Calculating the pH of a dilluted aqueous KOH solution Calculate $\mathrm{p}\ce{H}$ of an aqueous solution of $0.01\:\mathrm{nM}~\ce{KOH}$ in water.


I think that water also has $\ce{OH-}$ ions in concentration of $10^{-7}\:\mathrm{M}$ (since $\mathrm{p}\ce{H}=7$)

So the total concentration of $\ce{OH-}$ after mixing the solution with water would be:

$0.01\:\mathrm{nM} + 10^{-7}\:\mathrm{M} = 10^{-11}\:\mathrm{M} + 10^{-7}\:\mathrm{M} \implies 1.0001 \times 10^{-7}\:\mathrm{M}$

and given that $\mathrm{pH} = 14 - \mathrm{p}\ce{OH}$ 

$\implies 14 - \log(1.0001\times10^{-7})$

$\implies \mathrm{pH} \approx 7$

Does this make sense?
Yes, that's right, the concentration of $\ce{{}^{-}OH}$ from dissociation of water will be much greater than the amount added from $0.01\:\mathrm{nM}~\ce{KOH}$.  

$0.01\:\mathrm{nM}~\ce{KOH}$ will not significantly the affect $\mathrm{p}\ce{H}$ of water.