))) × ("1 mol NO"_3^"-")/(62.00 color(red)(cancel(color(black)("g NO"_3^"-")))) = "0.03 mol NO"_3^"-"`

Assume there is no volume change on dissolving. Then

`V = "250 mL = 0.250 L"`

`c = "0.03 mol"/"0.250 L" = "0.12 mol/L"`

(c) Mass percentage

`"Mass %" = "Mass of NO"_3^"-"/"Mass of solution" × 100 % = (1.8 color(red)(cancel(color(black)("g"))))/(253 color(red)(cancel(color(black)("g")))) × 100 % = 0.7 %`

Note: The answers can have only one significant figure because that is all you gave for the mass of potassium nitrate.