In a chemistry lab, you have two vinegars. one is 5% acetic acid, and one is 6.5% acetic acid. you want to make 200 ml of a vinegar with 6% acetic acid. how many milliliters of each vinegar do you need to mix together

Assume that the amount needed from the 5% acid is x and that the amount needed from the 6.5% acid is y.

We are given that:
The volume of the final solution is 200 ml
This means that:
x + y = 200
This can be rewritten as:
x = 200 - y > equation I

We are also given that:
The concentration of the final solution is 6%
This means that:
5%x + 6.5%y = 6% (x+y)
This can be rewritten as:
0.05 x + 0.065 y = 0.06 (x+y) > equation II

Substitute with equation I in equation II and solve for y as follows:
0.05 x + 0.065 y = 0.06 (x+y)
0.05 (200-y) + 0.065 y = 0.06 (200-y+y)
10 - 0.05 y + 0.065 y = 12
0.015y = 12-10 = 2
y = 2/0.015
y = 133.3334 ml

Substitute with the y in equation I to get the x as follows:
x = 200 - y
x = 200 - 133.3334
x = 66.6667 ml

Based on the above calculations:
The amount required from the 5% acid = x = 66.6667 ml
The amount required from the 6.5% acid = y = 133.3334 ml

Hope this helps :)

To answer, let x be the volume, in mL, of the of the vinegar with 6.5% acetic acid. With this representation, the volume of the vinegar with 5% acetic acid is equal to 200-x. Using the component mass balance for acetic acid, we will have the equation,

    (0.05)(200 - x) + (0.065)(x) = (0.06)(200)

The value of x from the equation is 133.3. The value of 200-x is therefore equal to 66.67.

5%-acetic acid vineger = 66.67 mL
              6.5% acetic acid vinegar = 133.3 mL