If [OH-]=10^-pOH and [OH-]= 0.003 then what does pOH equal? I know this is simple but I just can't figure out how to do this calculation. Any help would be appreciated.
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If $a = 10^{-b}$, then by taking the logarithm of both sides, we get $$b = -\log_{10} a$$
Since we know the value of $a$, you can just substitute.
MT_
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From $c\left[OH^{-}\right]=10^{-\rm{pOH}}$ and $c\left[OH^{-}\right]=0.003=3\times 10^{-3}$ we have $10^{-\rm{pOH}}=3\times 10^{-3}$.
Simplify it, we get $$-p\rm{OH}\cdot log_{10}{10}=log_{10}{\left(3\times 10^{-3}\right)}.$$
Hence $$-p\rm{OH}=log_{10}{3}-3.$$
So we get $$p\rm{OH}=3-log_{10}{3}=3-0.48=2.52.$$
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