How do I find the final concentration after an additional dilution?

Question

I took poor notes and didn't record what the final concentration of an amylase solution I made last year. I recorded that I took $\rm 0.2\ g/10\ ml$ of the amylase solution, and then took $\rm 1\ ml$ of that and diluted to $\pu{35 ml}$.

How would I solve for the final concentration? Would I use

$$ \begin{align} c_1V_1 &= c_2V_2\tag{1}\\ (\pu{0.2 g}/\pu{10 ml})(\pu{1 ml}) &= c_2(\pu{35 ml})\tag{2}\\ c_2 &= \frac{\pu{0.02 g}}{\pu{35 ml}}?\tag{3} \end{align} $$

Therefore, I would take $\pu{0.02 g}$ of amalyase powder and add up to $\pu{35 ml}$ with distilled $\ce{H2O}$ to achieve the final concentration that I would have previously had after dilution?

Answer 1

Check your math: It looks like you forgot to divide by $ 10\ \mathrm{ml} $ in finding $ c_2 $.

Also: the process you are describing in the first paragraph will not produced the same concentration that you started with. If you want $ 35\ \mathrm{ml} $ at the same concentration ($ 0.02\ \mathrm{g/ml} $) by taking some fraction of your original solution and diluting it, you will need to add more solute.

e.g. $$ \rm(0.02\ g/ml) \cdot (35\ ml) = 0.07\ g $$ $$ \rm(0.07\ g + \mathit{m}) / 35\ ml = (0.02\ g/ml) $$ $$ m \rm = (0.02\ g/ml) \cdot (35\ ml) - 0.07\ g $$

The first expression is the concentration of your fraction (which doesn't change from the original solution, assuming a homogenous solution).

The second expression is equating the concentration that you want (conc. of the original solution, for example, with the concentration of your diluted fraction with $ m\ \mathrm{g} $ of solute added (using the mass we just found).

Then I rearranged this to express the mass of solute needed to obtain this concentration.