0
$\begingroup$

Say I have five solutions of various concentrations such as the below:

A = 16%
B = 12%
C = 12.5%
D = 17%
E = 5%

Their quantities do not matter, only their concentrations. Say I want to mix them and the final spread between them should follow a ratio of 1:2:1:2:6. Is this enough information given to determine the final % of each solution in the final mixture? If so, how would I calculate the final %s of each solution?

$\endgroup$
1
$\begingroup$

Riista..based on your description, if you know the dilution ratios, then since each solution contains a different solute, each solution is a relative dilutant to the others. For example, the concentration of A after mixing with your ratios will be: $\frac{1\cdot16\%}{1+2+1+2+6=12}$. In general, the final % will be $\frac{ratio \cdot conc.\%}{12}$

$\endgroup$
1
$\begingroup$

What do you mean spread? Is the concentration for each solution representing the percentage of the same thing in each one, so A is $16\%$ something, E is $5\%$ the same thing and you want the final percentage if you mix the ratio you give? If so, you just need a weighted average of the percentages, weighted by the proportion of each solution. It would be $\frac {1 \cdot 16+2\cdot 12 + \dots}{1+2+\dots}\%$ I'll leave the dots to you

$\endgroup$
  • $\begingroup$ Each solution contains a different solute in them. I guess a way to simplify the question is if I have a 15% solution of A, and a 10% solution of B, if I were to mix the two with a ratio of 2:1, what the final % of A and the final % of B would be in the AB solution. $\endgroup$ – gator Nov 1 '13 at 14:10
1
$\begingroup$

May be, it could be simpler to forget percents. Considering your last post, say that there is 15 moles of A in a liter of one solution and 10 moles of B in a liter of another solution. Now, you mix two liters of the first solution (this makes 30 moles of A) and one liter of the second solution (this makes 10 moles of B). As a result, we have now three liters of a solution which contains 30 moles of A plus 10 moles of B. Then, one liter contains 10 moles of A and 10/3 moles of B. Now, go back to percents : 10% of A and 3.33% of B. Does this help ?

$\endgroup$

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.