Question 163214: To the nearest tenth, how much pure acid must be added to 12 liters of a 20% acid solution in order to obtain a mixture that is 45% acid? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website!
Let x= amount of pure acid needed
Now we know that the amount of pure acid in the 20% solution (0.20*12)plus tha amount of pure acid added (x) has to equal the amount of pure acid in the final mixture ((12+x)*0.45). So our equation to solve is:
0.20*12+x=0.45(12+x) get rid of parens (distributive law) and simplify
2.4+x=5.4+0.45x subtract 0.45x and also 2.4 from each side
2.4-2.4+x-0.45x=5.4-2.4+0.45x-0.45x collect like terms
0.55x=3 divide each side by 0.55
x=5.5 liters
CK
.20*12+5.5=0.45*17.5
2.4+5.5=7.875
7.9~~~7.875
Maybe an easier way:
Amount of otherstuff in the 20% solution (0.80*12) plus the amount of otherstuff in the pure acid added(0) has to equal the amount of otherstuff in the final mixture (0.55(12+x)). So our equation to solve is:
0.80*12=0.55(12+x) get rid of parens and simplify
9.6=6.6+0.55x subtract 6.6 from each side
9.6-6.6=0.55x collect like terms
0.55x=3 -----------------same as before
Hope this helps---ptaylor
