Logs of fractions in Hederson-Hasselbach equation

[BabberSher]

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U guys, i don't really remember this, could u help me out... ?
How do u guys/gals calculate logs of fractions (in the Hederson-Hasselbach equation) for buffers, in ur head?

pH=pKa + log[A-]/[HA]

so how would u do the log of 0.50/0.11 or something like that

[Its problem 13 in Chapter 10 of Acid-Base ?'s]

This might be easier than it looks to me, but im just not comfortable w/logs...

 

[tinman831]

U guys, i don't really remember this, could u help me out... ?
How do u guys/gals calculate logs of fractions (in the Hederson-Hasselbach equation) for buffers, in ur head?

pH=pKa + log[A-]/[HA]

so how would u do the log of 0.50/0.11 or something like that

[Its problem 13 in Chapter 10 of Acid-Base ?'s]

This might be easier than it looks to me, but im just not comfortable w/logs...

its impossible to calculate logs in your head. the dat will never ask you to do that.

 

[dent_girl]

they will never ask you that unless it's 1:1 ratio.

 

[Gasedo]

its impossible to calculate logs in your head. the dat will never ask you to do that.

So how do we calculate “Acid Base” and “Titration” problems that require the use of log? 😕

 

[arash]

rule 1=>-log of 10^-4 = 4
rule 2=>now when you have a number times 10^-4 , -log of that is going to up to 1 less than 4, a number between 3 and 4,
rule 3=> y x 10^-4 => The smaller y is (for example 2), the closer your answer will be to 4, and the bigger it is (for example 9), the closer the answer will be to 3.

Examples: -log 1.01x 10 ^ -4 = 3.99

- log 9.9 x 10 ^ -4= 3.004

I think this is all you need to know. So you can never find the exact number but you can guess the approximation with a good accuracy.

 

[Gasedo]

rule 1=>-log of 10^-4 = 4
rule 2=>now when you have a number times 10^-4 , -log of that is going to up to 1 less than 4, a number between 3 and 4,
rule 3=> y x 10^-4 => The smaller y is (for example 2), the closer your answer will be to 4, and the bigger it is (for example 9), the closer the answer will be to 3.

Examples: -log 1.01x 10 ^ -4 = 3.99

- log 9.9 x 10 ^ -4= 3.004

I think this is all you need to know. So you can never find the exact number but you can guess the approximation with a good accuracy.

Good explanation arash. When you calculate concentration, and get for example x² = 1.75x10^-6. How you get x? What is the easiest way to get the square of x? I am having troubles with these acid base problems.
Thanks

 

[arash]

Good explanation arash. When you calculate concentration, and get for example x² = 1.75x10^-6. How you get x? What is the easiest way to get the square of x? I am having troubles with these acid base problems.
Thanks

OK I will tell you how, but just to make your life easier, you should know that they never ask for this .what they do is that in the answers they have

x= radical of 1.75x10^-6

Anyways first forget the 10^-6 part cause that is easy, 10^-3 x 10^-3
For 1.75 you should see that it is close to 169 which is 13 x 13, so your answer should be sth like very close but bigger than 1.3 x 10-3. to be more exact it is 1.32 x 10^-3. This approach should make it easier. Dont stress over these alot. Trust me the actual DAT unlike kaplan or topscore don't give tricky numbers to work with and is usually very straight forward.