Question

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

Answer 1

To solve this question, first we will find the volume of acid and then we will calculate the volume of the solution, and then we will be able to find the volume of the pure solution by the target solution’s percent concentration.

Complete answer:
The idea behind this problem is that you're adding pure acid to a solution that is ${\mathbf{60}}\% {\text{ }}{\mathbf{v}}/{\mathbf{v}}\;$ 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:
$25L\,solution.\dfrac{{60L\,acid}}{{100L\,solution}} = 15L\,acid$
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 $x$
${V_{acid}} = 15 + x$
At the same time, the volume of the solution will also increase by $x$
${V_{sol}} = 25 + x$
This means that the target solution's percent concentration by volume will be equal to,
$\dfrac{{{V_{acid}}}}{{{V_{sol}}}}.100 = 75\%$
$\dfrac{{(15 + x)L}}{{(25 + x)L}}.100 = 75\%$
Solve this equation for $x$ to get
(15 + x).100 = (25 + x).75 \\\ \Rightarrow 1500 + 100x = 1875 + 75x \\\ \Rightarrow 25x = 375 \\\ \Rightarrow x = \dfrac{{375}}{{25}} = 15L \\\
So, if we add $15L$ of pure acid to $25L$ of $60\%$ $v/v$ acid solution, we will get $40L\,of\,v/v$ acid solution.

Note:
Volume percent is relative to the volume of solution, not the volume of solvent. For example, wine is about $12\% {\text{ }}v/v$ ethanol. This means there is $12{\text{ }}ml$ ethanol for every $100{\text{ }}ml\;$ of wine. It is important to realize liquid and gas volumes are not necessarily additive.