What volume of HCl was added if 20ml of 1M NaOH is titrated with 1M HCl to produce a pH=2
?
you can do this one easily by doing this:
set up a rxn first... H+ + OH- -----> h2o
1 mole of oh- reacts with 1 mole of h+ to make 1 marginal mol of h2o.
since they tell you ur starting off with 20 ml of 1 M NaOH, you gotta calculate how many moles of OH- that is. so, 20 ml, which is .02 L x 1M OH- gives you .02 moles OH-. Since the OH- reacts in a one-to-one ratio with H+, there will also need to be .02 moles of H+ to neutralize the rxn, or perhaps, cancel the reacants out to form the additional water.
anyways, now we know we have to add 20 ml of H+. But also, we're told we gotta get that ph down to 2. Since ph = log[H+], we plug in the values and we get [H+] = to .01
So our goal, basically is to get a final [H+] of .01
Since we already need .02 moles (what we calculated earlier) to neutralize the original OH- present, that would leave us with 0 [H+] after the neutralization.
And since we know we need a final concentration of .01, we add .01 moles more moles to get that.
In total, .02 moles H+ to start, + .01 additional moles of H+ means we have to use .03 moles of H+ to get that ph of 2, given the aforementioned OH situation.
Anywho, since they tell us we should use 1M concentrated H+, we multiply that by .03 moles ot get .03 L of H+. Essentially 30 ml.