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conjugate pairs - a level chemistry

For this question I got CH3COO- and HCOO-
But idk which is the acid or base because they're both negative
Screenshot 2021-02-17 at 15.15.19.png269.4KB

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Original post by dnejfn
For this question I got CH3COO- and HCOO-
But idk which is the acid or base because they're both negative


Think about carefully again. What substance donates a proton and what a substance accepts it?
Original post by Kallisto
Think about carefully again. What substance donates a proton and what a substance accepts it?

an acid donates a proton and a base accepts a proton.
Original post by dnejfn
an acid donates a proton and a base accepts a proton.

Yep and this is the obvious thing you can distinguish an acid from a base in a conjugated acid-base reaction.

But I realized too late that two acids are added with each other. :s-smilie: Normally it can't work, as there is just two proton donators, but no acceptors.
Original post by dnejfn
an acid donates a proton and a base accepts a proton.

... and in this case you use the ka values of the acids to decide which becomes a proton donor and which accepts the proton.
Original post by charco
... and in this case you use the ka values of the acids to decide which becomes a proton donor and which accepts the proton.

woulc ethanoic acid be Acid 1 because it has a lower ka value?
Original post by dnejfn
woulc ethanoic acid be Acid 1 because it has a lower ka value?

If it has a lower ka value it is a weaker acid!
Original post by charco
If it has a lower ka value it is a weaker acid!


how do i use the ka values to see which is an acid or base?
Original post by dnejfn
how do i use the ka values to see which is an acid or base?

The highest ka value is the most acidic and will protonate the other molecule.
Original post by charco
how do i use the ka values to see which is an acid or base?

As mentioned, you look at the Ka values. The lower the Ka value, the weaker the acid.

So if you have 2 Ka values, whichever is the lower will act as the base. While the higher one will act as an acid.
Original post by qwert7890
As mentioned, you look at the Ka values. The lower the Ka value, the weaker the acid.

So if you have 2 Ka values, whichever is the lower will act as the base. While the higher one will act as an acid.

so HCOOH = A1

CH3COOH = B1
CH3COO^- = B2
HCOO^- = A2
Original post by dnejfn
so HCOOH = A1

CH3COOH = B1
CH3COO^- = B2
HCOO^- = A2

From what I can see, yes.
Original post by dnejfn
so HCOOH = A1

CH3COOH = B1
CH3COO^- = B2
HCOO^- = A2


No.

HCOOH loses a proton and becomes the conjugate base
CH3COOH GAINS a proton and becomes the conjugate acid
Original post by charco
No.

HCOOH loses a proton and becomes the conjugate base
CH3COOH GAINS a proton and becomes the conjugate acid

But the other person said a lower ka value will act as a base...
Original post by charco
No.

HCOOH loses a proton and becomes the conjugate base

How?

HCOOH = acid
CH3COOH = base
HCOO- = conjugate base
CH3COO- = conjugate acid
(edited 3 years ago)
Original post by dnejfn
But the other person said a lower ka value will act as a base...

and the other person is quite correct.... as I also said, the highest ka behaves as the Brønsted Lowry acid

You have to shake loose the Arrhenius acid concept and embrace Brønsted Lowry.

In this equilibrium the ethanoic acid is made to behave as a BASE and accepts a proton to become the conjugate acid.
Original post by charco
You have to shake loose the Arrhenius acid concept and embrace Brønsted Lowry.

:rofl:

Arrhenius, Brønsted Lowry, Lewis... these scientists really need to get their handle on their concepts!! Also just curious, I assume here you're looking at the reaction from reverse?

Original post by charco
No.

HCOOH loses a proton and becomes the conjugate base
CH3COOH GAINS a proton and becomes the conjugate acid
Original post by charco
and the other person is quite correct.... As i also said, the highest ka behaves as the brønsted lowry acid

you have to shake loose the arrhenius acid concept and embrace brønsted lowry.

In this equilibrium the ethanoic acid is made to behave as a base and accepts a proton to become the conjugate acid.

hcooh = a1

ch3cooh = b2
ch3coo- = a2
hcoo- = b1
Original post by dnejfn
hcooh = a1

ch3cooh = b2
ch3coo- = a2
hcoo- = b1

You're just not reading my answers:

BL_example.png
Original post by charco
You're just not reading my answers:

BL_example.png

20210219_131334.jpg

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