How many oz of 5% acid and 20% acid must be mixed to get 10 oz of a solution that is 12.5% acid?

How to solve an algebra mixture word problem.
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[@kennethwright870]

(5%+20%)/2=12.5%, so the correct ratio is 1:1. So 5oz of each combined will result in 10oz of 12.5% solution

[@terry_willis]

Your answer is correct, but how did you derive the 1:1 ratio from the preceding equation? (Let's say the problem said 23% and 7% to yield 9% mixture with total volume of 28 oz.)

[@richardl6751]

@@terry_willis I got 24.5 oz of 7% and 3.5 oz of 23%. Sound right?

[@terry_willis]

@@richardl6751 Yes, that's what I got too. Although the numbers had to be rounded slightly - this was a little messier than using John's numbers.

[@frederickdvorak6502]

​​@@terry_willisBecause the volumes of acid are the same, as it is implied. Only the level of the acidity is different. If that isn't implied, even the convoluted demonstration from that maths educator can't be worked out.

[@frederickdvorak6502]

This maths educator likes complexity! (20% + 5%) / 2 = 12.5% ; 1 oz + 1 oz = 2. ; so, 10 oz / 2 = 5 oz ; thus 5 oz of each give 10 oz in total at 12.5% concentration.

[@michaelrobinson9952]

Great that your showing people new to maths the method of simultaneous equations, it's a great way to build linear algebra skills when solving word problems. Yes those in the know can see a dozen ways to solve this, but we all had to build foundations. Great content.

[@BruceKarger]

Why not .05x + .20(10-x) =12.5((10)? Don't need two equations.

[@DevilbyMoonlight]

Am no maths genius but I worked it out in my head like this, if equal quantities of the solution were mixed then the strength of it would be the sum divided by 2 as one is 5 and the other is 20 that would be 25/2 = 12.5 , am old in I guess this is why I enjoy your channel so much as it fills in the gaps I never learnt - thankyou

[@Astrobrant2]

Yes, that worked in this problem because the amount of each concentration was the same. But if they were different, you couldn't just take the average.

[@RobertCookcx]

Brings back memories of my ChE mass balance fundamentals. It complicated fast after that! Great video!

[@mike-yp1uk]

Totally awesome dude.

[@jeffburdette2025]

Basic chemistry question dealing with concentrations

[@danmike2305]

Nice problem. A good book is, how to solve algebra word problems. By William A. Nardi. I like the second edition.

[@MommaBird52]

Did that one in my head using simple logic. 12.5 is 7.5 more than 5 and 7.5 less than 20. So ounce for ounce need equal amts to hit the midpoint. We need 10 oz so 5 of each.

[@mikesullivan5219]

Easy one, quick in the headroom, 1 ounce of each, or any equal amount gives you the 12.5%. 1 ounce each is 2 ounces, 5 of each 10 ounce!

[@laurindapace2164]

5oz plus 5oz equal. 10oz. 10oz/2= x=5,y=5

[@sidexplores9001]

12.5% is halfway between 5% and 20% so you would need the same amount of each acid for the resulting concentration to be 12.5%. Say we wanted a resulting concentration of 17%. That is (4/5) of the way between 5% and 20%, so (4/5) of the solution would be the 20% acid, i.e. 8oz and the remaining 2oz would be the 5% acid. Using ratios in my opinion is a quicker way to do such problems.

[@Abedchess]

Well, you want a standard general way to do these problems. (Of course, this problem is very simple. But, you could have a problem where a chemical reaction is happening during the addition. Or maybe there are multiple products and reactants all coming in and out in pipes.) This is usually the general way. Just do conservation of a bunch of things. In this example, it's conservation of mass and volume

[@user-ew4qn1um2l]

Easy peasy. 5 Oz. each, splitting the difference between the two

[@lizardfirefighter110]

So I started out on the right track with x + y = 10 but tried to d substitute with the acidity relation but crashed and burned. So I went with what I knew. 5 oz of acid in 100 oz of water is 5% acidity for all practical purposes. And, 20 once’s of an acid in 100 ounces of water would be a 20% solution - the acid being used was of the same ph, like say a ph of 1. Well, if you add 100 oz of a 5 percent solution to 100 oz of a 20 percent solution then you will have 25 oz of acid in 200 oz of solution. So that means in every 100 oz of solution you will have have of 25 or 12.5 oz of aside in 100 oz of solution - by definition that is the percent. Imagine that! 12.5 is what the problem is asking for. So just mixing the solutions gives us the correct percent needed to solve the problem. But wait, the total amount must add up to 10. So for equal amounts to add up to 10 the there are 5 oz of each! 😂😂 stupid guy still solved the problem. I don’t know what I would have done if it was not equal amounts. So now I learned the smashing of equations trick😂😂😂

[@marlberg2963]

I realize that this is an ideal abstract word problem, however, chemistry doesn't work this way. The percentage yield depends on the reactants. Not all acids can be combined without adverse reactions or unexpected yields. There are in fact acids that when combined with other specific acids yield significantly less solution than the theoretical maximum and in addition produce more water (H2O) thus diluting the solution and posdibly producing a precipitate such as a salt or a polymer. To be clear. 5oz 20% solution of Sulfuric Acid (H2SO4) + 5oz of 5% solution of Acetic acid aka White Vinegar (CH3COOH) does not yield 10oz of 12.5% Solution of acid. Instead it yields CH3O4S + CO2^ + H2O. In layman's terms it yields dimethyl sulfate salt Carbon Dioxide given off as a gas and water. Ph is neutral after the reaction so acid percentage in solution is not 12.5% but 0 percent.

[@open_water2411]

Somehow my brain worked out that it’s 50-50 but I didn’t have the maths working.

[@MrMousley]

Well 20 + 5 = 25 and 12.5 = 25/2 so you know it's just the same amount of both which is 5oz of each to make 10oz of solution

[@tomtke7351]

more formally.. two equations in two unknowns: Unknowns--------- F = 5% acid quantity T = 20% acid quantity Givens---------- F is 5% strong T is 20% strong final amount = 10oz final strength = 12.5% Equations---------- eq.1 F + T = 10 oz eq.2 F(5%)+T(20%)=(F+T)(12.5%) 0.05F+0.2T=0.125F+0.125T 0.2T-0.125T=0.125F-0.05F 0.075T=0.075F T = F REPEATED F + T = 10oz eq.1 F = T eq.2 substitute eq.2 into eq.1 F + F = 10oz 2F = 10oz F = 5oz from eq.2 T = 5oz VERIFY by eq.1 F(5%)+T(20%)=(F+T)(12.5%) 5(5%)+5(20%)=?(5+5)(12,5%) 0.25+1.0=?10(12.5%) 1.25=❤1.25✔️

[@Sashie_Plays]

Boss I worked this out in my head in the space of 2 minutes before even watching your video. You made this way to complicated. My solution is to spit it into ratios. Solution A is 5:100 and solution B is 20:100 . When added together you get solution C 25:200 when converted to a persentage is 12.5%. 1oz of A plus 1 Oz of B gives 2 Oz of C then all you need is 5 Oz of both A and B to get 10 Oz of C. This however does Assume the the acid has the same density as water. But as the acid is not specified in the question I won't be able to add that into the equation.

[@Sashie_Plays]

No graphs I was just scrolling through KZbin and found this and solved it

[@Abedchess]

Good job getting the answer. Actually, when chemist say a solution is x% some acid, they actually mean mass/volume ratio. I.e. 5% HCl means 5g/ml of HCl in water Also, the method he showed is Abit complicated, But it's really conservation of mass and volume (Initial mass in acids = final mass in acids Initial volume added up = final volume)

[@Abedchess]

I'm going to explain the videos method Abit more. Let volume of the acids be x and y So conservation of mass (5% )x + (20%)y = (12.5%)(10) Conservation of volume x+y= 10 Then solve simultaneously. This method will work for all possible initial and final concentrations. Hopefully it's not complicated to understand

[@danieldennis9831]

5 oz of each.

[@dayd7527]

Awesome. You're really funny..

[@michaelrobinson9952]

Used camers rule matrix for giggles

[@bobcornwell403]

I came up with 5 oz. each.

[@russelllomando8460]

got it by logic. 1/2 of each is 1/2 of 25% = 12.5%

[@user-ri6rn7ti5h]

(0.25÷5)=5 0r (0 25÷10)= 2.5

[@user-ri6rn7ti5h]

(0 20+0 5)=0.25 (0.25÷10)=2.5

[@user-ri6rn7ti5h]

125%

[@johnnyllooddte3415]

5 each im guessing

[@lewisreiman8124]

5 oz 5%;5,oz 20%

[@user-ri6rn7ti5h]

Real 125%

[@omdhandharmeshomhindusdhan7516]

You don't need to break it down to solve. If you need to break it down you have to start all over again from beginning. ? So 10=100% and 5%20% is 25% and divided by 100% you get 12.5%.

[@bobcornwell403]

I used the substitution method. I let "A" be my variable, and I multiplied everything by 100 except the total amount of ounces. This gave me: (5 x A) + (20 x A) = 125 x 10. Solving this equation gives me the percentage of the 5 % acid I need as a whole number. I was quite surprised to end up with 50. But it all checked out.

[@JohnnyBradfield]

I must be getting slow... I did it in my head in about 10 seconds.

[@bluejedi723]

My first question is why do I want this? What am I doing with this solution?

[@jimo9555]

1. It's a solution that stops idiots coming into KZbin comment sections with stupid questions 2. Drink it 👍👍

[@anthon3373]

If u work in a chem or med lab we ise this kind stuff

[@OddawallWood]

@@jimo9555 The important idea here is not the utility of the solution, but the method for finding the solution. I worked on the assembly line at a company. I don't know how, but one of the scientists found out that I understood this kind of math, and they moved me from the line to the lab. Normally, they would not have given me an interview because I had no engineering or science degree, but by accident and the fact that I paid attention in high school math and science classes made the difference.

[@ronaldlynch9316]

One example: How much 25% antifreeze solution do I have to drain out of my radiator so that when I top it up with 100% antifreeze, I get a full radiator with a 50% solution?

[@bluejedi723]

my mechanics answer: Drain the radiator and top it off with fresh. Cost $10 either way.@@ronaldlynch9316

[@The_golden_charlie]

Who uses ounces in science? 😊

[@GFlCh]

How many oz of 5% acid and 20% acid must be mixed to get 10 oz of a solution that is 12.5% acid? a is qty (ounces) of 5% acid solution b is qty (ounces) of 20% acid solution a + b = 10 ounces Ounces of pure acid in 10 ounces of 12.5% acid solution: 10 ounces * 0.125 = 1.25 ounces 5% of (a) plus 20% of (b) is 1.25 ounces of pure acid: a * 0.05 + b * 0.2 = 1.25 Get in terms of a: a + b = 10 a + b - (b) = 10 - (b) a = (10 - b) Substitute and solve for b: a * 0.05 + b * 0.2 = 1.25 ((10 - b) * 0.05) + b * 0.2 = 1.25 (10 * 0.05)-(b * 0.05)+(b * 0.2) = 1.25 ( 0.5 )-(b * 0.05)+(b * 0.2) - (0.5) = 1.25 - (0.5) -(b * 0.05)+(b * 0.2) = 0.75 (b * 0.2) - (b * 0.05) = 0.75 b * (0.2 - 0.05) = 0.75 b * (0.15) = 0.75 (b * (0.15))/(0.15) = 0.75/0.15 (b) = 5 Solve for a: a + b = 10 a + 5 = 10 a = 5 a = 5 ounces b = 5 ounces

[@rodneybruce4550]

Yes! This is the same way I understood it as well

[@rodneybruce4550]

Yes! This is the same way I understood it as well

[@lastchance8142]

Content is great on this channel, but actual students don't have time for all the extraneous talk. Just get down to the math please! Talk at the end so we can move on if necessary.

[@pharmdiesel]

Allegation... sorry I'm a pharmacist

[@gary987]

same here :)