topscore2, chem:68

[jeezy]

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What volume of 1M HCl was added if 20ml of 1M NaOH is titrated to produce a ph of 2?

20mL yields a 40mL soln of .5M Na+Cl- + H2O (ph 7)

ph 2 = -log H
so, H = 10^-2 mol/L

then what?

 

[joonkimdds]

What volume of 1M HCl was added if 20ml of 1M NaOH is titrated to produce a ph of 2?

20mL yields a 40mL soln of .5M Na+Cl- + H2O (ph 7)

ph 2 = -log H
so, H = 10^-2 mol/L

then what?

HCl => H + Cl , so it's an acid
NaOH => Na + OH, so it's a base.
both of them are 1mol(of either acid/base) per liter
so if they have the same amount, they will be neutral since there are
same amount of mol of both acid and base present. (pH=7)

if you know this much, the next step is to make this pH=7 into pH=2.
To do this, you should make this neutral solution into acidic since pH lower
than 7 is acidic and our goal is to make it into pH=2.

the next step is using M1V1=M2V2. Depending on how much you add HCl
the neutral solution will slowly become an acid.(ph7=>6=>5=>4=>3=>2)
let's find out how much we should add HCl.
By the way, pH =2 is same as 10^-2 so the concentration of H(pH) or molarity of HCl is 1M.
(1M of HCl)(x) = (0.01)(40+x)

some people don't understand how we come up with (1M of HCl)(x) = (0.01)(40+x) so let me explain it for you.
When people think M1V1=M2V2, they usually think that before = after, and get confused.
This isn't about before = after.
This is about (depends on how much you put acid) = (the volume and molarity of the solution changes)


I went over the same problem like 10 times last week so i believe x is 0.4. I even memorized the answer 😀
so if the result solution has total of 40.4 mL and the question asks you how much HCl is added from 20mL of NaOH,
you subtract 20 from 40.4 and the answer is (20.4 mL of HCl is added to make the neutral into pH=2)

 

[jeezy]

the next step is using M1V1=M2V2.

ooooooh. thanks. that was tricky