Three solutions contain a certain acid. The first contains 10 % acid, the second 30 %, and the t

Three solutions contain a certain acid. The first contains...

Key Concepts

System of Equations

This concept involves setting up multiple equations to represent the different constraints imposed by a mixture problem. In such problems, one equation typically represents the total volume of the mixture while another accounts for the concentration of a specific component—in this case, the acid percentage—ensuring that the combined contributions of individual solutions match the desired target.

Mixture Problems

Mixture problems require combining two or more substances with known properties (like percentage concentrations) to achieve a final mixture with a specified characteristic. They demand careful analysis of the ratios and percentages involved, and they often rely on algebraic methods to determine the correct volumes of each component that will meet the target mixture specifications.

Weighted Averages

The principle of weighted averages is crucial when calculating the overall concentration of a mixture. It involves multiplying the concentration of each component by its respective volume and then dividing the total by the overall volume of the mixture. This method ensures that each solution’s contribution to the final percentage is accurately reflected based on its proportion in the mixture.

Proportional Relationships

Proportional relationships are used to relate the volumes of different components based on given ratios. When one component's volume is expressed as a multiple of another's, it simplifies the problem by reducing the number of independent variables in the system of equations. This relationship is fundamental in ensuring the appropriate balance between the substances in the mixture.

Transcript

00:01 So here we have a problem about three solutions with its corresponding acid content.

00:08 A has 10 % acid content.

00:11 B has 30.

00:12 C has 50.

00:13 And the end goal is to create a resulting solution that is 50 liters in quantity, which has 32 % acid content.

00:24 And also, the chemist in this problem wanted to use twice as much of the 50 % solution as of the 30%.

00:32 Okay, so from this given, we can actually generate the equation.

00:37 First one is assuming that abc variables, abc represents the quantity in litters of each of the solution, then we can say a plus b plus c is equal to 50.

00:49 And that should be our equation number one.

00:52 Now, adding up all the acid content from each of the solution, so that should be 10 % for a, 30 % for b.

01:03 Let me erase that and let's try to write it more legibly and then 50 % for c that should end up to a value equal to the 32 % of the 50 liter solution resulting from adding all these three solutions with different acid content.

01:26 And this should be our equation number two but before i label it as equation number two, let me try to simplify it.

01:38 So we should have, oops, this should be c, and 0 .32 times 50 is actually equal to 16.

01:50 And let's call it our equation number 2.

01:52 Now for the last equation, we can use this description right here that says, chemists wanted to use twice as much of the 50 % solution as of the 30 % solution.

02:04 So the 50 % solution is solution c.

02:09 It has a quantity of c.

02:13 The 30 % solution has a quantity of p.

02:16 So i'll say c should be equal to twice of p.

02:20 And that's equation number three.

02:23 Now we have a bunch of methods to solve this.

02:26 We can use matrices elimination or substitution.

02:30 And let me try to do some substitution first.

02:35 So i will do substitution.

02:38 I'll substitute equation number three to both one and three.

02:48 Okay.

02:49 So for equation number one, we'll have a plus b plus that substitute 2b to all variable c.