How many liters of pure acid should be added to 25 liters of a 60% solution of acid to obtain a 75% acid solution?

Aug 10, 2015

You need to add 15 L of pure acid.

Explanation:

The idea behind this problem is that you're adding pure acid to a solution that is 60% v/v acid, so the added acid will change the percent concentration of the starting solution by increasing the volume of solute and the volume of solvent by the same amount.

So, start by calculating how much acid you have in your starting solution

25color(red)(cancel(color(black)("L solution"))) * "60 L acid"/(100color(red)(cancel(color(black)("L solution")))) = "15 L acid"

Let's say that $x$ denotes the volume of pure acid that you must add to your starting solution. Since you're dealing with pure acid, the volume of acid will Increase by $x$

${V}_{\text{acid}} = 15 + x$

At the same time, the volume of the solution will also incrase by $x$

${V}_{\text{sol}} = 25 + x$

This means that the target solution's percent concetration by volume will be equal to

V_"acid"/V_"soL" * 100 = 75%

((15 + x)color(red)(cancel(color(black)("L"))))/((25 + x)color(red)(cancel(color(black)("L")))) * 100 = 75%

Solve this equation for $x$ to get

$\left(15 + x\right) \cdot 100 = \left(25 + x\right) \cdot 75$

$1500 + 100 x = 1875 + 75 x$

$25 x = 375 \implies x = \frac{375}{25} = \textcolor{g r e e n}{\text{15 L}}$

So, if you add 15 L of pure acid to 25 L of 60% v/v acid solution, you'll get 40 L of 75% v/v acid solution.