Acid/base antilog question

[hqt331]

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If the pKa1 value of H2CO3 is 6.4, what is the pH of a 10^-3 M solution of this acid?

3.0, 4.7, 6.3, 9.3

This question involves finding the antilog of 6.4 How do you do this to get close enough to finding the answer to this problem?

 

[loveoforganic]

pH of a weak acid can be approximated by pH = pka/2 - log[HA]/2

so 6.4/2 - log(10^-3)/2 = 3.2 - (-1.5) = 4.7

 

[sv3]

pH of a weak acid can be approximated by pH = pka/2 - log[HA]/2

so 6.4/2 - log(10^-3)/2 = 3.2 - (-1.5) = 4.7

is this the BR trick i've heard about that allows you to skip ICE tables?

 

[loveoforganic]

yes, avoid the ice tables, they always lead to the devil quadratic >.<

 

[thebillsfan]

LOO, are there situations where you cant use this br trick? ive never heard of it before

 

[loveoforganic]

It's only applicable to weak acids. Other than that, I don't think there will be any cases ON THE MCAT where it won't be applicable.

 

[lDanny]

the rule for that shortcut is that the pka has to be between 2 and 12, and the [ ] > Ka. If I remember correctly...

 

[loveoforganic]

Ah, that sounds familiar 🙂

 

[thebillsfan]

well then, with all due respect, why would that be helpful at all? youd either be given the Ka or pKa, but not both, so you'd have to figure out the other just to sure the rule applies. with Ka to pKa this is easy, but pKa to Ka is a lot more difficult...

 

[loveoforganic]

Then don't use it, enjoy your ice table and quadratic equation, lol

 

[thebillsfan]

haha, fair enough. i'll avoid the quadratic

 

[wanderer]

yes, avoid the ice tables, they always lead to the devil quadratic >.<

Not really. If you use the ICE table for an acid disassociation (which I do out of habit, even though it is inefficient) you'll get some number like [HA]= .2-x, but in reality x is so small that you just use .2. If x is ever large enough that you would need to use the quadratic equation, then your shortcut formula won't work either. In terms of efficiency your equation is better.



BTW I had to use the quadratic equation on my MCAT for an unrelated problem (in BS of all places lol).

 

[ezsanche]

Not really. If you use the ICE table for an acid disassociation (which I do out of habit, even though it is inefficient) you'll get some number like [HA]= .2-x, but in reality x is so small that you just use .2. If x is ever large enough that you would need to use the quadratic equation, then your shortcut formula won't work either. In terms of efficiency your equation is better.



BTW I had to use the quadratic equation on my MCAT for an unrelated problem (in BS of all places lol).

if you are using the quadratic equation on your mcat then you are doing something wrong.

 

[wanderer]

if you are using the quadratic equation on your mcat then you are doing something wrong.

I didn't do anything wrong. The set up was in the passage.