H2 chemistry ionic equilibria

limywv SG

Eyelessz SG

Hi, is the answer : pH = 3.8959 ~ 3.90

If yes, i will post the solution. If not, wait for other forumers to come help out.

UltimaOnline SG

the hydrogen phthlate ion, HC8H4O4- is a weak monobasic acid. A soln was prepared by dissolving 0.100 moles of HC4H4O4K in 250cm3 of 0.200mol/dm3 C8H4O4K2.

Ka of HC8H4O4- = 3.23×10^-7 mol/dm3

Calculate the pH of the solution formed?

Oh man this qn is killing me…help!

Since Ka > Kb and molarity of conjugate acid > molarity of conjugate base, the acid will dissociate to generate more aqueous protons.

Since this is a buffer system, the ICE table isn't necessary (since there are already significant amounts of the species in both conjugate acid and conjugate base forms, hence approximations will suffice).

Calculate the no. of moles and hence molarities of both conjugate acid and conjugate base forms in the solution.

Plug these values into the Ka expression, and make molarity of protons [H+] the subject. Therefore, find pH.

Reasonable answer check : pH should be only slightly acidic (ie. approximately pH 6), since this is a very weak (see Ka value) monoprotic acid buffer solution.

Eyelessz SG

Hi, how did you see a buffer in the solution? i was under the impression that HC4H4OK and C8H4O4K2 react together (albeit weird) to give the hydrogen phthlate ion, and used limiting reagent concepts + ICE table to derive my wrong answer.

And which is the corresponding conjugate base (Salt) to be used in the formula of pH = -log pka + lg [Salt/Acid]

Could you explain to me? Thanks.

dkcx SG

And which is the corresponding conjugate base (Salt) to be used in the formula of pH = -log pka + lg [Salt/Acid]

Could you explain to me? Thanks.

The 2 salts are relatively similar so its abit like the HPO42- and H2PO4- buffers.

Salt should be C8H4O4K2 and acid should be HC4H4OK since the H at the front is there to signify and acidic H+ if i not wrong and it was also mentioned in that question that HC4H4OK is an acid.

Eyelessz SG

Isn't the hydrogen phthlate ion, HC8H4O4- is a weak monobasic acid instead of HC4H4OK ? lol

Unless T.S wrote the formulas wrongly? Because i don't see a compound with 4 carbon less exhibiting a conj acid/ conj base relationship

dkcx SG

Isn't the hydrogen phthlate ion, HC8H4O4- is a weak monobasic acid instead of HC4H4OK ? lol

Unless T.S wrote the formulas wrongly? Because i don't see a compound with 4 carbon less exhibiting a conj acid/ conj base relationship

Opps, see wrong but then an acid would have the acidic proton written outside the forumla as with both cases of the acid and the salt is the 1 without the H in front unless TS type wrongly of cos

limywv SG

dkcx SG

the answer for this is pH 6.19.

I tried using pH = pKa + lg([salt]/[acid]).

I got the answer but dont understand at all...

What do u not understand?

limywv SG

dkcx SG

I mean like how do I get the [salt] and [acid]. They are from which equations?

Have you learnt about any other buffers like the phosphate buffer?

For acids, its normally the compound with the acidic H+ or the 1 with more H+ if both contains them while the salt will be the other compound in the mixture. That acidic H can be seen when you write out the equation. Theres a reason y the H is outside and not together with the H4 inside the compound.

Eyelessz SG

limywv, is it HC4H4O4K or HC8H4O4K?

limywv SG

aran SG

the question is wrong. it should be HC8H4O4K

equation would be

HC8H4O4 (-) <--> C8H4O4 (2-) + H30 (+)

[HC8H4O4 (-)] = 0.100 x 1000/250 = 0.400

[C8H4O4 (2-)] = 0.200 (given)

3.23 x 10^-7 = (0.200)[H+]/(0.400)

after making [H+] the subject and taking -log, pH is 6.19