What is the (OH-) in a solution with a pOH of 6.48

1 answer:

3 0

Through manipulation of equations, we are able to obtain the equation:

$-pOH= log [ OH^{-}]$

Then we can transform the equation into:

$[ OH^{-}]= 10^{-pOH}$

Then we are able to plug in the pOH and directly get [OH-]:

$[ OH^{-}] = 10^{-6.48}$

$[ OH^{-}]=3.31* 10^{-7} M$

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