Expert: Dr. Henry Boyter, Jr. - 8/19/2008

Question
I am studying for the MCAT and I am in fact doing a practice problem but I am not asking you to solve it for me.  I need help with a simple detail which I can't seem to grasp.  Thanks in advance!

Find the amount of HCl to be added to produce a pH=2.?
20mL solution of a neutralized solution .5M NaCl (from 10mL 1M HCl and 1M NaOH)

We want a [H ] to equal 1x10^-2. Let x be the amount of HCl added. So the total final volume will be (20 x). I get that.. since this is now a dilution problem we can use:

M1V1=M2V2
(1M)(x mL) = (.01 M) (20   x mL)

Now what I don't get.. is for the initial concentration and volume (M1 and V1) why is it 1M and why do we use x for V1?  I can't seem to understand the initial concentrations.

Answer
Either you copied it funny or it is written funny.  The problem is adding 1M HCl to 10 mL of 1 M NaOH to give a pH of 2.  It takes ten mL to neutralize the NaOH at a total volume of 20mL.  Now you keep adding HCl to get pH 2 (0.01M).  THe 1M HCl is what you are given to add.  You need to know how much more after neutralization.  That is what you are trying to fin, so it is x.  THe final volume is 20mL + x, since you are adding volume.  So the equation is

(1M)(x mL) = (.01 M) (20+x mL)

x = 0.2 + 0.01x

0.99x = 0.2

x=0.202 mL = 0.2

So the total amount of HCL added is 10 + 0.2 or 10.2 mL


Check my math