I HATE LOGs, how do you figure out this log

[TexasDDS]

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so i am having difficult time with logs in general..

without a calculator i dont feel there is a way to do it, but kaplan constantly gives you these problems so there has to be a way.. can someone please explain..

the question asks:

At equilibrium, a certain acid, HA, in solution yields 0.94M [HA] and 0.060 M [A^-]..

calculate pH...

so i get the problem is set up pH = -log [H^+] = -log (0.060)

but how do i do that without a freaking calculator please explain in thorough detail ... thank you

 

[iwannabadentist]

wel exactly the way its set up, you would convert .060 into 6.0 x 10^-2
now taking the -log (6.0 x 10^-2) you can do 2-log (6.0) which would be 1.2
If that is the correct answer, then thats how you do it. However, I doubt that is the correct answer for some reason because a yield of 0.06M H+ is only about 6.4% dissociation, which would mean its a very weak acid. A pH of 1.2 sounds very low of a weak acid such as this one. Are you sure this is the way to set up this problem?

 

[TexasDDS]

yeah, the answer is 1.22

but how did you get 1.2 from 2-log (6.0)?

for some reason my math isnt working out to well

wel exactly the way its set up, you would convert .060 into 6.0 x 10^-2
now taking the -log (6.0 x 10^-2) you can do 2-log (6.0) which would be 1.2
If that is the correct answer, then thats how you do it. However, I doubt that is the correct answer for some reason because a yield of 0.06M H+ is only about 6.4% dissociation, which would mean its a very weak acid. A pH of 1.2 sounds very low of a weak acid such as this one. Are you sure this is the way to set up this problem?

 

[Jongho Yoon]

Here is a way to estimate.

First of all, you know -log (1E-x) = x right? For ex) -log (1E-7) = 7

pH = -log (0.060) = -log (6e-2)

-log (1e-2) = 2
-log (10e-2) = -log (1e1 * e-2) = -log (1e-1) = 1

-log (10e-2) = 1 < -log (6e-2) < -log (1e-2) = 2

Since the number is 6, little bit bigger than 5, which is a middle number between 1 and 10, approximate estimate will be a bit smaller than 1.5. (Notice how -log (6e-2) is closer to -log (10e-2) as shown above)

Your estimate would be about 1.3-1.4...It should be good enough approximation for you to find the correct answer 1.2.

 

[iwannabadentist]

you should try to memorize some simple logs, at least 1-5:
log 1 = 0
log 2 = 0.3
log 3 = 0.5
log 4 = 0.6
log 5 = 0.7
log 6 = 0.8
i just memorized it. its easy for 3-6, you just add 2 and put a decimal in front. log1 = 0 is also easy to memorize. hope that helps