TBR and Logs

[Jay2910]

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Hello everyone,

Within the TBR Gen Chem . .I think Acids and Bases chapter, there exists this pretty log table. They say, "make an effort to work through Table 4.7". Does that mean . .to memorize it? understand the patterns? Are you all memorizing logs?

Also, do you think its a good idea to know all amino acid structures?

Thanks,
Jay

 

[EnginrTheFuture]

Don't memorize too many things... especially math related. This is what I would know

Log(2) = ~0.3
log(3) = ~0.5
all log rules, how and when to use them (practice them) since 3 and 2 can make a lot of numbers x in log(x)

fractions of 1/ (1 through 11)
1/6 = .16666
1/7 = .143
1/8 = .125
1/9 = .111
1/10 = obvious ;P
1/11 = .91

sin and cos of 0, 30, 45, 60, 90

 

[altitude]

Don't memorize too many things... especially math related. This is what I would know

Log(2) = ~0.3
log(3) = ~0.5
all log rules, how and when to use them (practice them) since 3 and 2 can make a lot of numbers x in log(x)

fractions of 1/ (1 through 11)
1/6 = .16666
1/7 = .143
1/8 = .125
1/9 = .111
1/10 = obvious ;P
1/11 = .91

sin and cos of 0, 30, 45, 60, 90

Although I'm sure it was just a typo,
1/11 = 0.091

And for those who are lazy,
sin 0 = 0
sin 30 = 0.5
sin 45 = 0.71
sin 60 = 0.87
sin 90 = 1

cos 0 = 1
cos 30 = 0.87
cos 45 = 0.71
cos 60 = 0.5
cos 90 = 0

 

[Godric]

Don't memorize too many things... especially math related. This is what I would know

Log(2) = ~0.3
log(3) = ~0.5
all log rules, how and when to use them (practice them) since 3 and 2 can make a lot of numbers x in log(x)

fractions of 1/ (1 through 11)
1/6 = .16666
1/7 = .143
1/8 = .125
1/9 = .111
1/10 = obvious ;P
1/11 = .91

sin and cos of 0, 30, 45, 60, 90

log(3=~ .4* (.47, Ive used .4 instead of .5 and I think its a better estimation)

But you should make an effort to practice the entire table, your going to get so bloody good at logs by the end of this chapter.

 

[Jay2910]

Thanks for all of your advice everyone!

Now, how do we figure out something like example 4.21 and 4.22 within a minute? I don't really understand their explanation.

For example in 4.21 . .the question was
What is the PH of .07562 M HCOOH with pKa=3.642?


Under the explanation part . .they have " If the acid concentration were .10 M then pH would be 2.321(half pKa+.5)" Where are they getting the .5 from?

I thought that the log(.1)= -1 . .am I missing something?

Thanks again!

 

[RGIII]

Thanks for all of your advice everyone!

Now, how do we figure out something like example 4.21 and 4.22 within a minute? I don't really understand their explanation.

For example in 4.21 . .the question was
What is the PH of .07562 M HCOOH with pKa=3.642?


Under the explanation part . .they have " If the acid concentration were .10 M then pH would be 2.321(half pKa+.5)" Where are they getting the .5 from?

I thought that the log(.1)= -1 . .am I missing something?

Thanks again!

The formula is pH = (1/2)pKa - (1/2)log[HA]. They rounded up the concentration to .10 from .07562. If this is the case pH = (1/2)pKa - (1/2)(-1) which would yield half pKa + .5

 

[Skizzard]

Thanks for all of your advice everyone!

Now, how do we figure out something like example 4.21 and 4.22 within a minute? I don't really understand their explanation.

For example in 4.21 . .the question was
What is the PH of .07562 M HCOOH with pKa=3.642?


Under the explanation part . .they have " If the acid concentration were .10 M then pH would be 2.321(half pKa+.5)" Where are they getting the .5 from?

I thought that the log(.1)= -1 . .am I missing something?

Thanks again!

pH=1/2pKa-1/2[HA]

Where pKa=3.6* and [HA]=.07*

In terms of the MCAT you need only estimate. Understanding that [log.1] = -1 and [log.01] = -2 you can figure out the range of your answer.

pH=1/2(3.6)-1/2(-1) = 1.8+.5 = 2.3
pH=1/2(3.6)-1/2(-2) = 1.8+ 1 = 2.8

So your answer should be 2.3 < x < 2.8

Since the original concentration given is closer to .1 than .01 the answer should be closer to 2.3.

Hope this helps.

 

[Jay2910]

The formula is pH = (1/2)pKa - (1/2)log[HA]. They rounded up the concentration to .10 from .07562. If this is the case pH = (1/2)pKa - (1/2)(-1) which would yield half pKa + .5

Yeah that helped a lot! Thanks for catching my over sight.

 

[Flapjacks]

What I did when I studied that table, I looked at the example for Log of 4 and Log of 5, then noticed how the pattern was developed and from there, I took all the values from the left hand side (5,6,7,8,9,1.33,1.5 etc.) and wrote them on a separate sheet of paper and did the calculations long hand to watch how the answers were arrived.

That completely solidified the logic of logs for me. Except, I hated how they got lazy on log 7.