pOH

[OU11BB]

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what is the pOH if pOH= - log [2x10^-4]. How can I tell without entering into my TI-85?
Thanks

 

[AndyK]

Look at the exponent 2x10^-**4**.

So the number is somewher between -log(1x10^-4) = 4, and -log(1x10^-3) = 3

Since [2x10^-4] is closer to [1x10^-4] than to [1x10^-3] (which is also [10x10^-4]) the pOH closer to 4 than to 3. Therefore, the pOH of [2x10-4] will be about 3.8

 

[OU11BB]

Look at the exponent 2x10^-**4**.

So the number is somewher between -log(1x10^-4) = 4, and -log(1x10^-3) = 3

Since [2x10^-4] is closer to [1x10^-4] than to [1x10^-3] (which is also [10x10^-4]) the pOH closer to 4 than to 3. Therefore, the pOH of [2x10-4] will be about 3.8

Thanks man, this will help me out. My gen chem is weak and i need to be refreshed on the little things.