Question

Pure acid is to be added to a 10% acid solution to obtain 90L of 22% solution. What amount of each solution should be used?

Answer 1

Answer:

Mix the following to obtain the desired solution:

• $\rm 12 \; L$ of the 100%-pure acid solution, and
• $\rm 78\; L$ of the 22% acid solution.

Assumption:

After mixing, the volume of the final solution is equal to the volume of the 22% solution plus the volume of the pure acid solution.

Step-by-step explanation:

Let the volume of the pure acid required be $\rm x$ liters.

The volume of the final solution is given to be 90 liters. If the assumption is true, the volume of the 22% solution has to be equal to $(90 - x)$ liters.

$\text{Solvent} = \text{Solution} \times \text{Concentration}$.

Amount of solvent from that $x$ liters pure acid solution:

$\displaystyle x \cdot 100\% = x \cdot \frac{100}{100} = x$.

Amount of solvent from that $(90 - x)$ liters of 10% acid solution:

$\displaystyle (90 - x) \cdot 10\% = (90 -x) \cdot \frac{10}{100} = 9 - 0.1 x$.

Solvent from the two solutions, combined:

$x + (9 - 0.1x) = 0.9x + 9$.

Concentration the mixed solution:

$\displaystyle \text{Concentration} = \frac{\text{Solvent}}{\text{Solution}} = \frac{0.9x + 9}{90} = 0.01x + 0.1$.

This concentration is expected to be equal to

$\displaystyle 22\% = \frac{22}{100} = 0.22$.

In other words,

$0.01x + 0.1 = 0.22$.

Solve for $x$:

$x = 12$.

That is: $12$ liters of the pure acid is required. Another $90 - 12 = 78$ liters of the 22% solution will be required.