# Mixture Problems

Q:  How many mL of a 3% acid solution should be mixed with a 7% acid solution to create 120 mL of a 5% acid solution?

There is a generic formula we can use for mixture problems:

(quantity of solution A)*(% of A) + (quantity of solution B)*(% of B) = (final quantity)*(% of final)

So, for this problem, we know the final quantity is 120 mL.

Let’s use “x” to represent the amount of solution A that we need.  Since we need 120 total, we know that the amount of solution B can be represented by “120 – x” (the left-over).

So, plugging the values and variables into the equation:

(quantity of solution A)*(% of A) + (quantity of solution B)*(% of B) = (final quantity)*(% of final)

(x)(.03)+(120-x)(.07) = (120)*(.05)

Now we need to simplify and do the math to solve for x:

.03x + 8.4 – .07x = 6 [I distributed through the .07 on the left side and multiplied the numbers on the right side of the equation]

Now, combine the x’s on the left side:

-.04x + 8.4 = 6[now solve for x]

-.04x = -2.4

x = 60 mL

So, we should use 60 mL of solution A.

And, we should use 120 – x of solution B.  Since x = 60, we need 120 – 60 of solution B.  This is also 60 mL.

So, we should use 60 mL of solution A (the 3% solution) and 60 mL of solution B (the 7% solution) to get 120 mL of a 5% solution.

Note:  You could also solve this problem by observation:  Since 5% is exactly in between 3% and 7%, we know we need an equal amount of each solution for a perfect balance.  Since there is 120 mL total needed, we would need half to be 3% and half to be 7%

This type of logic only works when the numbers are “nice” like in this problem.  The work, shown above, can be used no matter what the numbers are.