Find the hydroxide concentration with the pOH solution 6.00?

Question

I need to find the hydroxide concentration of the 6.00 pOH solution. I know that I have the equation: 6.00= -log[OH-]
but I don't know where to go from there. If someone could explain it to me easily, I'd appreciate it.

Answer 1

You want to exponentiate your logarathim. For example, a=log x is the equivalent of 10^a=x.

In your problem, first multiply both sides by negative one to move the negative sign to the other side. Then, -6.00=log[OH-] becomes 10^(-6)=[OH-]. Your concentration of OH- is 10^-6 M.

Answer 2

"PhysicsPat" is correct.

pOH = -log ([OH-])

Remember what the log() function actually means.
For example, lets say that,
Log_z (x) = y
This means that some base (z [underscore indicates subscript]) to the power of y equals x,
z^y = x
In the case of an ordinary "Log" written without a base, a base of 10 is implied...which is what is used on the pH/pOH scale.

So if the -log of something equals 6.00, then the log of something equals -6.00.
-log ([OH-]) = 6.00
log([OH-]) = -6.00
Which means that 10^-6.00 equals [OH-].

[OH-] = 1 E-6 = 1 x 10^-6.00 = .000001 Molar