Relating ph and pKa

[Vindication]

If pH + pOH = 14 and pka and pkb = 14 then can pH + pOH = pKa + pkb?

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[RogueUnicorn]

well yes, since they're both 14?

 

[latrala2300]

pH + pOH does NOT always equal 14--that's only valid at a particular temp.

 

[bullsfan1986]

good point. pH + POH = 14 is only valid at 25C.

 

[Rabolisk]

pH + pOH = pKa + pKb always. At 25 degrees, they're 14. As temperature goes up, they go down. By they I mean pH + pOH or pKa + pKb.

 

[MCAT guy]

pH + pOH does NOT always equal 14--that's only valid at a particular temp.

True.

This is the type of stuff I learned in BR chem, stuff that is completely useless in MCAT prep.

The MCAT has questions on pH that ask you... if pH is ___ then what is the concentration of H+ or OH-?

I learned so much unnecessary chem trivia in my MCAT prep.

 

[ModerateMouse]

True.

I learned so much unnecessary chem trivia in my MCAT prep.

Don't say that: there can always be a passage based on this, and knowing it in advance will help you save time and/or get something right that you might not have otherwise. 😉

Sure, the chance is small, but it's always there.

 

[latrala2300]

pH + pOH = pKa + pKb always. At 25 degrees, they're 14. As temperature goes up, they go down. By they I mean pH + pOH or pKa + pKb.

this seems correct; the "proof" for this is probably more involved than setting them equal to 14. if i had to venture a guess, i'd say the henderson-hasselbach is used. somehow.

the first thing i thought of when i read your question was that pH=pKa+log (base)/(acid). i'm guessing that's enough for this grueling beast of an exam, but what do i know...

 

[SimplyDave]

Sorry my initial explanation was incorrect.

The answer is yes because all you're doing is relating hydrogen ion concentrations when you go from pH to pKa. As you know, pH and pOH are simply ways to express the hydronium/hydroxide concentrations in solution while pKa and pKb are related to hydronium/hydroxide concentrations in this way:

Ka x Kb = [H+][A-][OH-][HA]/[HA][A-] = [H+][OH-] = Kw

pKa + pKb = pKw = 14

Now if you want to calculate pH from pKa or vice versa, I think this equation applies:

.pH = ½pKa – ½log [HA] ..
.

 

[ElliottB]

Is the HH equation necessary on the MCAT?

 

[Rabolisk]

Is the HH equation necessary on the MCAT?

Absolutely. It is useful in titrations as well as generally to describe relative concentration of an acid and its conjugate base at a pH.

 

[coyfish]

Yeah just don't forget that the pH =1/2pkA - log [HA] only applies to completely dissociated situations.

HH applies to mixes.

 

[BerkReviewTeach]

Don't say that: there can always be a passage based on this, and knowing it in advance will help you save time and/or get something right that you might not have otherwise. 😉

Sure, the chance is small, but it's always there.

Good point and great attitude. The perfect example is physiogical conditions, where pH + pOH is a little less than 14.

The point that the BR book was making is that when the pH and pOH are the result of autoionization of water, then pH = pOH.

 

[BerkReviewTeach]

Now if you want to calculate pH from pKa or vice versa, I think this equation applies:

.pH = ½pKa – ½log [HA] ..
.

How did you do that really cool ½?

By the way, that equation can be made easier by noting the following:

pH = ½pKa – ½log [HA] = ½(pKa – log [HA]) = ½(pKa + pHif HA were to fully dissociate).

The pH is an average of the pKa and the pH if the acid were strong instead of weak.

So a 0.02 M HOCl solution with pKa = 7.5 would be an average of 7.5 and 1.7, which equals 4.6. It's the same equation as before, just easier to use.

What is the pH of 0.05 M HOAc (pKa = 4.74)?
A. 1.30
B. 3.02
C. 4.74
D. 6.04

What is the pH of 0.30 M HF (pKa = 3.2)?
A. 0.50
B. 1.85
C. 3.20
D. 3.70

What is the pH of 0.10 M uric acid (pKa = 3.9)?
A. 1.0
B. 2.5
C. 3.9
D. 9.3

 

[TieuBachHo]

pH = ½pKa – ½log [HA] = ½(pKa – log [HA]) = ½(pKa + pHif HA were to fully dissociate).

If HA were to fully dissociate, wouldn't that mean p[H] = p[HA] assuming that's a monoprotic acid? Is pH = ½pKa + ½p[HA] just simply a mathematic workout of a weak acid that's negligible in dissociation?

 

[sugarbabee0]

Yeah just don't forget that the pH =1/2pkA - log [HA] only applies to completely dissociated situations.

HH applies to mixes.

Doesn't -log[HA] mean pH? Then how does that equation make sense?

 

[dextor2003]

How did you do that really cool ½?

By the way, that equation can be made easier by noting the following:

pH = ½pKa – ½log [HA] = ½(pKa – log [HA]) = ½(pKa + pHif HA were to fully dissociate).

The pH is an average of the pKa and the pH if the acid were strong instead of weak.

So a 0.02 M HOCl solution with pKa = 7.5 would be an average of 7.5 and 1.7, which equals 4.6. It's the same equation as before, just easier to use.

how would you get the 1.7 without having a calculator on the real thing ?
(im assuming you did -log .02 to get 1.7)

 

[BerkReviewTeach]

If HA were to fully dissociate, wouldn't that mean p[H] = p[HA] assuming that's a monoprotic acid? Is pH = ½pKa + ½p[HA] just simply a mathematic workout of a weak acid that's negligible in dissociation?

Exactly! It's just a math trick to say that the pH of a weak acid is greater than the pH of a strong monoprotic acid with same molarity but less than the pKa. The math tells us that the pH of the weak acid is approximately an average of its pKa and the pH of a strong acid of the same molarity.

Doesn't -log[HA] mean pH? Then how does that equation make sense?

I think it's simply a typo by coy, where it should read like SimplyDave typed: pH = ½pKa – ½log [HA]

how would you get the 1.7 without having a calculator on the real thing?
(im assuming you did -log .02 to get 1.7)

Another math trick:

-log (2 x 10exp-2) = - (log 2 + log 10exp-2) = - (log 2 + (-2)) = 2 - log 2

This means we can skip this math in the future.

-log (4 x 10exp-3) = 3 - log 4
-log (6 x 10exp-1) = 1 - log 6
-log (5.1 x 10exp-5) = 5 - log 5.1
-log (7.3 x 10exp-4) = 4 - log 7.3
etc...

Back to the question:

-log (2 x 10exp-2) = 2 - log 2 = 2 - 0.3 = 1.7.

It can be done without a calculator. If you happen to have the BR book, it's explained much better than I did here.