Question 641405: What quantity of pure acid must be added to 500 mL of a 40% acid solution to produce a 50% acid solution? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=quantity of pure acid needed
Now we know that the amount of pure acid that exists before the mixture takes place (x+0.40*500) has to equal the amount of pure acid that exists after the mixture takes place (0.50(500+x)). Soooo
x+0.40*500=0.50(500+x) simplify
x+200=250+0.50x subtract 0.50x and also 200 from each side
x-0.50x=250-200
0.50x=50
x=100 mL quantity of pure acid needed
CK
100+200=0.50*600
300=300
Hope this helps--ptaylor
