Question

In a solution with a pH of 4, the [OH-] is?

Answer 1

I'm gonna write some theory then you do the calculation:
Water protolytic properties make the following true for pure water and dilute aqueous solutions at 25 ° C:
\[[H _{3}O ^{+}]=[OH ^{-}]=1,0 * 10^{-7} M <-> [H _{3}O ^{+}]*[OH ^{-}]=1,0*10^{-14} M\]
pH is now defined as negative logarithm of the concentration of H 3 O +, while pOH is defined as the negative logarithm of the OH - concentration:
\[pH=-\log([H ^{+}])\]
\[pOH=-\log([OH ^{-}])\]
We take the logarithm on both sides in the first written equation:
\[1,0*10^{-14} M = K _{w}=[H _{3}O ^{+}]*[OH ^{-}] <-> \log(K _{w})=pH*pOH\]
In other terms there following most apply at 25 ° C:
\[pH*pOH=14,0\]

Answer 2

You know where to begin or need further help?

Answer 3

I give you some hints. Start by using the last equation pH*pOH=14,0 to find pOH

Then use the pOH=-log([OH-]) equation... solved for [OH-] it is going to look like this:
\[[OH ^{-}]=10^{-pOH}\]

Answer 4

sorry meant:
\[pH+pOH=14,0\]