Question 994765
If you wish to increase the percent of acid in 50 mL of a 15% acid solution in water to a 25% acid, how much pure acid must you add? 
:
let x = amt of pure acid required
:
A typical mixture equation based on the amt of acid in the water
.15(50) + x = .25(x+50)
7.5 + x = .25x + 12.5
x - .25x = 12.5 - 7.5
.75x = 5
x = 5/.75
x = 6{{{2/3}}} mL of pure acid required