To produce a solution that is 4% hydrochloric acid, one must add 120 mL of water to the current solution.
To start to solve this problem, we should try to figure out what we know. We know that we are going to have 3 different solutions, the one we have now, one that is completely water, and one that is both water and our current solution mixed together.
Our current solution is 16% hydrochloric acid and is 40 ml.
The mixture of the water and current solution would be 4% hydrochloric acid and would be x ml (since we don't know how many).
The water solution would be 0 % hydrochloric acid (because it's just water) and it would be x - 40 ml (since the result solution is the sum of the two other solutions). Now, we can start to solve.
Since the result is 4% hydrochloric acid and has a total of x mL, to find out how much hydrochloric acid there is inside, we can multiply x by 4% or 4/100, which is 1/25x. In addition, since the substance we have now is 16% hydrochloric acid, and has a total volume of 40mL, to find out how much hydrochloric acid is inside of it, we can multiply 40 by 16/100, which gives us 6.4 mL. Finally, our water has 0 mL of hydrochloric acid (because it's pure water).
Since we know that the result is the sum of what's in the water and substance we have now, we can find out what x is equal to using how much hydrochloric acid is inside of each thing. The known substance has 6.4 mL, and the water substance has 0 mL. 6.4 + 0 mL = 6.4 mL. Since the result has 1/25 x mL of hydrochloric acid, we know that 6.4 mL = 1/25 x. Now we can solve our equation.
6.4 mL = 1/25x
Multiplying each side by 25, we get that x = 6.4 * 25 = 160 mL.
But, that's not the answer. Since the question is asking us for how much water is needed, we must use substitution into how much water we are using. We are going to add x - 40 mL of water (made from our previous calculations), and x = 160, we are going to add 160 - 40 = 120 mL of water. All in all, there is no formula necessary but you have to know the method to solving these kinds of problems.
