Search 70,565 tutors

# What quantity of 70 per cent acid solution must be mixed with a 30 per cent solution to produce 600 mL

What quantity of 70 per cent acid solution must be mixed with a 30 per cent solution to produce 600 mL of a 50 per cent solution?

mL of acid solution =

x = Volume of 70% acid

y = Volume of 30% acid

Since the volume of the 70% solution (x) plus the volume of the 30% solution (y) will equal the total volume of the final solution:

x + y = 600         Equation 1

The volume times concentration percentage of the first solution (.7x) plus the volume times the concentration percentage of the second solution (.3y) will equal the volume times the percentage of the final solution (.5 *600).

.7x + .3y = .5*600

.7x + .3y = 300                  Equation 2

Now that we have 2 equations in 2 variables, I can use substitution to solve the system.

x + y = 600                                 Equation 1

x + y - y = 600 -y                        Subtract y from each side

x = 600 - y                                 Simplify

.7x + .3y = 300                           Equation 2

.7(600 - y) + .3y = 300                Substitute the value of x from equation 1 into equation 2

420 - .7y +. 3y = 300                   Distribute the .7

420 - .4y = 300                           Simplify

420 - .4y - 420 = 300 - 420          Subtract 420 from each side

-.4y = -120                                 Simplify

-.4y/(-.4) = -120/(-.4)                 Divide both sides by -.4

y = 300                                      Simplify

x = 600 - y                                  Equation 1 solved for x

x = 600 - 300                              Substitution

x = 300

You will need 300 mL of 70% solution and 300 mL of 30% solution.