Question #a7737

Parts per billion, or ppb, are used to express the concentration of a solution that contains very, very small quantities of solute, i.e. trace amounts of solute.

A #"1 ppb"# solution will contain #1# part solute for every #10^9# parts of solvent. When working with grams, a #"1 ppb"# solution will contain

#"1 ppb" = "1 g solute"/(10^9"g solvent")#

Consequently, a #"1.5 ppb"# solution will contain #"1.5 g"# of solute for every #10^9"g"# of solvent.

If you take the density of water to be equal to #"1.0 g mL"^(-1)#, you can say that your solution will contain

#2 color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "2 g" -># the mass of the solvent

You can thus say that in order to have a concentration of #"1.5 ppb"#, this sample must contain

#2.0 color(red)(cancel(color(black)("g water"))) * "1.5 g amoxicillin"/(10^9color(red)(cancel(color(black)("g water")))) = 3 * 10^(-9)"g amoxicillin"#

This is equivalent to having a mass of

#3 * 10^(-9) color(red)(cancel(color(black)("g"))) * (10^9"ng")/(1color(red)(cancel(color(black)("g")))) = "3 ng"#

of amoxicillin. As far as I know, this mass of amoxicillin is far beyond the capability of scales used in regular Chem labs or even in most research labs.

We do have scales that work on the nanoscale, but I don't think that you're supposed to assume that you have one available.

So you definitely can't simply weigh out that much amoxicillin and add it to #"2 mL"# of water. Instead, you are going to have to work with dilutions.

Here's one example of how you can make this solution. Start by dissolving #"0.18 mg"# of amoxicillin in #"1 L" = 10^3"mL"# of water. This will get you a solution that has

#"0.18 mg amoxicllin"/(10^3 "mL water") = 1.8 * 10^(-4)"mg amoxicllin/mL water"#

Make sure that the solution is stirred properly. Next, use a micro-drip set to isolate a micro drip of this solution. Micro-drip sets deliver #60# drops per milliliter of solution, so take a single drip of this solution.

This micro drip will have a volume of

#1 color(red)(cancel(color(black)("drip"))) * "1 mL"/(60 color(red)(cancel(color(black)("drips")))) = "0.016667 mL"#

and contain

#0.016667 color(red)(cancel(color(black)("mL"))) * (1.8 * 10^(-4)"mg amoxicllin")/(1color(red)(cancel(color(black)("mL")))) = 3 * 10^(-6)"mg amoxicllin"#

Now all you have to do is add enough water to get the final volume to #"2 mL"#. The resulting solution will have a volume of #"2 mL"# and contain

#3 * color(blue)(cancel(color(black)(10^(-6)))) color(red)(cancel(color(black)("mg"))) * (color(blue)(cancel(color(black)(10^(-6))))"ng")/(1color(red)(cancel(color(black)("mg")))) = color(darkgreen)(ul(color(black)("3 ng amoxicillin")))#