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 =

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 =

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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.

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