SOLUTION: A chemist has two large containers hydrochloric acid (HCl) solution. The concentrations of the acid is different in the two containers. She blends 10 mL of the first solution with
Question 224231: A chemist has two large containers hydrochloric acid (HCl) solution. The concentrations of the acid is different in the two containers. She blends 10 mL of the first solution with 90 mL of the second solution to obtain a solution that is 15.15.% acid. She blends 100 mL of the first solution with 150 mL of the second solution to obtain a solution that is 15.60.% acid. What are the concentrations of hydrochloric acid in the original containers?
You can put this solution on YOUR website! The concentrations of the acid is different in the two containers.
He blends 10 mL of the first solution with 90 mL of the second solution to obtain a solution that is 15.15.% acid.
He blends 100 mL of the first solution with 150 mL of the second solution to obtain a solution that is 15.60.% acid.
What are the concentrations of hydrochloric acid in the original containers?
:
Use decimal equiv of the percent;
:
Let x = acid concentration of the 1st container
Let y = acid concentration of the 2nd container
:
"He blends 10 mL of the first solution with 90 mL of the second solution to obtain a solution that is 15.15.% acid."
10x + 90y = .1515(100); resulting amt is the sum of the two solutions
10x + 90y = 15.15
:
"He blends 100 mL of the first solution with 150 mL of the second solution to obtain a solution that is 15.60.% acid."
100x + 150y = .1560(250)
100x + 150y = 39
Simplify, divide by 10
10x + 15y = 3.9
:
Subtract this from the 1st equation:
10x + 90y = 15.15
10x + 15y = 3.9
--------------------subtraction eliminate x, find y
75y = 11.25
y =
y = .15 or 15% is the concentration of the 2nd container
:
find x using equation: 10x + 15y = 3.9
10x + .15(15) = 3.9
10x + 2.25 = 3.9
10x = 3.9 - 2.25
10x = 1.65
x =
x = .165 or 16.5% is the concentration of the 1st container
:
:
Check solution in the statement:
He blends 100 mL of the first solution with 150 mL of the second solution to obtain a solution that is 15.60.% acid."
.165(100) + .15(150) = .156(250)
16.5 + 22.5 = 39
39 = 39
