GC question

[BAJackson16]

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What volume of HCL was added if 20 ml of 1M NAOH is titrated with 1 M HCL to produce a ph=2?

 

[sailinx]

This is a little complicated question and I think I have the answer. Take a look,

Given: 20ml of 1M NaOH, find Volume of 1M HCl to make pH = 2 solution.

Lets assume, it requires X ml of HCl to make pH = 2.
It means, at the end [H+] = 10^-2 (from pH= -log[H+])

Because, HCL and NaOH are strong acid and base, therefore, they completely dissociate.

Now, number of moles in 20 ml NaOH = 20 x 10^-3 x 1 mole/L
= 20x10^-3 moles of NaOH
number of moles in X ml of HCl or H+ = X x 10^-3 x 1 mole/L

We know the solution is acidic, therefore the remaining moles of H+ in solution = X x 10^-3 - 20 x 10^-3

Now the molarity of this solution = number of moles/total vol of solution

[H+] = (X - 20) x 10^-3/(X + 20) x 10^-3

and we know for pH = 2, [H+] = 10^-2

therefore,

10^-2 = (X - 20) x 10^-3/(X + 20) x 10^-3

solve for X and it is 20.4 ml

One thing to note, see how quickly solution become so acidic. It would have required 20ml of 1M HCl to neutralize 20 ml of NaOH and pH would have been 7. By just adding another 0.4 ml, the solution pH = 2.

I hope this helps.

Sailinx

 

[jk5177]

Nice solution. (excuse my pun)

 

[sailinx]

Here is another solution that might be less confusing.

I simply used the pH = [H+]/[HA]

so 10^-2 = [H+]/[.040]

[H+] = .040 * 10^-2

[H+] = .004

You then can add that to the volume of HCl needed to neutralize the NaOH, so .020 + .004 = .024mL

I don't see how it can be the solution to the problem. First where are you getting this formula pH=[H+]/[HA] (Remember HCl is a strong acid and NaOH is a strong base. So you can't use the method for weak acid/base titration.) Secondly, I am assuming you really wanted the answer as L (it can't be 0.024mL, it has to be 0.024L or 24mL) but if that is the case, then 24 ml will make the pH = 1 (not 2)

why pH = 1 ?
# of extra moles of H+ = (24 - 20)x 10^-3
Vloume of Sol = (24 + 20) x 10^-3
[H+] = moles/Vol
= 4x10^-3/44x10^-3
[H+] = 0.09

pH= -log[H+] = -log[0.09] = 1

That is why I emphasized at the end of my solution that a small bit (just extra 0.4ml) of HCl would significantly make the solution acid.

 

[bpenly]

Good point. I am sadly mistaken. The closer I get to taking the DAT, the more new formulas I create. Many thanks for poking your finger in my eye, it will help me avoid making my own mistake on the real test. I am removing that last piece of crap to prevent anyone from falling into my error.

In addition, your solution is very nice.