Does NaOH have a higher OH- concentration than NH3?

In my book it's a multiple choice question with only one correct answer. 


There are two solutions of $\ce{NaOH}$ and $\ce{NH\_3}$ that have similar concentrations and volumes. 


These two are among the options:


1. $\ce{NaOH}$ has a higher $\ce{pH}$ than $\ce{NH\_3}$
2. $\ce{NaOH}$ has a higher $\ce{OH^-}$ concentration than $\ce{NH\_3}$


Wouldn't these *both* be correct?




Given that the solutions have equal concentrations, a solution of $\ce{NaOH}$ (a strong base) will be more basic than one of $\ce{NH3}$ (a weak base), since $\ce{NH3}$ only dissociates partially:


$$\ce{NH3 + H2O <<=> NH4+ + OH-}$$


Loosely speaking, the pH of a solution is defined to be


$$\mathrm{pH} = -\log\_{10}[\ce{H+}]$$


and from the ionic product of water, we know that $[\ce{H+}][\ce{OH-}]$ is always a constant (equal to $10^{-14}\ \mathrm{M^2}$ at $25\ ^\circ\mathrm{C}$).


Therefore, a higher pH implies a lower concentration of $\ce{H+}$, and therefore a higher concentration of $\ce{OH-}$. This means that the truth of option 1 implies the truth of option 2, and vice versa, i.e. option 1 is [logically equivalent](https://en.wikipedia.org/wiki/Logical_equivalence) to option 2.


My first suspicion would therefore be a typo in your book. In any case, your reasoning is correct.


 5