So I know that if you want to dilute a stock solution to a smaller concentration you simply use the $M_1V_1 = M_2V_2$ equation where the $M_1$ and $V_1$ are the molar concentrations and volumes of the stock solution and $M_2$ and $V_2$ are the molar concentrations and volumes of the resulting dilute solution. This is assuming if you are diluting with water.

My question is, if you have a stock solution of say para-nitrophenol and you wanted to dilute it with $\pu{0.100 M}$ $\ce{NaOH}$ solution, what modifications would you use to calculate how much of the stock solution you would need to get a dilute para-nitrophenol solution?

What I'm thinking is that the density of the $\ce{NaOH}$ solution would come into play here but I don't know how I would go about it.

  • $\begingroup$ First, you didn't tell us the whole story and what you have done so far. Your sought compound, p-nitrophenol is a solid. Thus, I assume you already have a solution of it (your stock solution). Is it in $\mathrm{0.100M}$ $\ce{NaOH}$ as well? Or, in some other solvent? In any solvent, what is the concentration of your stock solution? You want to the find final concentration of $\ce{NaOH}$ or sodium p-nitrophenolate? (Remember, p-nitrophenol reacts with $\ce{NaOH}$) $\endgroup$ – Mathew Mahindaratne Apr 11 '18 at 16:18
  • $\begingroup$ Ok, so I really asked this as a general question as in if I were to dilute a solution with something other than water . But I'm doing a lab using a 96-well plate to do spectroscopy to create a standard curve for pNP. I have a 100 micro-molar solution of stock pNP (paranitrophenol, given to us so I assume its dissolved in water). The stock solution is probably colorless and will change after I dilute it because the instructions say that the solution will turn yellow after diluting with NaOH. So the current goal is to create dilute solutions of 10, 20, 30, 40, 50, 60, and 70 micro-molars. $\endgroup$ – Mayuresh Patel Apr 11 '18 at 17:23
  • $\begingroup$ What is your expected final volume in each well? $\pu{300 \mu L}$? $\endgroup$ – Mathew Mahindaratne Apr 11 '18 at 20:14
  • $\begingroup$ The final volume for each standard solution should be 1.00 mL. $\endgroup$ – Mayuresh Patel Apr 12 '18 at 1:59

Since you'd like to use the equation, $\mathrm{M_1V_1 = M_2V_2}$, I'll show you here, how to do dilution for your extreme concentrations. You can follow same method to do the rest.

Suppose, your final volume of each of dilute solutions is $\pu {V \mu L}$. Then, for all calculations, $\mathrm{V_2 = V}$ and $\mathrm{M_1 = 100\space\mu M}$ (concentration of stock solution). Note that, concentration of $\ce{NaOH}$ does not play a role here, since: (1) it is $1000\times$ more than that of the stock solution (pNP) and hence, dialution of itself is minimal; and (2) it acts just as any other solvent, except for reacting with pNP to give the desired color.

Now, let the volume of $\ce{NaOH}$ solution needed to make your highest concentrated pNP solution is $\pu {V_{1} \mu L}$. Then, using the equation $\mathrm{M_1V_1 = M_2V_2}$, you get: $$\mathrm {100\space\mu M \times V_1 = 70\space\mu L \times V\space\mu L}$$ $$\mathrm {V_1 = \frac {70\space\mu M \times V\space\mu L}{100\space\mu M} = \frac {7}{10}V \space\mu L}$$ Thus, you dilute $\pu {0.7V \mu L}$ stock solution with $\pu {0.3V \mu L}$ $\ce{NaOH}$ solution to make $\pu {70 \mu M}$ pNP solution.

To see if there are enough $\ce{NaOH}$ in $\pu {0.3V \mu L}$ portion, calculate $\pu {\mu mol}$ of $\ce{NaOH}$: $$\mathrm {100\times 1000\space\mu M \times 0.3V\space\mu L = 3\times10^{-2}\times V\space\mu mol}$$ Similarly, you can calculate $\pu {\mu mol}$ of pNP in final solution: $$\mathrm {100\space\mu M \times 0.7V\space\mu L = 7\times10^{-5}\times V\space\mu mol}$$ Thus, all p-nitrophenol will be converted to sodium p-nitrophenolate in the highest concentrated solution. Therefore, it is safe to assume that rest of the solutions would undergo the same conversion.

By similar calculations, we can show you must dilute $\pu {0.1V \mu L}$ stock solution with $\pu {0.9V \mu L}$ $\ce{NaOH}$ solution to make $\pu {10 \mu M}$ pNP solution. Since you know value of $\pu {V \mu L}$, you can get numerical values.


Density changes when mixing liquids is not only applicable to NaOH solutions. Even when you dilute a stock solution with water you may find that the volume is not additive. The factors that matter are the starting concentration and the concentration change.
At the concentrations you're using, you can probably safely ignore density changes and assume volumes are additive, unless you need extreme precision (but then you could use volumetric glassware and do the dilutions properly, instead of just adding volumes).

You are correct: when you dilute your $100 \ \mu M$ pNP solution with $0.1 \ M$ NaOH, it will turn yellow, because you form the pNP anion, which is intensely yellow.
However, for the spectroscopy part to work correctly, you must make sure you always have at least the stoichiometric amount of NaOH, i.e. you must convert all pNP to its anion.
This will be the case in all your diluted solutions, simply because the concentration of NaOH is 1000 times larger than the pNP concentration ($0.1 \ M = 100 \ mM = 100000 \ \mu M$), so whenever the volume of NaOH solution you add to the pNP solution is greater than 1/1000 times the volume of the latter, you're already in excess of NaOH.

For the rest, the formula you mentioned works correctly. For instance, to make the $70 \ \mu M$ solution of pNP anion: $V_2 = \frac {100 \cdot V_1} {70} \approx 1.43 \cdot V_1$, so you could add $0.43 \ mL$ of $0.1 \ M$ NaOH to $1 \ mL$ of $100 \ \mu M$ pNP. You can see that you are in huge excess of NaOH, so there is no risk to leave any non-ionised pNP.

You can achieve the same result by a 'quick' subtraction method, which also works when both solutions contain the substance of interest.
In this case, your stock is $100 \ \mu M$ in the substance of interest (pNP). You dilute it with a solution that contains no pNP, so it's $0 \ \mu M$. You want to obtain a $70 \ \mu M$ solution.
Subtract the final desired concentration from the concentration of each solution you're mixing, take the absolute values and swap the results.
So, pNP stock minus final desired solution is $|100 - 70| = 30$. NaOH minus final desired solution is $|0 - 70| = 70$. Swap the two: 30 applies to NaOH and 70 applies to pNP.
This means that, to obtain the desired final solution, you need to mix 30 'parts' of NaOH with 70 'parts' of pNP. You can easily verify that indeed $100/70-1 = 30/70$.

This may sound more complicated than the other formula, but in fact if you write the concentrations of the two 'starting' solutions on the top left and bottom left corner of an ideal square, respectively, and the final concentration in the centre of the square, and then do the subtractions 'across the diagonals', you will get in the right corners of the square the 'parts' to mix for each starting solution.
E.g. if you had a $10 \ \mu M$ solution (A) and a $90 \ \mu M$ solution (B) and wanted to obtain a $40 \ \mu M$ solution by mixing them, this method would quickly tell you that you need to mix 50 'parts' of A and 30 'parts' of B.
Obviously any mixing-based dilution will work only if the final desired concentration is 'bracketed' by the concentrations of the solutions being mixed. In these examples, $0<70<100$ and $10<40<90$.

  • $\begingroup$ Thanks. I actually just asked my professor and she just wanted a percentage of pNP to NaOH. So like 100 uL of pNP and 900 uL NaOH for the 10 uM solution. This was just theoretical and the actual concentrations should be close to these values. I kind of over thought about this :) $\endgroup$ – Mayuresh Patel Apr 12 '18 at 19:31

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