Pure acid is to be added to a 10% acid solution to obtain 90L of 58% solution. What amounts of each should be used? How many of 100% pure acid should be used to make the solution?
Pure acid =100% acid = 1
So...let x be the amout of pure acid to be added and we have that
1x +.10 *( 90 - x) = ( .58)(90)
x + 9 - .10x = 52.2
.90x = 52.2 - 9
.90x = 43.2 divide both sides by .90
x = 43.2 / .90 = 48 L
