composition problem

[Dani]

I want to make 300ml of a 80% acid concentration from two acid concentrations a) 50% concentration and b) 100% concentration. what volumes of a and b would be required to make the 300 ml of 80% acid concentration?

 

[lookagain]

I want to make 300ml of a 80% acid concentration from two acid concentrations a) 50% concentration and b) 100% concentration. what volumes of a and b would be required to make the 300 ml of 80% acid concentration?

Dani,

please post your work first so we make help you with where you are stuck.

 

[Dani]

Dani,

please post your work first so we make help you with where you are stuck.

Let X be ml of 50% concentration
Y be ml of 100% concentration

X + Y= 300ml or Y=300-X ------> eq 1

required final concentration 80%

.5X+Y=.8(X+Y) -------eq 2

substituting Y from eq 1 to eq 2

.5X+300-X=.8(X+300-X)

.5X-.8X+.8X =300-300
.5X=0

X=0

Please help this does not make sense?

 

[soroban]

Hello, Dani!

Your set-up is excellent.

Let \(\displaystyle x\) = ml of 50% concentration
\(\displaystyle y\) = ml of 100% concentration

\(\displaystyle x + y \:=\: 300\quad\Rightarrow\quad y \:=\:300 - x\;\;[1]\)

Required final concentration 80%: .\(\displaystyle 0.5x+y\:=\:0.8(x+y)\;\;[2]\)

Substitute [1] into [2]: .\(\displaystyle 0.5x+300-x\:=\:0.8(x+300-x)\)

. . \(\displaystyle 0.5x-0.8x+0.8x \:=\:300-300\) . ?

You had: .\(\displaystyle 0.5x + 300 - x \:=\:0.8(x + 300 - x)\)

Then: .. . \(\displaystyle 0.5x + 300 - x\:=\:0.8x + 240 - 0.8x\)

Got it?

 

[Dani]

Hello, Dani!

Your set-up is excellent.



You had: .\(\displaystyle 0.5x + 300 - x \:=\:0.8(x + 300 - x)\)

Then: .. . \(\displaystyle 0.5x + 300 - x\:=\:0.8x + 240 - 0.8x\)

Got it?

Thanks what a silly mistake I was making.