math question - solving ln

[fbradley99]

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how are you guys solving questions that involve ln ???

Is there any tricks to this?

 

[Bernoull]

do you have an example? U can also convert to ln to log since base 10 is easier to manipulate? I can't think of where u'll need ln, maybe nernst eq?

 

[PRATFO]

ln(x) = 2.3log(x)


Or close enough for the MCAT at least

 

[capostat]

I have a question as well...

how do you go from pka to Ka?

if pka = 6.4

then,

pka = -log Ka, and Ka= 10^-pka

but Ka is not 1 x 10^-6.4.


All I know is that if you had it's, __ x 10^-7.

But how do you estimate that number correctly?

my kaplan answer says 4.

 

[crazybob]

ln(x) = 2.3log(x)


Or close enough for the MCAT at least

log (x) / log (e) = ln (x)

log (x) / log (e) = 2.3 log(x)

1 / log (e) not exactly equal to 2.3

in this case, x = 1, so ln (x) = 0 and log (x) = 0

0 / log (e) = 2.3 * 0

 

[EyEnStein 07]

I have a question as well...

how do you go from pka to Ka?

if pka = 6.4

then,

pka = -log Ka, and Ka= 10^-pka

but Ka is not 1 x 10^-6.4.


All I know is that if you had it's, __ x 10^-7.

But how do you estimate that number correctly?

my kaplan answer says 4.

wow...that is pretty complicated without a calc/computer and i cant tell you the answer. What i can tell you is technically Ka IS 1*10^-6.4 ...
because you can rewrite that as 1 / 10^6.4 ....and estimate that to be
____*10^-7 ...4 is probably the closest estimation but i cant tell you exactly how to get a definite one

 

[thebillsfan]

well it is tough. but think about how you calculate pH from an H concentration. Let's say you had 6*10^-7 as you h concentration. you know the pH is between 6 and 7. let's say 6.4

now, do the same thing backwards: pH is 6.4 how do you get the H concentration? well it has to be between 1*10^-6 and 1*10^-7. you dont need anything more specific than that.

 

[crazybob]

does the kaplan book give any reasoning on why it's 4 X 10^-7?

 

[capostat]

wow...that is pretty complicated without a calc/computer and i cant tell you the answer. What i can tell you is technically Ka IS 1*10^-6.4 ...
because you can rewrite that as 1 / 10^6.4 ....and estimate that to be
____*10^-7 ...4 is probably the closest estimation but i cant tell you exactly how to get a definite one

yeah-- cause 10 to the power of a rational number doesnt work, but I'll go with my gut. I would have said "5" is that number. i just hope the answer choices dont cut it close with the numbers...

well it is tough. but think about how you calculate pH from an H concentration. Let's say you had 6*10^-7 as you h concentration. you know the pH is between 6 and 7. let's say 6.4

now, do the same thing backwards: pH is 6.4 how do you get the H concentration? well it has to be between 1*10^-6 and 1*10^-7. you dont need anything more specific than that.

i'll come up with some rule system like maybe if the pka is 8.5, i'll guess that the Ka is 3 or 4 is that number x 10^-9. I'll try to be very conservative.

 

[capostat]

does the kaplan book give any reasoning on why it's 4 X 10^-7?

that's the funny thing.... the solution paragraph says that there's a neat trick to estimate antilogs-- but then they dont even say how they got to choosing that "4".

they just said if the log of X is 2, then X is equal to 100, and if the log of X is -4, then X is equal to 0.0001, and so on.

they didnt explain how they took 6.4 and got that "4". was really bugged me...

 

[EyEnStein 07]

were all the choices ___*10^-7?
if that was the case....very unfair IMO.

 

[capostat]

were all the choices ___*10^-7?
if that was the case....very unfair IMO.

no it wasnt like that. you needed the Ka to calculate the pH.

They gave you the concentration of the weak acid and the pKa, and you just had to set up the equilibrium expression. With the answer choices were:

pH = 3, 4.7, 6.3, 9.3

the answer was 4.7.

i dont think you would have really gotten to pH=3 or 6.3, if you really kept your Ka guess around 4-5.5 x 10^-7. I didnt try it with "5" to verify.

The question was:
pka=6.4 of 10^-3M H2CO3. Find the pH.


yeah it's kind of tricky, becuase i had forgot entirely about knowing how to get the Ka from pKa...

 

[eeboy]

10^-6.4 = 10^-7 * 10^0.6 // -7+.6=-6.4
10^0.6 = 4 //approx