# solution concentrations [closed]

I think I'm making this more complicated than it needs to be, but here it is:

I have 7 drug solutions, all 1 mg/mL which need to have the following concentration in urine:

A: 25,000 ng/ml

B: 25,000 ng/ml

C: 100,000 ng/ml

D: 100,000 ng/ml

E: 10,000 ng/ml

F: 1000 ng/ml

G: 1000 ng/ml

How much of each 1 mg/ml solution do I need to add to urine to get a stock with the above concentrations? My ultimate goal is to take several 75 ul aliquots of urine for testing. Any suggestions?

## closed as off-topic by jerepierre, Klaus-Dieter Warzecha, bon, Martin - マーチン♦, Geoff HutchisonJun 3 '15 at 10:45

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• To answer your question, you also need to know how much of your final solution you want to make. – MarcoB Jun 2 '15 at 15:01

Since you did not tell us how much of this final solution you would like to make, let's imagine that you want to make a generic $V_f\ \ce{mL}$ of it.
Take $\text{A}$ as an example:
$$n_A \ \text{in final solution} = 25000\ \ce{ng/mL} \times V_f\ \ce{mL}$$
$$V_{A}\ \text{stock to take} = \frac{n_A}{c_{A\ stock}} = \frac{25000\ \ce{ng/mL} \times V_f\ \ce{mL}}{1\ \ce{mg/mL} \times 10^6\ \ce{ng/mg}} \frac{1000\ce{\ \mu L}}{\ce{mL}}= \frac{25000}{10^3}\ V_f\ \ce{\mu L}$$
So let's suppose that you want to make $10\ \text{mL}$ of such a solution containing $\text{A}$ at $25,000\ \text{ng/mL}$, then you need to measure out $25000 \times 10\ /\ 1000\ \ce{\mu L} = 250\ \ce{\mu L}$ of the $\text{A}$ stock and bring it up to the final volume with urine.