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G chem.
Started by Danny289
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assuming this is a weak monoprotic acid, the concentration of H+ in the solution is going to be equal to the concentration of the conjugate base. Therefore, it should be 10^-2.7 which is close to 2 * 10^-3.
I would pick B.
you mean C harry, 2x10^-3 is answer choice C...
I actually agree with you, because the dissociation assuming its a monoprotic acid like you said will make up H+ and the conjugate base, therefore the H+ concentration should equal the conjugate base concentration...
you mean C harry, 2x10^-3 is answer choice C...
I actually agree with you, because the dissociation assuming its a monoprotic acid like you said will make up H+ and the conjugate base, therefore the H+ concentration should equal the conjugate base concentration...
Yeah, I meant choice C. Sorry about that. It was just a typo. Danny, you mean the answer is not C?
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you are rightttttttttttttttttttt> king of chem beautiful resoningassuming this is a weak monoprotic acid, the concentration of H+ in the solution is going to be equal to the concentration of the conjugate base. Therefore, it should be 10^-2.7 which is close to 2 * 10^-3.
I would pick C.
B sounds reasonable to me! for the exact same reason Osims said!
HCl -> H+ + Cl-
So if [H+] is around 2 x 10^-3, then [Cl-] = .1M - (2x10^-3)
Pooya, what you are calculating is the concentration of HCl left.
you mean C harry, 2x10^-3 is answer choice C...
I actually agree with you, because the dissociation assuming its a monoprotic acid like you said will make up H+ and the conjugate base, therefore the H+ concentration should equal the conjugate base concentration...
if the PH of the weak acid is 2.7, then the [H+] = 2 x 10^-3M , but I dont understand why that would equal the conjugate base?
Pooya, what you are calculating is the concentration of HCl left.
Whoops, you're right!
you are rightttttttttttttttttttt> king of chem
Lol, what do you mean I'm very close? You mean it is not C yet? I meant 2 * 10^-3 at the begining, But I though it was choice B.
Lol, what do you mean I'm very close? You mean it is not C yet? I meant 2 * 10^-3 at the begining, But I though it was choice B.
I edit my post the answer is C ecatly for the reason you explained no need for calcalution!!!
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if the PH of the weak acid is 2.7, then the [H+] = 2 x 10^-3M , but I dont understand why that would equal the conjugate base?
AHHH... nm. The conjugate base is H30+.
for a weak acid
HA + H20 <---> H30+ + A-
so given pH= 2.7, then H3O+ = 2x10^-3M
if the PH of the weak acid is 2.7, then the [H+] = 2 x 10^-3M , but I dont understand why that would equal the conjugate base?
if x amount of HA dissociates , you will have x amount of H+ and x amount of A- [the conjugate base]. Therefore, they will be equal.
AHHH... nm. The conjugate base is H30+.
for a weak acid
HA + H20 <---> H30+ + A-
so given pH= 2.7, then H3O+ = 2x10^-3M
nope! H3O+ is the same as H+. You can simplify and right it as HA <---> H+ + A-
A- is the conjugate base.
So basically its like this:
HAC ----> AC- + H+
pH= [H+] = 2x10^-3 which would also be the same concentration for AC-
yes bro, you got it. you didn't realy need to calculate. just should know the simple concept.
but if it was a diprotic acid it would still be like this correct:
H2AC ----> HAC- + H+
it would still be the same concentration tho for H+ and HAC-
Yeah, but if you consider this H2AC ---> 2H+ + AC - then it would not. Anyway, that question should have been more specific.
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Harry,would tell me so when do we use Henderson-Hasselbach Equation.I'm very confuse about this subject?if x amount of HA dissociates , you will have x amount of H+ and x amount of A- [the conjugate base]. Therefore, they will be equal.
Harry,would tell me so when do we use Henderson-Hasselbach Equation.I'm very confuse about this subject?
We could have used Henderson-... if we had the pka of the acid.
PH - pka = log A-/HA
This is really simple. if 1 mole of 10 moles of HA dissociates, you will have 1 mole of H+ and 1 mol of A-. concentration of H+ and A- will be the same.
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