TBR weak acid equation

[GRod18]

---

Members don't see this ad.

Hi, are there any TBR students here who could help answer this question?
I was wondering if the TBR weak acid formula could be used for all scenarios. I am very slow on doing ICE problems so I feel like this equation could save me a lot of time. The derivation part is pretty weird so I just memorized it.

PH = 1/2 Pka - 1/2 log [HA]
Pka b/w 2 and 12
[HA] > Ka

I was just wondering can this be used for all weak acid problems that are within these limitations? If an weak acid problem was outside these boundaries what would it mean? Would I have to use ICE instead?



Thank you for all your help!


On a side note: i'm working on kaplan topicals and came across this problem i'm pretty sure I'm not making calculation errors but I'm just not getting what the anwer says I'm wondering if thats an errata on their part.

If the Ka of HClO is 33 × 10^–8, the Kb value of
ClO– is:

A . 3 × 10^–7
B . 3.3 × 10^-8
C . 3.3 × 10^-6
D . 14

they said answer is A, I'm pretty sure it is B

Based on Ka x Kb = 1 x 10^-14

 

[MCAT guy]

Hi, are there any TBR students here who could help answer this question?
I was wondering if the TBR weak acid formula could be used for all scenarios. I am very slow on doing ICE problems so I feel like this equation could save me a lot of time. The derivation part is pretty weird so I just memorized it.

PH = 1/2 Pka - 1/2 log [HA]
Pka b/w 2 and 12
[HA] > Ka

I was just wondering can this be used for all weak acid problems that are within these limitations? If an weak acid problem was outside these boundaries what would it mean? Would I have to use ICE instead?



Thank you for all your help!


On a side note: i'm working on kaplan topicals and came across this problem i'm pretty sure I'm not making calculation errors but I'm just not getting what the anwer says I'm wondering if thats an errata on their part.

If the Ka of HClO is 33 × 10^–8, the Kb value of
ClO– is:

A . 3 × 10^–7
B . 3.3 × 10^-8
C . 3.3 × 10^-6
D . 14

they said answer is A, I'm pretty sure it is B

Based on Ka x Kb = 1 x 10^-14

B . 3.3 × 10^-8

is right. That equation is Ka equation just reworked, so yeah you can use it anytime. The assumption is that the concentration of the weak acid is so large as to make the X (change/dissociation) insignificant.

If you look over the derivation again it really is just reworking the Ka = (h30)^2/(HA) ... since the A- is equal to h30 you can just say h30^2 in the numerator.

After that it is just logs and bring down square roots, remember if you have a log square rooted, you can bring down 1/2 in front of it.

 

[GRod18]

B . 3.3 × 10^-8

is right. That equation is Ka equation just reworked, so yeah you can use it anytime. The assumption is that the concentration of the weak acid is so large as to make the X (change/dissociation) insignificant.

If you look over the derivation again it really is just reworking the Ka = (h30)^2/(HA) ... since the A- is equal to h30 you can just say h30^2 in the numerator.

After that it is just logs and bring down square roots, remember if you have a log square rooted, you can bring down 1/2 in front of it.

hey thanks a lot man, working weak acid problems with this equation is so easy I love it, only thing gotta remember is some log values lol..