Question

# Pure acid is to be added to a 20% acid solution to obtain 28 L of a 40% acid solution. What amounts of each should be used?

Probability and combinatorics

Pure acid is to be added to a $$20\%$$ acid solution to obtain 28 L of a $$40\%$$ acid solution. What amounts of each should be used?

2021-06-25

Let x be the amount of pure acid to be added so that 28-x is the amount of $$20\%$$ acid solution, both in L.
In terms of percentage, a pure acid corresponds to $$100\%$$. So, we write: $$20\%$$ of $$28-x L+100\%$$ of $$x L = 40\%\ of \ 28 L$$
$$0.2(28-x)+1(x)=0.4(28)$$
Solve for $$x: 5.6-0.2x+x=11.2$$
$$5.6+0.8x=11.2$$
$$0.8x=5.6$$
$$x=7$$
So, we need 7 L of pure acid solution and 21 L of $$20\%$$ acid solution.