# How many pounds of water must be added to 40 pounds of a 10% boric acid to reduce it to a 4% solution?

Oct 13, 2015

$\text{60 lbs.}$

#### Explanation:

Think about it like this. Your current solution is 10 parts boric acid and 90 parts water.

You need to figure out how much water you must add in order to reduce this ratio to 4 parts boric acid and 96 parts water.

The important thing to keep in mind here is that the amount of boric acid remains unchanged because you're only adding water to the initial solution - you're essentially diluting this solution.

So, the initial solution contains

40color(red)(cancel(color(black)("lbs. solution"))) * "10 lbs. boric acid"/(100color(red)(cancel(color(black)("lbs. solution")))) = "4 lbs. boric acid"

Let's say that $x$ represents the amount of water that you need to add. This means that you can write

(4color(red)(cancel(color(black)("lbs."))))/((40 + x)color(red)(cancel(color(black)("lbs.")))) xx 100 = 4%

Solve this equation for $x$ to get

$4 \cdot \left(40 + x\right) = 400$

$160 + 4 x = 400$

$4 x = 240 \implies x = \frac{240}{4} = \text{60 lbs.}$

Therefore, if you add $60$ pounds of water to the initial solution, the percent concentration of boric acid will decrease from 10% to 4%.