Titration question

[aspiringdent]

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Can someone please explain the following question from TopScore:

What volume of HCl was added if 20mL of 1 M NaOH is titrated with 1 M HCl to produce pH = 2?

I understand that its a strong acid and strong base and titrating them together will give a pH of 7 at the equivalence point, so in this case it would be 20 mL of HCl. But I am clueless on what volume of HCl would be needed to reach a pH of 2. Please help!!

 

[herkulease]

this came up before I'm gonna copy and paste what I posted last time. I'm lazy to edit but I explained everything back then.

You have a strong base and a strong acid. so you would get a neutral product right?

But here the pH is 2 so that tells us we have added a bit more of the HCl to make the solution acidic.

so with a pH of 2 that means our [H+] concentration in the final solution is 0.01. (10^-2)

our total volume will be 20ml of naoh, 20 ml of HCl plus more of HCl which we'll call X.

why 20 of each? again strong acid + strong base = neutral so you need to have the same volume in order to neutralize it. So the final volume will be 40mL + X

Now we'll use the equation M1V1 = M2V2

M1 is out concentration of HCl
V1 is the volume of extra HCl needed

1M(x) = 0.01(40+x)
x = 4 + 0.01x
0.99x = 0.4
x = to about 0.4

so now we know that the extra amount of HCl added was 0.4 ml

But it wants the Total volume of HCl added.

so its 20 + 0.4 = 20.4 ml of HCl.

 

[aspiringdent]

Ok, I get it! Thanks! I think you just made a small mistake in your explanation, its supposed to be x = .4 + .01x

I was clueless on how to do this but I tried to rationalize it in my head and because the final pH was so acidic I would think a much larger amount than 20 mL would be needed but I was way off.