Stoichiometry problem- confused.
Started by virtualmaster999
I'm curious what the answer is...I would have done M1V1 = M2V2 and solved, which gives me A.
The answer is C but I find it strange that u can't use the formula
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Saying M1V1=M2V2 is pretty much saying that the mols of KOH and H2SO4 are equal at neutralization.
If you take an example of 1mol of KOH and 1 mol of H2SO4, you can see that there is only 1mol of OH-, but it needs to combine with 2 mols of H+ in order to neutralize.
For these sort of neutralization acid/base problem, it's more appropriate to use normality instead. N1V1=N2V2.
This would give you the normality of H2SO4. But they are asking for molarity, so you would have to convert it to molarity.
So you can just think of it logically with mv=mv, or you can do normality-molarity conversions with nv=nv.
If you take an example of 1mol of KOH and 1 mol of H2SO4, you can see that there is only 1mol of OH-, but it needs to combine with 2 mols of H+ in order to neutralize.
For these sort of neutralization acid/base problem, it's more appropriate to use normality instead. N1V1=N2V2.
This would give you the normality of H2SO4. But they are asking for molarity, so you would have to convert it to molarity.
So you can just think of it logically with mv=mv, or you can do normality-molarity conversions with nv=nv.
Why is the normality of H2SO4 = 2? H2SO4 is a strong diprotic acid, but I thought only the first H+ completely dissociates. After that point the second H is weakly acidic and only dissociates to a small extent, if my understanding is correct. For titrations do we just count the number of H's it contributes regardless?
I remember thinking the exact same thing, and I asked my teacher about it long time ago, but unfortunately, I don't remember the answer. In short he said it's a little more complicated than that. I think you are safe to just count the number of H's for these neutralization problems.
Yeah that's basically the formula I have always used. n = # of moles of Hydrogen ions (or OH), if you had for example Ba(OH)2
n1: KOH = 1 mole of OH-
v1: 35.4ml = .0354L
m1: .125M
n2: H2SO4 = 2 Moles H+
v1: 50ml = .05L
m1: X
(1)(.0354)(.125)=(2)(.05)(X)
X=.0443
n1: KOH = 1 mole of OH-
v1: 35.4ml = .0354L
m1: .125M
n2: H2SO4 = 2 Moles H+
v1: 50ml = .05L
m1: X
(1)(.0354)(.125)=(2)(.05)(X)
X=.0443
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Alternative solution. First find the moles of sulfuric acid needed to reach neutralization.
35.4mL KOH (0.125 mol/10000mL KOH)(1 mol H2SO4/2 mol KOH) = 0.0022125 mol H2SO4
Then find the molarity 0.0022125 mol/0.050L = 0.04425 M H2SO4
When dealing with acids and bases, use this formula
nMaVa = nMbVb, taking account the moles of protons or OH ions present
35.4mL KOH (0.125 mol/10000mL KOH)(1 mol H2SO4/2 mol KOH) = 0.0022125 mol H2SO4
Then find the molarity 0.0022125 mol/0.050L = 0.04425 M H2SO4
When dealing with acids and bases, use this formula
nMaVa = nMbVb, taking account the moles of protons or OH ions present
Ahhh I forgot about the n in the equation... Thanks everyone!
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So would this change our answer if it was asking about pH? Because I just did a chad's video quiz question using H2SO4 and he ignored the 2nd mol of H+, saying what you said about the 2nd H+ being weakly acidic and only dissociating to a small extent. When should we treat it as a weakly acidic H versus treating it as a whole mol? Are there any other areas that could be situational such as this?
It seems that for everything but titrations, we ignore the second H+. Pretty confusing!
it isn't confusing if we look at differently : we are dealing with a strong acid and a strong base here , so at the titration point , the moles of acid = moles of base. using the 35.4 ml and .125 M , we can easily calculate the m ( mole ) for KOH which is 0.04425 ,,, so you would assume that we should have 0.4425 mole of H2SO4 ..BUT here we have 2 moles of KOH to begin with so we divide the 0.04425 by 2 and .0221 is the number of mole for H2SO4... now 0.221/0.05 = .442 M for H2so4
I think it's just confusing to us because they usually don't treat the 2nd mol of H as a mol of H equal to a strong acid's. Like in your example KOH makes sense to have 1 mol because it dissociates completely, but they (Chad) say that the 2nd mol in H2SO4 does not dissociate completely (which wouldn't be the same as KOH). We understand the math behind it just not so much the contradiction between 2nd mol dissociating completely in stoich problems and 2nd mol being treated as a weak acid (ie not dissociating completely) in pH problems.
I think we are mistaking the dilution with titration here ! lets say theoretically , 1 mole KOH neutralize 1 mole of the H2SO4 in question . we are left with hso4- and another mole of KOH which is a classic case of SB/WA titration ! and from your post I assume you have reviewed chad's videos , and if you recall, he mentions that when dealing with SA/WB or WA/SB , number of moles for both SB and WA are still the same but the M for the WA is higher than the M of the SB. that's why when you set up the up the n1m1v1=n2m2v2 equation , you put the 2 for h2so4 and compensate with M and volume for titration to complete . otherwise, you could never neutralize HSO4- !
oh shoot you're right. I completely forgot about Chad mentioning that moles are the same for SA/WB or WA/SB! I just checked my notes and he said, "Whether you titrate a strong acid or weak acid, if you have the same number of moles of strong acid or weak acid, it will take the SAME number of moles of NaOH to titrate it." I even put it in starts lol. And ya it makes sense about the HSO4- still being there so it's another weak acid that needs a whole mol to titrate. But can you explain the M for WA higher than M for SB part real quick?
1 mole of KOH dissociate 100% while 1 mole of HSO4- doesnt , so you need more concentrated HSo4- , hence a higher molarity.
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