Question

how many liters of a 40% acid solution must be added to 12 liters of a 20% solution to obtain a 25% solution​

Answer 1

To achieve a 25% acid solution, you must add 4 liters of a 40% acid solution to 12 liters of a 20% acid solution.

Step-by-step explanation:

To solve the problem of mixing two different concentrations of an acid solution to achieve a certain volume and concentration, you can use the formula for the conservation of mass of the solute acid. Specifically, the mass of the acid before mixing is equal to the mass of acid after mixing. You want to obtain a 25% acid solution after mixing a 40% acid solution with a 20% acid solution.

Let x be the number of liters of the 40% solution that needs to be added. The total volume of acid in the 40% solution is 0.4x liters (since 40% of the solution is acid), and the total volume of acid in the 12 liters of a 20% solution is 2.4 liters (since 20% of 12 is 2.4). After mixing, the total volume of the solution is x + 12 liters, and the volume of acid should be 0.25(x + 12) liters to make a 25% solution. The equation is set up as follows and solved for x:

0.4x + 2.4 = 0.25(x + 12)

Solving for x gives:

0.4x + 2.4 = 0.25x + 3

0.15x = 0.6

x = 4

So, you would need to add 4 liters of the 40% acid solution to the 12 liters of the 20% acid solution to achieve a 25% solution.

Answer 2

The answer is 4 liters.

Step-by-step explanation:

We want a final 25% solution, which means that the solvent should constitute 25 percent of all the liters of solution.

Let us call
`x` the litres of 40% solution, so when they are added to the 12 litres of 20% solution, the total litres of the new solution are:

`x+12`

of these liters, 25% must be the solvent:

`N=0.25(x+12)`

Now for the 40% solution, the liters of the solvent are
`0.4x`, and for the 20% liter solution, the liters of solve are
`0.2*12=2.4` liters. Thus the total liters
`N` of the solvent are

`N=0.4x+2.4`

Therefore

`N=0.4x+2.4=0.25(x+12)`

`\therefore x=4\:litres`