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Danny289

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  1. Pre-Dental
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If the pH of a 0.10 M weak solution is 2.70, what is the concentration of the conjugate base?

A. 0.10 M
B. 2.0 x 10^-2 M
C. 2.0 x 10^-3 M
D. 2.0 x 10^-4 M

what you think? and why
 
hmm i think how you do this is you take the pH= 2.7 = about 2 x10^-3 (H+ concentration aka Ka) and then subtract that from .1M and you should get something x10^-2 so i would go with B
 
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 C.
 
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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 all are wrong! pure conceptual! rethink you are in right track but.....
 
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)
 
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?
 
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 C.
you are rightttttttttttttttttttt> king of chem beautiful resoning
 
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.
 
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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?
 
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!!!
 
😀😀OH ok. I'm glad we are finally done with this😀😀 Bro, once you said I'm very close, my heart stopped.
 
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.
 
So basically its like this:

HAC ----> AC- + H+

pH= [H+] = 2x10^-3 which would also be the same concentration for AC-
 
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.
 
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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-
 
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.
 
yah that .1M weak solution threw me off. If only if they asked what's the concentration of H+. Bam C.
 
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.
 
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?
 
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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