Destroyer pH log functions???

[Walleye]

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I understand things like pKa=-log Ka and how to calculate the pH with pOH=-log, etc. But can anyone help with estimating or another way to calculate log or -log? Several pH problems in Destroyer involve this type of function (such as #117). Before Destroyer, the only way I ever really did this kind of prob was with a calculator...so ANY tips would help out a great deal.
Side question: Has anyone ever seen a pH question on the actual DAT using -log functions?

Please Help!!!

 

[PooyaH]

You mean estimating what is the pH of or pOH of 8.6x10^-3M without using a calculartor?

 

[Walleye]

Yeah...that type of question.

 

[Tina324]

yup, i think that's what he means. i would like to know this too! anyone?

 

[PooyaH]

You need to know that pH scale goes up and down by a factor of 10. So when you have 1x10^-3 or 1x10^-n you should know right away that the pH is 3 or n respectively! Now for 8.6x10^-3 since we know the pH goes up and down by a factor of 10, and 8.6 is closer to 10 than to 1 so our pH should be closer to 2 than to 3. So I guess the pH should be around 2.2.

Now, let's look at 3.5x10^-5M:

we know our pH should be around 5. But, 3.5 is closer to 1 than to 10, so our actual pH should be closer to 5 than to 4, probabely 4.7!

Hope it makes sense!

 

[Tina324]

pooyah..u r my hero..thanks man

 

[PooyaH]

pooyah..u r my hero..thanks man

haha, anytime

 

[Xtian]

here's a calc someone showed me and it works

ie: 3.5x10^-5M

so you'd say 5-0.35=4.65

 

[Walleye]

Thanks pooyah! Great answer.
You way is great too Xtian.

 

[doc3232]

here's a calc someone showed me and it works

ie: 3.5x10^-5M

so you'd say 5-0.35=4.65

Well, not exactly, the best way to do it is to memorize them
Just remember that 3x10^-5 will definitely be between 4 and 5 and from there, you just need to remember that in this case it will be 4.5
If you see a 3 estimate it to be in the middle.
As previous poster said, the closer you get to 10 it becomes 4 and the closer to 1 gives you 5.
Remember 3 (times) 10^-(some number) will be (some number-1) + 0.5
If that makes sense

 

[mddang]

log(1) = 0
log(2) = ~.3
log(3) = ~.5
log(4) = ~.6
log(5) = ~.7
log(6) = ~.8
log(7) = ~.8
log(8) = ~.9
log(9) = >.9



For the log of 3,4,5, and 6, I just looked at the number that I was taking the log of, added 2, and moved the decimal to the left one place. (Now, I obviously didn't look at it this way and do the stupid little calculations in my head... Just note the trend!)

From then, I could figure that the log(2) is obviously going to be less than the log(3), and the log(7) is going to be greater than log(6).

 

[HoangDDS]

If you just understood this thread, try this other thread for extra practice
http://forums.studentdoctor.net/showthread.php?t=543723

Good luck guys.