Logarithmic tricks

[going2breakdown]

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Here's the problem, gastric juice has a pH of about 1. What is the pH when 10mL of the stomach juice and 10 mL. of water are mixed.
I know from deduction that it is going to still be about 1.
However, in the solutions manual it says that -log(.05)=log(20) what is the trick there? When I take the inverse of .05 I get 20. Can I do that with all logarithmic problems? i.e. just take the inverse of what you're taking the negative log of?

 

[shaboobly]

yes it's a "trick" and you can inverse the ratio in a log, just be sure to take the opposite of the log.

 

[going2breakdown]

thanks! I kept checking it on my calculator too. I just wanted to make sure that was all there was to it. Thanks!

 

[circulus vitios]

-log(0.05)
= -log(1/20)
= log(1) - (-log(20))
= 0 + log(20)
= log(5*2*2)
= log(5) + log(2) + log(2)
= ~ 0.7+0.3+0.3
= ~1.3

 

[shaboobly]

yea there are other ways to simplify logs.. you can probably find them on wikipedia or something, but i dont think there is a need to memorize all of it, unless you just wanna get good at logs or something.

 

[ebasappa]

log(.05)
log(5x10^-2)

2- log 5
2 -(log10/log 2)
2-(log 10-log 2)
2-(1-.3)
2-.7
1.3



^^^ another way, just for the general public lol

 

[Singhp03]

how did you guys get the .05? sorry i suck at ph problems

 

[circulus vitios]

how did you guys get the .05? sorry i suck at ph problems

pH of gastric acid is given as 1.
pH is defined as -log([H+])
-log([H+]) = 1 implies that [H+] is 0.1 M

Remember that M1V1=M2V2

(10 mL gastric acid)(0.1 M)=(M2)(10 mL gastric acid + 10 mL water)
M2 = 0.05 M = [H+] of the new gastric acid solution.
-log(0.05) ~ pH of 1.3

 

[Singhp03]

pH of gastric acid is given as 1.
pH is defined as -log([H+])
-log([H+]) = 1 implies that [H+] is 0.1 M

Remember that M1V1=M2V2

(10 mL gastric acid)(0.1 M)=(M2)(10 mL gastric acid + 10 mL water)
M2 = 0.05 M = [H+] of the new gastric acid solution.
-log(0.05) ~ pH of 1.3

awesome, thanks!

 

[Sandstorm]

awesome, thanks!

Quick shortcut...

-log [a x 10^-b] = b - 0.a

Example... [h+]= 3 x 10^-6

pH = about 5.7 (6 - 0.3)

Should get you close enough to choose your answer