How much water in mL must be added to 40mL of a 10% acid solution to reduce its concentration to 4%?

Answer 1

x = 10% = 10 / 100 = 0.1

10% acid solution of 40 mL =

0.1 * 40 mL = 4 mL of acid

Volume of new acid solution = 40 mL + x

where x = volume of added wather

4 mL represent 4 %

4 % = 4 / 100 = 0.04

Now 4 mL of acid / new volume = 4 % =0.04

4 mL / ( 40 mL + x ) = 0.04

In mathematical form :

4 / ( 40 + x ) = 0.04 Multiply both sides by 40 + x

4 = 0.04 * ( 40 + x )

4 = 0.04 * 40 + 0.04 * x

4 = 1.6 + 0.04 x Subtract 1.6 to both sides

4 - 1.6 = 1.6 + 0.04 x - 1.6

2.4 = 0.04 x Divide both sides by 0.04

2.4 / 0.04 = x

60 = x

x = 60 mL

Proof :

New volume = 40 mL + 60 mL = 100 mL

4 mL of acid / new volume = 4 mL / 100 mL = 0.04 = 4 %

Answer 2

0.0x + .1(40) = .04(x+40)
0.0x + 4 = .04x + 1.6
.04x = 2.4
x = 60

60 ml of water must be added

Answer 3

or, you are reducing the concentration by a factor of 4/10, so you must increase the volume by a factor of 10/4.

10/4 * 40 = 100

So, you must wind up with 100ml, by adding 60ml of water.