Question

How could I prepare a $ 1\;ppm $ solution of ammonium chloride in water?

Answer 1

Hint: $ ppm $ , which is known as parts per million, is one part of solute present per million parts of solution. In terms of defining equations, we can write:
 $ ppm\left( {\dfrac{m}{m}} \right) = \dfrac{{mass\;of\;solute}}{{mass\;of\;solution}} \times {10^6} $
 $ ppb $ , which is known as part per billion, is one part of solute present per billion parts of solution. In terms of defining equations, we can write:
 $ ppb\left( {\dfrac{m}{m}} \right) = \dfrac{{mass\;of\;solute}}{{mass\;of\;solution}} \times {10^8} $ .

Complete answer:
Let us take $ 4\;mg $ of ammonium chloride in $ 1\;L $ of water. We know $ 1\;ppm $ means one gram of solute present one million gram of solution. When working with aqueous solutions, let us assume that the density of such a dilute solution is $ 1.00\;gm{L^{ - 1}} $ . Thus, it is common to equate $ 1\;mL $ of solution with $ 1\;g $ of solution.
Assuming that we want to make $ 1\;L $ of solution. Then,
 $ \Rightarrow 1000\;mL\;solution \times \dfrac{{1\;g}}{{1\;mL\;solution}} \times \dfrac{{4g\;N{H_4}Cl}}{{{{10}^6}g\;solution}} $
 $ \Rightarrow 0.004g\;N{H_4}Cl $
 $ \Rightarrow 4mg\;N{H_4}Cl $
Hence, we could dissolve $ 4\;mg $ of $ N{H_4}Cl $ in $ 1\;L $ of water in a volumetric flask.
But it is difficult to measure $ 4\;mg $ of $ N{H_4}Cl $ accurately. It is easier to dissolve $ 400\;mg $ of $ N{H_4}Cl $ in $ 1\;L $ . This would give you a concentration of $ 400\;ppm $ . Then we can dilute $ 10\;mL $ of this solution to $ 1\;L $ in another volumetric flask to get a final solution with a concentration of $ 4\;ppm $ .

Note:
$ ppm $ is a special concentration unit which is used only when solute is present in negligible amounts in the solution and it is defined as the number of parts or weight of solute present in per million parts by weight of solution. It is independent of temperature as the mass of solute and solution do not vary with respect to temperature. Similarly $ ppb $ is also independent of temperature for the same reason.