Question

a teaching assistant at a university needs 28% acid solution for her classs lab experiment. there isnt 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? write your answer as a whole number or as a decimal rounded to the nearest tenth.

Explanation:

Step1: Set up the equation

Let $x$ be the volume (in liters) of the 32% acid - solution to be added. The amount of acid in the 22% solution is $0.22\times50$ liters, the amount of acid in the 32% solution is $0.32x$ liters, and the total volume of the final solution is $(50 + x)$ liters with a desired concentration of 28%. So the equation based on the conservation of the amount of acid is $0.22\times50+0.32x=0.28\times(50 + x)$.

Step2: Expand the equation

$0.22\times50+0.32x=0.28\times50+0.28x$.
$11 + 0.32x=14+0.28x$.

Step3: Solve for $x$

Subtract $0.28x$ from both sides: $11 + 0.32x-0.28x=14+0.28x-0.28x$.
$11 + 0.04x=14$.
Subtract 11 from both sides: $0.04x=14 - 11$.
$0.04x=3$.
Divide both sides by 0.04: $x=\frac{3}{0.04}=75$.

Answer:

75