How many liters of each should be mixed to give the acid needed for the experiment if it calls for one liter of sulfuric acid at a 15% concentration, but the supply room only stocks sulfuric acid in concentrations of 10% and 35%?

Dec 10, 2014

Let $X$ be the amount of 10% sulfuric acid that is required
Let $Y$ be the amount of 35% sulfuric acid that is required

Since the resulting 15% sulfuric acid should be 1 L,

$Y = 1 - X$

Equate the amount of pure sulfuric acid resulting from the parts and from the whole

$0.10 X + 0.35 \left(1 - X\right) = 0.15 \left(1\right)$

$\implies 100 \left[0.10 X + 0.35 \left(1 - X\right) = 0.15 \left(1\right)\right]$

$\implies 10 X + 35 \left(1 - X\right) = 15 \left(1\right)$

$\implies 10 X + 35 - 35 X = 15$

$\implies - 25 X = 15 - 35$

$\implies - 25 X = - 20$

$\implies X = \frac{20}{25}$

$\implies X = 0.80$

$Y = 1 - X$

$\implies Y = 1 - 0.80$

$\implies Y = 0.20$

0.80 L of 10% Sulfuric Acid and 0.20 L of 35% Sulfuric Acid is required