A teaching assistant at a university needs $28\%$ acid solution for her class's lab experiment. There isn't any of this concentration in stock, but the lab has $50$ liters of $22\%$ acid solution, as well as a lot of $32\%$ acid solution. How much of the $32\%$ acid solution should the teaching assistant add to the $22\%$ acid solution to obtain a solution with the desired concentration? $\newline$ Write your answer as a whole number or as a decimal rounded to the nearest tenth. $\newline$ ____ liters

Full solution

Q. A teaching assistant at a university needs

28\%

acid solution for her class's lab experiment. There isn't any of this concentration in stock, but the lab has

50

liters of

22\%

acid solution, as well as a lot of

32\%

acid solution. How much of the

32\%

acid solution should the teaching assistant add to the

22\%

acid solution to obtain a solution with the desired concentration?

\newline

Write your answer as a whole number or as a decimal rounded to the nearest tenth.

\newline

____ liters

Denote Acid Solution Volume: Let's denote the amount of $32\%$ acid solution that needs to be added as $x$ liters. The total volume of the new solution will be $50$ liters + $x$ liters.
Calculate Pure Acid in $22$ % Solution: The amount of pure acid in the $50$ liters of $22\%$ solution is $0.22 \times 50$ liters. $\newline$ Calculation: $0.22 \times 50 = 11$ liters of pure acid.
Calculate Pure Acid in $32$ % Solution: The amount of pure acid in the $x$ liters of $32\%$ solution is $0.32 \times x$ liters. $\newline$ Calculation: $0.32 \times x = 0.32x$ liters of pure acid.
Calculate Total Pure Acid: The total amount of pure acid in the final solution should be $28\%$ of the total volume, which is $28\%$ of $(50 + x)$ liters. $\newline$ Calculation: $0.28 \times (50 + x) = 14 + 0.28x$ liters of pure acid.
Set Up Equation: To find the value of $x$ , we set up the equation that the sum of the pure acid from both solutions equals the pure acid in the final solution: $11 + 0.32x = 14 + 0.28x$ .
Solve for x: Now we solve for x: $\newline$ $0.32x - 0.28x = 14 - 11$ , $\newline$ $0.04x = 3$ , $\newline$ $x = \frac{3}{0.04}$ , $\newline$ $x = 75$ liters.