Question

How many milliliters of 60% carbonic acid must be mixed with how many milliliters of 15% carbonic acid to make 650 milliliters of a 38% carbonic acid solution

Answer 1

348.9 mL of the 60% solution and 251.1 mL of the 15% solution.

Step-by-step explanation:

First, we calculate how many mililiters of pure carbonic acid are there in 650 mL of a 38% solution:

• 650 mL * 38/100 = 247 mL

Then we can express the sum of both initial solutions as:

• 1) x * 60/100 + y * 15/100 = 247

for the volume of carbonic acid; and

• 2) x + y = 600 mL

For the volume of the solutions.

We now have a system of two equations and two unknowns (x is the volume of the 60% solution and y is the volume of the 15% solution).

We express x in terms of y in equation 2):

• x = 600 - y

And replace x in equation 1):

• (600 - y) * 60/100 + y * 15/100 = 247

• 360 - 0.6y + 0.15y = 247

• -0.45y = -113

• y = 251. 1 mL

Finally we calculate x using equation 2):

• x + 251.1 = 600

• x = 348.9 mL