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 $35 ml$.

How would I solve for the final concentration?

Would I use :

$$c_1\ V_1 = c_2\ V_2$$

$$\rm(0.2\ g/10ml) \cdot (1\ ml) = (\mathit{c_2}) * ( \mathrm{35\ ml})$$

$$c_2\rm = \frac{0.02\ g}{35\ ml} $$

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


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.

  • $\begingroup$ I desired the final concentration of when I took the 1ml from the (0.2/10ml) solution and added up to 35ml. Is the equation above assuming I want a final concentration of (0.2/10ml)? $\endgroup$ – Ro Siv Feb 21 '16 at 21:37

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