Rigorous proof of concentration of solution given concentrations of parent solutions

Question

Suppose I have two jars, each with some concentration of a chemical. We also have an empty container.

Example: Suppose Jar A contains $50\%$ concentration and Jar B contains $0\%$ concentration. We can produce a child solution with a concentration taking one value in the range $[0,50]\%$. We know this because it's impossible to get a concentration greater than $50\%$ and we can create any concentration lower than $50\%$ by adding arbitrary amounts of Solution B.

Now what if we have variable concentrations $A$ and $B$? Without loss of generality, assume that $A$'s concentration is greater than or equal to $B$'s. Of course, dilution is not allowed (i.e. we are only allowed to pour A and B; we don't have a jug of water at our disposal, and we don't have a means of removing liquid.)

I'm inclined to believe that the new container's concentration can take any value in the range $[B,A]$, but how can we rigorously prove this?

Answer 1

The weighted sum calculation applicable to the problem is

$$w_1 c_1+w_2 c_2=w_3 c_3\tag{1}$$

where $w_1$ and $w_2$ are the positive percentage proportions of the total liquid added with concentration $c_1$ and $c_2$ respectively. [By definition $0\le w_n \le 100$]. Without loss of generality let us define $c_2\ge c_1.$

If we define the resultant total liquid $w_3$ to be $100\%$ we have $w_1+w_2=w_3=100\%$.

we can then rewrite (1) as

$$w_1 c_1+w_2 [(c_2-c_1)+c_1]=w_3c_3,$$

$$(w_1+w_2) c_1+w_2 (c_2-c_1)=w_3c_3.$$

Giving

$$w_3 c_1+ w_2 (c_2-c_1)=w_3 c_3.$$ Thus finally by substituting $w_3=100\%$ we have $$100\% \times c_1+ w_2 (c_2-c_1)=100\% \times c_3\tag{2},$$

and since by definition $0\le w_2 \le100$, we immediately find from (2) that $c_1 \le c_3 \le c_2$

Answer 2

So we are talking of mixtures of some elements $A$ and $B$, of different concentrations, let's say of "mass concentrations".
Let's also agree that speaking of a concentration of $q$ we refer to that of the element $A$, so that $$0 \le mass(A)/(mass(A)+mass(B)) = q \le 1$$

When from a jar of concentration $q$, you pick $x$ grams of mixture, you are picking $qx$ grams of $A$ and $(1-q)x$ grams of $B$.

If then you pick $y$ grams from a second jar of concentration $p$, and put together, you get a mixture with:
$qx+py$ grams of $A$
$(1-q)x+(1-p)y$ grams of $B$
$x+y$ grams in total

which gives a concentration of $$ c = {{qx + py} \over {x + y}} $$

From here I think you can proceed alone