Log problems

[ZoeMarkson]

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pOH = -log([OH–])
pOH = -log(0.01)
pOH = 2

How do you do this without a calculator?

 

[rdub]

1. You got to the point of having pOH = -log(0.01)
2. Now change the 0.01 to scientific notation, which is 1x10^-2. So you have pOH = -log(1x10^-2)
3. The negative log of 1x10^-2 is 2 (this is a principle you must memorize and get used to. Whenever you have 1x10^[negative exponent], then the negative log of it is just the exponent but positive)

Make sense? If not, watch Chads Videos.

 

[ZoeMarkson]

1. You got to the point of having pOH = -log(0.01)
2. Now change the 0.01 to scientific notation, which is 1x10^-2. So you have pOH = -log(1x10^-2)
3. The negative log of 1x10^-2 is 2 (this is a principle you must memorize and get used to. Whenever you have 1x10^[negative exponent], then the negative log of it is just the exponent but positive)

Make sense? If not, watch Chads Videos.

Yeah, yay thank you it makes much sense now

 

[Don Kim]

Another quick trick with logs that will help tremendously in Gchem: any H+ concentration with 3.16 x 10^x M will have a pH of x-0.5. For example, 3.16 x 10^5 M of H+ will have pH of 4.5. 3.16 seems to be the dividing line between x and x-1. So using this, you can estimate a lot of other arbitrary logs, like 6.5 x 10^8 M of H+ will have a pH of ~ 7.2 (pH is closer to 7 than 8) since 6.5 is around half way between 3.16 and 10.