5
$\begingroup$

Original Question

A solution of dye, Amido Black 10B dye ($10~\mathrm{mM}$, $6.165~\mathrm{g\over L}$), as-provided absorbs light much too strongly for measurement of an absorbance value. It is therefore necessary to dilute it before proceeding with an experiment to construct a standard curve of absorbance vs. concentration or amount of dye. The stock $10~\mathrm{mM}$ Amido Black solution needs to be diluted $200\times$ using purified (RO) water. To achieve the dilution, it is recommended that serial dilutions be used.

How would you dilute the Amido Black 10B $200\times$ to a total volume of $3~\mathrm{mL}$? Assume you only have pipettes that cannot dispense less than $20~\mu\mathrm L$.

Comments

We've already been through this question together in class but despite asking a lot of questions I still didn't understand what my tutor was trying to explain. I already have the answers, but I want to know how to figure it out for myself. My tutor's way was this:

  1. Adding $50~\mu\mathrm L$ of dye into $50~\mu\mathrm L~\ce{H2O}$ gives a $1$ in $2$ dilution.
  2. Take $30~\mu\mathrm L$ of the new solution and add $2970~\mu\mathrm L~\ce{H2O}$, giving a further $1$ in $100$ dilution.

I do not understand where she got $50~\mu\mathrm L$ of dye and $50~\mu\mathrm L$ of $\ce{H2O}$ if it's $1\!:\!2$. I know she split $1\!:\!200$ into $1\!:\!2$ first because the pipettes can't dispense less than $20~\mu\mathrm L$. I also don't understand where she got the $30~\mu\mathrm L$ from in the second step.

$\endgroup$
  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution. I am sure you can figure out how much the 200 fold dilution of your mother solution is. Just figure out how much water you would need to dilute it in one go, include it in the post, and we'll reopen it to cover the rest. $\endgroup$ – Martin - マーチン Mar 15 '16 at 12:32
  • $\begingroup$ It is impossible to answer this since you don't specify what pipettes and volumetric flasks you have to use. It would seem reasonable to assume that you have more than enough stock solution to prepare the 3ml of dilute solution. $\endgroup$ – MaxW Mar 15 '16 at 16:16
1
$\begingroup$

To your questions:

I do not understand where she got $50~\mu\mathrm L$ of dye and $50~\mu\mathrm L$ of $\ce{H2O}$ if it's $1\!:\!2$.

It's always problematic expressing dilutions as ratios, because it's ambiguous as to what the numbers are referring to. Here, the "$1$" is referring to the volume of the solution to be diluted, and the "$2$" is referring to the final volume after dilution. Thus, the $50~\mu\mathrm L$ ("$1\times$") is being diluted to $100~\mu\mathrm L$ ("$2\times$"), and in order to do that you need to add $100~\mu\mathrm L-50~\mu\mathrm L=50~\mu\mathrm L$ of the RO $\ce{H2O}$. This could just as validly be expressed as a $1\!:\!1$ dilution; thus the ambiguity.

I also don't understand where she got the $30~\mu\mathrm L$ from in the second step.

Per the same $\mathit{Initial~Volume : Final~Volume}$ ratio convention, a $1\!:\!100$ dilution to a final volume of $3~\mathrm{mL}=3000~\mu\mathrm L$ will require starting with $3000\div 100 = 30~\mu\mathrm L$ of the intermediate solution.


More generally:

I prefer to express dilutions using an "n-fold" notation, e.g.:

  1. Dilute $50~\mu\mathrm L$ of dye solution $2$-fold with RO $\ce{H2O}$
  2. Dilute $30~\mu\mathrm L$ of the solution from step ($1$) $100$-fold with RO $\ce{H2O}$

Though there is still room for ambiguity in this phrasing, to me diluting $n$-fold indicates much more clearly that the final volume $V_f$ should be $n\times$ the initial volume $V_i$, and thus that the volume of diluent needed is:

$$ V_d = V_f-V_i = nV_i-V_i = V_i\left(n-1\right) $$

$\endgroup$
0
$\begingroup$

The whole problem is contrived to test you knowledge of dilutions.

(1) Assuming that you had a 15 μL pipette and a 3ml volumetric flask then you'd just take 15 μL from the stock solution and add that to 3ml volumetric flask and fill with water to the mark. Done...

(2) The solution using a 50 μL pipette and then a 30 μL pipette is sort of stupid. You don't need two different pipette sizes. You could mix 30 μL of stock solution with 30 μL of water to get a solution with half the concentration. Taking 30 μL of the 1/2 concentration solution and diluting to 3ml would give you another 100 fold reduction in concentration. So overall the two steps would reduce concentration to 1/200th of the original.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.