A chemist has prepared two acid solutions, one of which is 4% acid by volume, and the other is 11% acid. How many cubic centimeters of each should the chist mix together to obtain 50cm^3 of a 6.94% solution.

Hi Melissa,

We have a system of equations that needs to be solved. One acid solution has 4% acid, and the other has 11% acid. Say we have x cm

^{3}of the first solution (4%) and y cm^{3}of the second solution (11%). When mixed, we want to have a total of 50 cm^{3}, sox + y = 50

The resultant solution must have 6.94% acid, and the solution is 50 cm

^{3}, so0.04x + 0.11y = 50*0.0694

or just

0.04x + 0.11y = 3.47

Since x + y = 50, y = 50 - x. Substitute this into the above to obtain

0.04x + 0.11(50 - x) = 3.47

Multiplying out yields

0.04x + 5.5 - 0.11x = 3.47

Combine like terms, and move the 5.5 over to obtain

-.07x = -2.03

Dividing by -.07 yields

x = 29

Since

**x = 29, y = 50 - 29 = 21**