Advice on solving mixture problems?

[Dababy]

I'm having a lot of difficulty solving mixture problems and the biggest problem I think is that I just can't wrap my head around how to even set up a problem. An example problem I'm having a lot of trouble with is something like this.

How much water should be added to 115 ml of 55% acid solution to dilute it to a 45% acid solution.

A very clear step by step explanation to solve problems like this would be most appreciated thank you

 

[Deleted member 4993]

I'm having a lot of difficulty solving mixture problems and the biggest problem I think is that I just can't wrap my head around how to even set up a problem. An example problem I'm having a lot of trouble with is something like this.

How much water should be added to 115 ml of 55% acid solution to dilute it to a 45% acid solution.

A very clear step by step explanation to solve problems like this would be most appreciated thank you

What is being asked to "find"? - How much water should be added

Assign a variable name to that "find".

Let the volume of water to be added = W

What is given? original volume of solution = 115 ................... after adding extra water the final volume of the solution = 115 + W

Amount of acid in original solution = 115 * 0.55 ......... (this amount does not change due to addition of water)

Amount of acid calculated after adding water = (115 + W) * 0.45

Continue....

 

[lev888]

Explain what 45% solution means.

 

[HallsofIvy]

Certainly, the first thing you should think about is lev888's question- "what does '45% solution (or 55%) mean'? You are told that you have "115 ml of 55% acid solution". So how many ml of that is acid and how many ml are water? Diluting to 45% by adding water, the amount of acid doesn't change. You want to add enough water that that same amount of acid is now 45% of the whole (both acid and water) amount.

 

[Dababy]

Okay I kind of understand how I should look at it now and I tried setting up the problem I tried to follow this formula I saw (C1V1+C2V2=C3V3) and ended up with this.
.55(115)+.45(115)=(115+x).45
The answer I got was 140.5 but I don't think that's even close

 

[ksdhart2]

This provides a perfect illustration of the importance of understanding in math, as opposed to just blindly trying to follow some formula you found somewhere. In your case, that formula can be used (though it's probably not the most direct way of approaching the problem), but care has to be taken to keep track of what each variable represents.

• $C_1$ represents the [acid] concentration of solution #1; $V_1$ represents the volume of that solution
• $C_2$ represents the [acid] concentration of solution #2; $V_2$ represents the volume of that solution
• $C_3$ represents the [acid] concentration of solution #3; $V_3$ represents the volume of that solution

Solutions #1 and #2 are the two "ingredients" you're mixing together and solution #3 is the resultant mixture. According to the numbers you plugged into the formula, this would imply that you're mixing 115 milliliters of 55% acid and 115 milliliters of 45% acid... but does that make any sense? Does that align with the problem text and with your goal?

Another huge red flag you should have noticed is that your proposed solution requires new volume to mysteriously pop into existence. If you mix together 115 milliliters of one solution and 115 milliliters of another solution, this must produce 230 milliliters of liquid. However, your solution of $\displaystyle x = \frac{13915}{99} = 140.\overline{5} \approx 140.5$ says that you'll end up with $255.\overline{5}$ milliliters of liquid. Where did the extra 25-ish milliliters come from?

Returning to the drawing board: The problem text tells you that you start with 115 milliliters of 55% acid. This is your solution #1. Now what is your solution #2? Well, you're diluting with water. What is the acid concentration of water? Further, you don't yet know how many milliliters of water you need to add - that's the ultimate goal of the exercise. If you try it again with this new line of thinking, things should become a lot clearer (and you'll also probably see what I meant about this particular formula not being the most direct way).

 

[Dababy]

Thank you for all the responses I apologize for how long this is taking me to understand but I think I kind of understand now
so solution 1 is equivalent to (.55)(115) but I don't know solution 2 and I don't know how much water is needed to dilute it to the final solution so I figured that's equal to x(0) and then the final solution is of course (115+x).45
So the setup would look like this
(.55)(115)+x(0)=(115+x).45
When I finished solving it I got 25.5 is this correct?

 

[tkhunny]

How much water should be added to 115 ml of 55% acid solution to dilute it to a 45% acid solution.

25.5555... Right. Careful with rounding. Sometimes, the problem statement specifies how to do it.

In ALL cases there are two ways to go about it. You must equate SOMETHING.

x = Amount of Water added (Always define your variables explicitly. This will keep you from forgetting what they are, later.)

Equate Acid

Starting Acid + Added Acid = Ending Acid
115 * 0.55 + x * 0 = (115 + x) * 0.45

Equate NonAcid

Starting NonAcid + Added NonAcid = Ending NonAcid
115 * (1 - 0.55) + x * (1 - 0) = (115 + x) * (1 - 0.45)

Try it and see that you get the same value for x.

A 55% Acid solution, is also a 100% - 55% = 45% NonAcid solution.
Water is a 0% acid solution.

It is usually most convenient to pick the one with the zero, if there is one. That was the solution you made. Good work. Some problems will present other information and it will be important for you to think about both ways, using good judgment to pick the best way to go about it.

 

[Dababy]

Thank you very much!