PH values

Methanoic acid, HCOOH, has a Ka value of 1.58 x 10-4 moldm-3. What ratio of methanoic acid and sodium methanoate would give a buffer of pH = 4?

Reply 1

charco

Study Forum Helper

18

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Methanoic acid, HCOOH, has a Ka value of 1.58 x 10-4 moldm-3. What ratio of methanoic acid and sodium methanoate would give a buffer of pH = 4?

Can someone please explain the above question to me and how to answer it? I have it sort of worked out for me and the answer (1:1.58) but I really don't understand it. Help Please. :confused:

All buffer questions are based on the acid dissociation equation:

ka = [H+][A-]/[HA]

you want a specific pH (4) so from this you can get the [H+] because:

pH = -log₁₀[H+]

therefore [H+] = 10^-pH

You also have the value for the ka of the acid.

So all you have to do is rearrange the equation to get the ratio of the acid and the salt:

ka = [H+][A-]/[HA]

therefore [HA]/[A-] = [H+]/ka

Reply 2

interact

OP

15

charco

All buffer questions are based on the acid dissociation equation:

ka = [H+][A-]/[HA]

you want a specific pH (4) so from this you can get the [H+] because:

pH = -log₁₀[H+]

therefore [H+] = 10^-pH

You also have the value for the ka of the acid.

So all you have to do is rearrange the equation to get the ratio of the acid and the salt:

ka = [H+][A-]/[HA]

therefore [HA]/[A-] = [H+]/ka

Can you please put some values in because my tutor showed me a trial and method method. I've tried rearranging the equation but I get the a value for the ratio of both the acid and salt..

OP

Anyone?

Ka = 1.58 x 10^-4 and [H+] = 1 x 10^-4

so [A-]/[HA] = 1.58 (1.58 x 10^-4/1 x 10^-4)

finally [A-]:[HA] is 1.58:1

:smile:

Reply 5

hp707

10

Original post by EierVonSatan

Ka = 1.58 x 10^-4 and [H+] = 1 x 10^-4

so [A-]/[HA] = 1.58 (1.58 x 10^-4/1 x 10^-4)

finally [A-]:[HA] is 1.58:1

:smile:

Bit late but could someone please explain how you're rearranging to get A/HA? Can;t for the life of me figure it out, im getting ka X ha= a- and h+ but nothing more than that?