– A CHEMIST NEEDS TO MAKE TWO

LITERS OF A 15% ACID SOLUTION FROM HER 10% ACID SOLUTION

AND A 35% ACID SOLUTION. SO WE’RE MIXING

THE 10% AND 35% SOLUTION TO CREATE TWO LITERS

OF 15% SOLUTION. WE WANT TO KNOW HOW MANY LITERS

OF EACH SHOULD SHE MIX TO GET THE DESIRED SOLUTION. SO WE’RE GOING TO NEED

TWO VARIABLES HERE AS WELL AS TWO EQUATIONS. LET’S LET X EQUAL THE NUMBER

OF LITERS OF THE 10% SOLUTION. AND WE’LL LET Y EQUAL THE NUMBER

OF NUMBER OF LITERS OF THE 35% SOLUTION. NOW SEE IF WE CAN WRITE

OUR TWO EQUATIONS. WE KNOW AFTER MIXING THESE

TWO AMOUNTS, WE’LL HAVE EXACTLY TWO LITERS

OF THE 15% SOLUTION. SO OUR QUANTITY EQUATION WILL BE

X + Y MUST EQUAL TWO LITERS. THE SECOND EQUATION IS GOING DEAL WITH THE

CONCENTRATIONS OF EACH SOLUTION. SINCE X IS EQUAL TO THE NUMBER

OF LITERS OF THE 10% SOLUTION AND Y IS EQUAL TO THE NUMBER

OF LITERS OF THE 35% SOLUTION, 10% x X, OR 0.10 x X + 35% x Y,

OR 0.35 x Y MUST EQUAL 15% OR 0.15 x THE TOTAL AMOUNT

OF THE SOLUTION, WHICH IS 2 LITERS. SO HERE’S OUR SYSTEM OF

EQUATIONS THAT WE HAVE TO SOLVE. THE FORM OF THE FIRST EQUATION

WILL BE GOOD TO WORK WITH, BUT LET’S TRANSFORM

THE SECOND EQUATION BEFORE WE SOLVE THE SYSTEM. LET’S START BY ELIMINATING

THE DECIMALS. IF WE MULTIPLY EVERYTHING

BY 100, THAT’S THE SAME AS MOVING

THE DECIMAL POINT TO THE RIGHT TWO PLACES. SO WE CAN WRITE THIS

AS 10X + 35Y EQUALS– AND THIS WOULD BE 15 x 2,

WHICH IS EQUAL TO 30. SO NOW WE’LL SOLVE THE SYSTEM

OF EQUATIONS USING THESE TWO EQUATIONS. SO AGAIN WE HAVE X + Y=2

AND THEN WE HAVE 10X + 35Y=30. NOW WE CAN SOLVE

THIS ALGEBRAICALLY USING ELIMINATION

OR SUBSTITUTION. IF WE CALL THIS EQUATION 1

AND THIS EQUATION 2, WE COULD VERY EASILY SOLVE

THE FIRST EQUATION FOR X OR Y. TO SOLVE THIS FOR X, WE WOULD

SUBTRACT Y ON BOTH SIDES. SO USING EQUATION 1,

WE WOULD HAVE X=2 – Y. AND THEN WE CAN PERFORM

SUBSTITUTION INTO EQUATION 2. IF X=2 – Y,

WE CAN REPLACE X WITH 2 – Y. NOW EQUATION 2

AFTER SUBSTITUTION, WOULD BE 10 x 2 – Y,

INSTEAD OF TIMES X, + 35Y=30. AND NOW WE HAVE ONE EQUATION

WITH ONE VARIABLE, SO NOW WE’LL SOLVE FOR Y. WE HAVE 20 – 10Y + 35Y=30. WELL HERE WE HAVE 2Y TERMS SO

WE CAN COMBINE THESE TWO TERMS. -10Y + 35Y=25Y,

SO WE’LL HAVE 20 + 25Y=30. SUBTRACT 20 ON BOTH SIDES WOULD

GIVE US 25Y=10. DIVIDE BOTH SIDES BY 25. AND WE HAVE Y=10/25

WHICH DOES SIMPLIFY. THERE’S A COMMON FACTOR OF 5

HERE. SO WE HAVE Y=2/5. REMEMBER Y IS THE NUMBER

OF LITERS OF THE 35% SOLUTION. SO AS AN ORDERED PAIR WE KNOW

THAT Y=2/5 WE STILL HAVE TO GO BACK

AND DETERMINE X, BUT WE DO KNOW FROM BEFORE

X=(2 – Y). SO WE’D HAVE 2 – 2/5,

WHICH IS 1 3/5. SO WE HAVE 1 3/5 LITERS

OF THE 10% SOLUTION. WE MAY WANT TO CONVERT

THESE TO DECIMALS. 3/5 IS THE SAME AS 0.6. SO THE MIXTURE CONTAINS 1.6

LITERS OF THE 10% SOLUTION AND 2/5 AS A DECIMAL WOULD BE

0.4 LITERS OF THE 35% SOLUTION. OKAY. I HOPE YOU FOUND

THIS HELPFUL.

Thanks for posting this!!

neat and to the point.

thanks this really helped. please check our videos @ https://www.youtube.com/channel/UCKfUqyk2gJHeKuAdy_8vBtA

Thanks for making the video!

I've been trying to find a video that went into depth on how to do these kind of problems, and they always skipped over something, or cheated, and since I have to show work, I couldn't do that. This helped me learn how to do these kinds of problems. Thanks so much for creating this, you just got a like and a sub! <3

What if it's like this? "You need a 50% alcohol solution. On hand, you have a 45 mL of a 40% alcohol mixture. You also have 65% alcohol mixture. How much of the 65% mixture will you need to add to obtain the desired solution?"