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Chem question
Started by Zerconia2921
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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.
do you have the answer to this zerc? i want to know if im right, cuz if im not, then i got studying to do.
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Because HCl is strong acid and NaOH is strong base . 20ml of 1M NaOH will neutralize 20mL of 1M HCl . The total volume at pH = 7 is 40mL . However , the final volume has pH=2 . That means there is more H+ than OH- .
For HCl :
CiVi = CfVf
Ci = 1M
Vi= x
Cf = 0.01 (because pH= 2)
Vf= 40 +x
(1M)(x)= (0.01M)(40+x)
x=0.4 ml
Then the volume of HCl needed to add = 20+0.4= 20.4 ml
ya ur right. im too stupid lol
dang, thanks for correctin me man. its so clear now. i forgot to think that the final volume after neutralization is 40, not 20. thanks
