hi everyone....has anyone seen acid base questions, like ph=pka+log....or the one above or anything?
im worried about -log stuff...any tips? anything.
messenger.
im worried about -log stuff...any tips? anything.
messenger.
-
New SDN Article: “The Boards”: USMLE Overview
messenger said:hi everyone....has anyone seen acid base questions, like ph=pka+log....or the one above or anything?
im worried about -log stuff...any tips? anything.
messenger.
use n-log x and you can sort of estimate stuff out.
Don't know if this helps or not... but the - log stuff can sometimes be solved with this trick.
if something has a H+ of 1.0 x 10^-4, you know you just take the power of it and thats yours pH. If its something bigger than 1 and less than 10, then you take the power, subtract it by one, and whatever you are multiplying just subtract that number by 10.
like 4.0 x 10^-8... subtract that 8 by 1 so you get 7. something... subtract 4 from 10 and you get 6. so your answer is pH = 7.6, or something close enough to that.
if you are given the pH, you do the opposite. so if the pH is 11.4, you know the exponent will be 10^-12, and the 4 part subtracted by 10 is going to be 6. so your H+ should be 6 x 10^-12.
works the same for [OH-], pOH, Ka's and pKa's. if the Ka is 1.9 x 10^-6, then the pKa is going to be 5.81
hope that helps.
if something has a H+ of 1.0 x 10^-4, you know you just take the power of it and thats yours pH. If its something bigger than 1 and less than 10, then you take the power, subtract it by one, and whatever you are multiplying just subtract that number by 10.
like 4.0 x 10^-8... subtract that 8 by 1 so you get 7. something... subtract 4 from 10 and you get 6. so your answer is pH = 7.6, or something close enough to that.
if you are given the pH, you do the opposite. so if the pH is 11.4, you know the exponent will be 10^-12, and the 4 part subtracted by 10 is going to be 6. so your H+ should be 6 x 10^-12.
works the same for [OH-], pOH, Ka's and pKa's. if the Ka is 1.9 x 10^-6, then the pKa is going to be 5.81
hope that helps.
- Thread Starter
- #5
very sweet, thank you for the tips!Pilot Mike said:
great short cuts, that i hadnt seen!
No prob.
Don't know if this helps or not as well, but here's another quick trick for things like Ksp.
If you are given the Ksp and need to find the molar solubility, as long as its not a common ion effect its almost always these things.
x^2
4x^3
27x^4
so on...
For example... Ksp of AgCl is 4.0 x 10^-8. whats molar sol of [Cl-]?
you know AgCl --> Ag+ + Cl-. so Ksp =[x][x] or 4.0 x 10^-8 = x^2. so x or [Cl-] = 2 x 10^-4. if its like Mg(OH)2 then its just [x][2x]^2 or 4x^3.
Most of the time its gonna be one of these things, so you can throw that ICE table away.
Good luck guys....
Don't know if this helps or not as well, but here's another quick trick for things like Ksp.
If you are given the Ksp and need to find the molar solubility, as long as its not a common ion effect its almost always these things.
x^2
4x^3
27x^4
so on...
For example... Ksp of AgCl is 4.0 x 10^-8. whats molar sol of [Cl-]?
you know AgCl --> Ag+ + Cl-. so Ksp =[x][x] or 4.0 x 10^-8 = x^2. so x or [Cl-] = 2 x 10^-4. if its like Mg(OH)2 then its just [x][2x]^2 or 4x^3.
Most of the time its gonna be one of these things, so you can throw that ICE table away.
Good luck guys....
since its 1.9 x 10^-6, subtract 6-1 to get 5. thats the first part. for the 1.9 part, subtract it from 10, so 10-1.9 is 8.1. so the 5 goes first, then the 8.1 follows after... turns into 5.81.
Pilot Mike said:
I dont truly understand this method! i would love to understnad it, because it seems like a nice short cut! can you PM me, or maybe write where the exponenets and the coefficients are coming from? i dont understand where the 4x, or the 17x comes from as well as the exponents.
Know this. The log(1) = 0, log(3.16..) = 0.5 and log(10) = 1. So if you have a question that is what is pH of "N x 10^-M", the general rule is M-log(N) From here you can estimate, if the ratio given in the log funtion is less than 3.16 than the pH will be closer to the exponent. For example, 1.8x10^-2 you know the number you will have to subtract from 2 will be less than 0.5 and the pH will then most likely be really close to 2. So you can estimate what will be the pH. Also if the number is above 3.16 like 8 or 9 you know this will be far from 2 and approaching 1. I hope this helps. Goodluck.
Can someone re-explain this? I think I understand it, but at the same time, I don't.
Ok I seem to understand.
For example... -log6.4x10^-2...
I take 2-1 and I get 1. Okay, so far so good.
However, 10-6.4=3.6.. The answer, however, is 1.2.
Is there a more accurate way to find the number after the decimal?
I'm concerned because what if the numberafter the decimal is over 5 when I subtract it from 10, but the actual aswer is under 5?
If I round up then I'll get the wrong answer because of false calculations.. :/
----------
Also, what's the easiest way to divide decimals? Like... .02/.030
For example... -log6.4x10^-2...
I take 2-1 and I get 1. Okay, so far so good.
However, 10-6.4=3.6.. The answer, however, is 1.2.
Is there a more accurate way to find the number after the decimal?
I'm concerned because what if the numberafter the decimal is over 5 when I subtract it from 10, but the actual aswer is under 5?
If I round up then I'll get the wrong answer because of false calculations.. :/
----------
Also, what's the easiest way to divide decimals? Like... .02/.030
I was in the same situation, I sucked at doing logs in my head. This is from another person on this forum, not me, but this is the method I use and it works everytime:
When you see log problems, it should be instant free points 
Just have to memorize this...
log 2 = .3
log 3 = .5
log 4 = .6
log 5 = .7
log 6 = .8
log 7 = .85
log 8 = .9
log 9 = .95
As an example: -log(3.2x10^-3) **approximate 3.2 as 3
= 3 - log 3.2
= 3 - .5
= 2.5
As another example: -log(5.7 x 10^-7) **approximate 5.7 as 6
= 7 - .8
= 6.2
Another example: -log(8.1 x 10^-5)
= 5 - .9
= 4.1
Another example: -log(6.9 x 10^-2)
= 2 - .85
hope that helps!
wow.. i just saw this thread was from 06 lol
When you see log problems, it should be instant free points 
Just have to memorize this...
log 2 = .3
log 3 = .5
log 4 = .6
log 5 = .7
log 6 = .8
log 7 = .85
log 8 = .9
log 9 = .95
As an example: -log(3.2x10^-3) **approximate 3.2 as 3
= 3 - log 3.2
= 3 - .5
= 2.5
As another example: -log(5.7 x 10^-7) **approximate 5.7 as 6
= 7 - .8
= 6.2
Another example: -log(8.1 x 10^-5)
= 5 - .9
= 4.1
Another example: -log(6.9 x 10^-2)
= 2 - .85
hope that helps!
wow.. i just saw this thread was from 06 lol