TBR: Henderson-Hasselbach equations clarification

[lostnconfused]

TBR has 2 equations listed for finding the pH.

One is TBR's short-cut equation: pH=1/2pKa - 1/2log[HA]
And the other the Henderson-Hasselbach equation: pH=pKa+log([A-][HA])

Can somebody please clarify when we use one over the other? Everything made perfect sense when I was doing practice problems from TBR but when I turned to AAMC questions, I got confused...

For example, for passage 9 in AAMC's chemistry question pack, question 47 asks "Which of the following mixtures, with each component present at a concentration of 0.1M, has a pH closest to 7?" The answer is HClO (aq) and NaClO (aq), and you are given the Ka of HClO as 3.2*10^-8.


I understand the solution and how the Henderson-Hasselbach equation gives you the correct answer. However I don't understand why we cannot use TBR's equation in this case. (Using TBR equation would give you roughly pH~ 1/2pKA+0.5, which is much too small since we are dividing the pKa by 2.)

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[OrdinaryDO]

http://forums.studentdoctor.net/threads/tbr-chem-determining-ph-for-weak-reagents-shortcut.643012/

 

[BerkReviewTeach]

TBR has 2 equations listed for finding the pH.

One is TBR's short-cut equation: pH=1/2pKa - 1/2log[HA]
And the other the Henderson-Hasselbach equation: pH=pKa+log([A-][HA])

Can somebody please clarify when we use one over the other? Everything made perfect sense when I was doing practice problems from TBR but when I turned to AAMC questions, I got confused...

For example, for passage 9 in AAMC's chemistry question pack, question 47 asks "Which of the following mixtures, with each component present at a concentration of 0.1M, has a pH closest to 7?" The answer is HClO (aq) and NaClO (aq), and you are given the Ka of HClO as 3.2*10^-8.


I understand the solution and how the Henderson-Hasselbach equation gives you the correct answer. However I don't understand why we cannot use TBR's equation in this case. (Using TBR equation would give you roughly pH~ 1/2pKA+0.5, which is much too small since we are dividing the pKa by 2.)

The BR shortcut equation is for a weak acid solution, which is why there is only an [HA] in the equation. On the other hand, the HH equation is for a buffer, which is why there is both an [HA] term and a [A-] term in the HH equation.

For the question you are asking about, you have HClO (a weak acid) and ClO- (its weak conjugate base). That results in a buffer, so you need the HH equation. Because [HA] = [A-], the log of [A-]/[HA] = log 1 = 0, so the pH will be equal to pKa. This is where another BR shortcut is handy. For getting pH from [H+], pOH from [OH-], pKa from Ka, or pKb from Kb, you take -(power of 10) - log (unit number). If Ka is 3.2 x 10^-8, then its pKa is -(-8) - log 3.2 = 8 - log 3.2 ≈ 8 - 0.50 = 7.50.

The section on pH is BR gives by far the best strategies and tricks anywhere, so you should make sure to do everything in that section. If you do, you may never miss another acid-base question.

 

[tturchi51]

weak acid pH = sqrt (Ka*HA)

weak base pOH = sqrt (Kb*A-)

 

[BerkReviewTeach]

weak acid pH = sqrt (Ka*HA)

weak base pOH = sqrt (Kb*A-)

I think what you might have meant is: [H+] for a weak acid is sqrt (Ka x [HA]) and [OH-] for a weak base is sqrt (Kb x [A-])

The shortcut equations (version 1.0) are:

weak acid pH = 1/2 pKa - 1/2 log [HA] and weak base pOH = 1/2 pKb - 1/2 log [A-]​

Version 2.0 is a trick taught in the titration curves lecture (not too much different, but more useful).