Question

Suppose that x mL of 40% acid are added to 100 mL of 50% acid:

a) Express the concentration C(x) of the new mixture as a function of x.
b) How much 40% acid should be added to 100 mL of 50% acid to obtain a mixture that is 48% acid?

Answer 1

The concentration of the acid mixture, C(x), can be expressed by the equation 0.40x + 0.50(100) = (x + 100)C(x). To obtain a mixture that is 48% acid, one should add 25 mL of 40% acid to the 100 mL of 50% acid.

Step-by-step explanation:

To find the concentration C(x) of the new mixture when x mL of 40% acid are added to 100 mL of 50% acid, we can use a mass balance equation. The amount of pure acid in the new solution is the sum of the pure acid in each component before mixing:

(40% of x) + (50% of 100 mL) = C(x)% of (x + 100 mL)

Solving for C(x):

0.40x + 0.50(100) = (x + 100)C(x)

To make a 48% acid mixture, we set C(x) to 48% and solve for x:

0.40x + 50 = 0.48(x + 100)

0.40x + 50 = 0.48x + 48

Rearrange to isolate x:

50 - 48 = 0.48x - 0.40x

2 = 0.08x

x = 25 mL

Thus, to obtain a mixture that is 48% acid, 25 mL of 40% acid should be added to the 100 mL of 50% acid.