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How do you do log(0.5), log(0.3), etc?
Started by pfaction
So TBR says log (2x10^-3) is 3 - log (2), which is less than 3. Okay.
I go to do log (0.5) which means log (5x10^-1) = 1-log5
How is this a negative answer?!
because log 10^-1 is -1 not 1, and multiplying logs is adding not subtracting.
so it's log5 + log 10^-1 = log5 -1
best way to do log(.5)
log(1/2) = log(1) - log(2) = 0 - .3 = -.3
logs of decimals in between 1 and 0 are made easier by making it into a very very close fraction and using the division rule.
log(1/2) = log(1) - log(2) = 0 - .3 = -.3
logs of decimals in between 1 and 0 are made easier by making it into a very very close fraction and using the division rule.
Page 256 of Gen Chem 1:
What is the pH of 0.002M HCl?
pH = 3-log2 = 2.7
I guess I was wrong about 5x10^-1? I still don't get it since 0.002 is 2x10^-3; and 0.5 is 5x10^-1?
pH=-log([H]) - note the minus in front of the log.
You are correct pH of 0.5 is -log(0.5)=-log(5*10^-1)=-(log5-1)=1-log5>0
What bothers you about that result?
Last edited:
Page 256 of Gen Chem 1:
What is the pH of 0.002M HCl?
pH = 3-log2 = 2.7
I guess I was wrong about 5x10^-1? I still don't get it since 0.002 is 2x10^-3; and 0.5 is 5x10^-1?
pH is - log(acid concentration)
You really really have to pay attention to - signs due to - exponents and definitional - signs like pH = - log(stuff)
You are correct pH of 0.5 is -log(0.5)=-log(5*10^-1)=-(log5-1)=1-log5>0
Because its below 0. http://www.google.com/search?q=log%280.5%29&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US😳fficial&client=firefox-a
Okay so let's go through this. I'm on page 257.
What is the pOH of 0.050M KOH?
pOH = -log[OH]
-log(5x10^-2)
-[log5 + log 10^-2]
-log5 - -2
-log5+2
2-log5
--> 2 - log 5
So that works. What if I asked the pOH of 0.5 KOH?
-log (5x10^-1)
-[log 5 + log 10^-1]
-log5 - -1
--> 1 - log 5
1-0.7 is 0.3
That is incorrect....🙁
Because its below 0. http://www.google.com/search?q=log%280.5%29&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US😳fficial&client=firefox-a
Okay so let's go through this. I'm on page 257.
What is the pOH of 0.050M KOH?
pOH = -log[OH]
-log(5x10^-2)
-[log5 + log 10^-2]
-log5 - -2
-log5+2
2-log5
--> 2 - log 5
So that works. What if I asked the pOH of 0.5 KOH?
-log (5x10^-1)
-[log 5 + log 10^-1]
-log5 - -1
--> 1 - log 5
1-0.7 is 0.3
That is incorrect....🙁
log(0.5) is < 0
pOH is -log(0.5) and will be positive.
There is nothing incorrect in what you've typed.
So TBR says log (2x10^-3) is 3 - log (2), which is less than 3. Okay.
I go to do log (0.5) which means log (5x10^-1) = 1-log5
How is this a negative answer?!
pH = - log [H+]
you are correct with 1-log5, but remember that there's a negative sign in front of the log, so you reverse it. - (1-log5) = log5-1
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