Mixing 2 liquids of different concentration

Hi,

If I had 500 litres of a solution that is 10% concentration and I want to add another solution to it which is 20% concentration to have a total concentration of 14%, how much of the 20% solution do I need to add? I think the answer is 334 litres but I am not sure how I came to this answer can anyone teach me a formula?

Thanks

Re: Mixing 2 liquids of different concentration

Quote:

Originally Posted by

**bronson** Hi,

If I had 500 litres of a solution that is 10% concentration

So you have 500(.10)= 50 litres of concentrate.

Quote:

and I want to add another solution to it which is 20% concentration to have a total concentration of 14%, how much of the 20% solution do I need to add? I think the answer is 334 litres but I am not sure how I came to this answer can anyone teach me a formula?

334 litres of a 20% solution would give .2(334)= 66.8 litres of concentrate. Adding that to the original 50 litres gives 116.8 litres of concentrate in a total of 500+ 334= 834 litres. That is a concentration of 116.8/834= .14 or 14%. Looks good!

Let "x" be the amount added. That will be .2x of concentrate giving a total of 50+ .2x concentrate in 500+ x litres. That is a concentration of (50+ .2x)/(500+ x)= .14. Solve that for x.

Re: Mixing 2 liquids of different concentration

Hi,

Thanks for the reply, your answer looks great but without trying to sound dumb or ask for all the answers how would I go about solving that equation?

Re: Mixing 2 liquids of different concentration

Hello, bronson!

Quote:

I have 500 litres of a solution that is 10% acid.

I want to add another solution which is 20% acid to have a total concentration of 14% acid.

How much of the 20% solution do I need to add?

I think the answer is 334 litres . Quite close!

but I am not sure how I came to this answer. . If you don't know, we sure can't guess!

We start with 500 liters of a solution which is 10% acid.

. . It contains: .$\displaystyle (0.10)(500) \:=\:50$ liters of acid.

We add $\displaystyle x$ liters of a solution which is 20% acid.

. . It contains: .$\displaystyle 0.20x$ liters of acid.

Hence, the mixture will contain: .$\displaystyle 50 + 0.20x$ liters of acid.

But we know that the mixure will be: $\displaystyle 50+x$ liters which will be 14% acid.

. . The mixture will contain: .$\displaystyle 0.14(50+x)$ liters of acid.

We just described the amount of acid in the final mixture in *two** *__ways__.

There is our equation! . . . .$\displaystyle 50 + 0.20x \;=\;0.14(50 + x)$

Got it?

Re: Mixing 2 liquids of different concentration

Hi,

thanks for your reply, I understand how both equations where figured out but now I am having problems solving them.

I cannot find any info on how to solve (50+ .2x)/(500+ x)= .14

and the info i found on solving 50 + 0.20x = 0.14(50 + x) gives me the wrong answer here is what I did:

first I simplified the equation

50 + 0.20x = 0.14(50 + x)

50 + 0.20x = (50 * 0.14 + x * 0.14)

50 + 0.20x = 7 + 0.14x

Add -0.14x

50 + 0.20x + -0.14x = 7 + 0.14x + -0.14x

Combine like terms:

50 + 0.06x = 7 + 0.14x + -0.14x

50 + 0.06x = 7 + 0.00

50 + 0.06x = 7

Add -50

50 + -50 + 0.06x = 7 + -50

Combine like terms:

0 + 0.06x = 7 + -50

0.06x = 7 + -50

0.06x = -43

Divide by 0.06

x = -716.6666667

Re: Mixing 2 liquids of different concentration

Quote:

If I had 500 litres of a solution that is 10% concentration and I want to add another solution to it which is 20% concentration to have a total concentration of 14%, how much of the 20% solution do I need to add?

500(10) + x(20) = (500+x)(14)

5000 + 20x = 7000 + 14x

6x = 2000

x = 333 L

Re: Mixing 2 liquids of different concentration

Thanks to everyone that has helped, I can now easily work this out using the method that skeeter posted.

I would still be interested in finding out how to solve it using the equation from HallsofIvy's post which is:

(50+ .2x)/(500+ x)= .14