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- Thread starter Danny289
- Start date

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?

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!!!

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.

Harry,would tell me so when do we use Henderson-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-HasselbachEquation.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.