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

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?

## 1 Answer

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.