Algebraic Word Problem: Chemist

Question

A chemist needs 60 milliliters of a 23% solution but has only 19% and 31% solutions available. How many milliliters of each should be mixed to get the desired solution? I am having trouble setting up this problem, do I need to set up 2 equations for this? And also with word problems like these, how do you know when to set up two equations?

Answer 1

Yes, you need two equations. One equation that expreses that however you mix these up, the total volume is 60 milliliters, and one equation that expresses that the resulting concentration is 23%.

In general, each "piece of information" that you're given is an equation. Here we are given four pieces of information:

• The concentration of solution 1
• The concrentation of solution 2
• The final volume of the mixture
• The final concrentation of the mixture

And each one, you can write up as its own equation. So we get four equations. However, the two first pieces of information here are so simple to deal with algebraically that they usually aren't even written up explicitly. So only the two last pieces of information actually give us something that needs to be solved.

In full, we get $$ \cases{c_1 = 0.19 & Concrentation 1\\ c_2 = 0.31 & Concrentation 2\\ v_1 + v_2 = 60 & Volume of final mixture\\ c_1v_1 + c_2v_2 = 0.23\cdot(v_1+v_2) & Final concrentation} $$ where $c_1$ and $c_2$ are the concentrations of the two mixtures you start with, and $v_1$ and $v_2$ are the volumes you take of each of them. And then you solve this using your favourite method of simultaneous equation solving.

As mentioned above, however, the first two equations here are already "solved", so most people would just insert those values into the remaining two equations and say that this is a system of two equations: $$ \cases{v_1 + v_2 = 60 & Volume of final mixture\\ 0.19v_1 + 0.31v_2 = 0.23\cdot (v_1+v_2) & Final concrentation} $$

Answer 2

Here are the two equations: one for content and the other for volume

$$0.19x+0.31y=0.23\times 60$$

$$x+y=60$$

which yields

$$x=40,\>\>\>y=20$$

Answer 3

$$\begin{array}{c|c|c}31&&4\\\hline&23&\\\hline19&&8\end{array}$$ giving 12 parts of 5 ml each. so 4 parts is 20 ml for the 31% and 8 parts is 40 ml for the 19%. The way I was taught originally was through active ingredient calculations using equivalent fractions (at least for standard dilution questions):$${23\over100}={h\over60}\to h={69\over5}={60\cdot23\over100}\tag{active ml}$$ you for dilution with a solution with no active ingredients the rest is diluent. But the graphic above is just using the simple oberservation that 31% adds 8 percent more, than 23% which takes 2 portions of being less than 23% by 4% like 19% is. This leads to the fact that mixing two solutions the overages need to cancel the amounts less than. This does not take volume contraction, or expansion, upon mixing into account.