Question

How many millimeters of 30% acid solution need to be added to 60mL of 55% acid solution to get a solution that is 45% acid

Answer 1

so.. first off.. let's keep in mind that if 30% of acid is there in a solution, say the solution is some amount of liquid and 30% of that total liquid only is the acid, so if the amount of the liquid is say "x", then 30% of that is just (30/100) * x, or 0.3x.

now, we'll do the same for the 55% and the 45%, use the decimal format to get how much quantity in the substance is of acid.


`\bf \begin{array}{lccclll} &\stackrel{mL}{amount}&\stackrel{acid~\%}{concentration}&\stackrel{amount}{concentration}\\ &------&------&------\\ \textit{30\% sol'n}&x&0.3&0.3x\\ \textit{55\% sol'n}&60&0.55&33\\ ------&------&------&------\\ mixture&y&0.45&0.45y \end{array}`

now, whatever "x" and "y" are, we know that x + 60 = y, and we also know that 0.3x + 33 = 0.45y.


`\bf \begin{cases} x+60=\boxed{y}\\ 0.3x+33=0.45y\\ ----------\\ 0.3x+33=0.45\left( \boxed{x+60} \right) \end{cases} \\\\\\ 0.3x+33=0.45x+27\implies 33-27=0.45x-0.3x \\\\\\ 6=0.15x\implies \cfrac{6}{0.15}=x\implies \stackrel{mL}{40}=x`