How many liters of a 90% acid solution must be added to 6 liters of a 15% acid solution to obtain a 40% acid solution?

1 Answer
Sep 24, 2016

Answer:

You must add 3 L of the 90 % acid.

Explanation:

For this problem, we can use the relation

#color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2 = "amount of solute" color(white)(a/a)|)))" "#

Let the #"volume of 90 % acid" = xcolor(white)(l) "L"#

Then, after mixing, we have #(6 + x") L of 40 % acid"#.

This is made up of #"6 L of 15 % acid"# and #xcolor(white)(l) "L of 90 % acid"#.

#"Moles before = moles after"#

#c_1V_1 = c_2V_2 + c_3V_3#

#40 color(red)(cancel(color(black)(%))) × (6 + x) color(red)(cancel(color(black)("L"))) = 15 color(red)(cancel(color(black)(%))) × 6 color(red)(cancel(color(black)("L"))) + 90 color(red)(cancel(color(black)(%))) × x color(red)(cancel(color(black)("L")))#

#240 + 40x = 90 + 90x#

#50x = 150#

#x = 150/50 = 3#

So, we add #"3 L of 90 % acid"# to #"6 L of 15 % acid"# and get #"9 L of 40 % acid"#.

Check:

#3 × 90 + 6 × 15 = 9 × 40#

#270 + 90 = 360#

#360 = 360#

It checks!