Expert: Sherry Wallin - 11/1/2010

Question
What volume, in millilitres, of a 60% hydrochloric acid solution must be added to a 100 ml of a 30% hydrochloric acid solution to make a 36% hydrochloric solution?

Answer
Shelly~

This is a problem in the category of rate, just like d = rt and you set it up in a similar way, a table usually works well. You need to label a variable and since you are looking for the amount of 60% acid solution to add to the 30% acid solution let x = amount of 60% acid solution

------------------------------------------------------------------------

initial conditions          rate %                     quantity (ml)
------------------------------------------------------------------------

beginning                   60%                          x
beginning                   30%                         100
------------------------------------------------------------------------
end                            36%                         100 + x

given 60% is .6 and 30% is .3 and 36% is .36 the sum of the beginning
conditions has to equal the end condition:

.6x + .3(100)    = .36(x+100) =>
.6x + 30           = .36x + .36(100) =>
.6x + 30           = .36x + 36
.6x +30 - 30     = .36x + 36 - 30 =>
.6x                   = .36x + 6  =>
.6x - .36x         = .36x - .36x + 6 =>
.24x                 = 6 =>
.24x/.24           = 6/.24 =>
x                      = 25

So you will need to add 25ml of 60% acid solution to 100ml of 30% acid solution in order to obtain 36% acid solution.

Math Prof