How many milliliters of 35% acid solution need to be added to60mL of 65% acid solution to get a solution that is45% acid?

Question

How many milliliters of 35% acid solution need to be added to60mL of 65% acid solution to get a solution that is45% acid?

Answer 1

If x = number of ml of 35% acid solution and 60 = number of ml of 65% acid solution, then x + 60 = number of ml of the mixture.

 

Acid content of first ingredient + acid content of second ingredient = acid content of the mixture

 

So, 0.35x + 0.65(60) = 0.45(x + 60)

     0.35x + 39 = 0.45x + 27

                 12 = 0.10x

                120 = x

 

ANSWER:  mix 120 ml of 35% acid solution with 60 ml of 65% acid                      solution to obtain 180 ml 0f 45% acid solution 

Answer 2

The amount of pure acid in the existing 65% solution is 65% of 60 mL, which is 33 mL of acid.

 

Let x be the amount of 35% solution that needs to be added.  The amount of pure acid that will be in that amount will be 35% of x, or .35x

 

When the mixture is done, the total amount of pure acid divided by the total amount of solution will be 45%, or .45

 

So the amount of pure acid in the solution at the end will be 33 mL from the original 65% solution plus .35x from the 35% solution that will be added, for a total of 33 + .35x.

 

The total amount of solution at the end will be 60 mL from the original 65% solution plus x from the 35% solution that will be added, for a total of 60 + x.

 

So (33 + .35x) / (60 + x) will be the percentage acid in the final mixture, which needs to be 45%.

 

So (33 + .35x) / (60 + x) = .45

 

Let's solve this for x:

 

Multiply both sides by 60 + x:

33 + .35x = .45  (60 + x)

 

Simplify

33 + .35x = 21 + .45x

 

Subtract 21 from both sides:

12 + .35x = .45x

 

Subtract .35x from both sides:

12 = .10x

 

Divide both sides by .10:

120 = x

 

So the answer is that 120 mL of 35% solution will need to be added.