Some math word problems in Algebra should be familiar to the student with chemical mixing word problems. These are often divided into two types (for lack of a better title), “type A” and “type B.” In this article I will discuss how to set up and solve type B problems of the word chemical mixture (The material is covered in Saxon Algebra II, Lesson 61.)
With type B chemical mixing word problems we start with a given solution and then add or remove something from the solution to get the mixing solution we want. In order to keep them simple and tractable, these solutions are supposed to consist of only two, the second of which is usually water. For example, we will start a sodium solution and we want to know how much water we need to increase the concentration of sodium. Or we would do the opposite, such that the amount of water needed to evaporate from the solution increases the concentration of sodium.
The best way to understand these math problems is to simply work through one solution. Let’s start by looking at where something (water) needs to be added.
Sample Chemical Mixture Problem #1:
A physicist found that 10 gallons of water he studied contained 2% impurities. How much pure water should the scientist add to dilute the impurity to 0.5%?
The Saxon way to solving-skills”> the word maths to solve these types suggests that A represents how much to approach Unlike type A chemical mixture problems, which require a system of two equations, type B chemical mixture problems seem to be necessary for us;
.98(10) + A = .995(10 + A)
How did we come up with this equation? If you look closely, you have three parts, the original solution (98% water, 10 gallons), added water (100% water, A gallons) and the final solution (99.5% water, 10 + gallons). Where the student may be confused is recognizing that 100% represents the number 1 and that “A” is really just “A”.
Solving this equation gives;
9.8 + A = 9.95 + .995A
.005A = .15
A = 30
The physicist must add 30 gallons of pure water.
We could also set up an equation looking at the percentage of impurity as opposed to the percentage of water.
.02(10) + 0A = .005(10 + A)
Or in a simpler form;
.02(10) = .005(10 + A)
.2 = .05 + .005A
.15 = .005A
A = 30
Where the student may be confused, he recognizes that the second term “0*A” represents that no impurity is received in the added water, so it can be omitted.
According to our example, let’s look at when something needs to be removed, as in the case where water needs to evaporate.
Sample Chemical Mixture Word Problem #2:
How much water must be evaporated from 200 gallons of a 5% sodium chloride solution to get a 25% sodium chloride solution? Muria is another name for salt.
The Saxon way of solving these types of math word problems is to let E represent the amount of water evaporated (removed). Unlike the first example, with one equation not by adding terms, but by subtracting.
We were able to set up two equations, both of which would do the job for us. Let us first look at what the equation would look like if they were concentrated in water;
.95(200) – E = .75(200 – E)
Or we could focus on the wall.
.05(200) = .25(200 – E)
Both equations, once solved, would give us the answer E = 160. It is necessary to evaporate 160 gallons of water to obtain a 25% solution of muriatic acid.
To wrap up this discussion, let’s look at the last exchange where something is added to the solution. Instead of simply adding water, we can also add the solution we are looking for in the same set, but in a different concentration.
Sample Chemical Mixture Word Problem #3:
The professor had 250 mL of a 15% potassium solution. How much of a 50% potassium solution does he add to a solution that is 22% potassium?
We set up our equation in a very similar way to how we set it up for #1. The only difference is that the percentage for our second term is different from 1 or 0.
May pain be in heaven for our equation.
.15(250) + .50A = .22(250 + A)
Solving for A gives A = 62.5. The professor must add 62.5 mL of a 50% potassium solution to make a new 22% potassium solution.
Source
John H. Saxon, Jr. Algebra 2. Third edition.
