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.