What is the $\left[{\mathrm{OH}}^{-}\right]$concentration of $\mathrm{NaOH}$ solution with a $\mathrm{pH}$ of $9.4$ ?

$[ {\mathrm{OH}}^{-} ] = 2.5\times {10}^{- 5 }\mathrm{M}$

For starters, you know that an aqueous solution kept at room temperature has

$\mathrm{pH} + \mathrm{pOH} = 14$

So use this equation to find the pOH of the solution.

$\mathrm{pOH} = 14 - 9.4 = 4.6$

Now, the pOH of the solution is given by the concentration of hydroxide anions, OH−

$\mathrm{pOH} = - \log ( [ {\mathrm{OH}}^{-} ] )$

To find the concentration of hydroxide anions starting from this equation, rewrite it as

$\log ( [ {\mathrm{OH}}^{-} ] ) = - \mathrm{pOH}$

${10}^{ \log ( [ {\mathrm{OH}}^{-} ] )} = {10}^{- \mathrm{pOH}}$

This will get you

$\left[{\mathrm{OH}}^{-}\right]={10}^{-\mathrm{pOH}}$→ remember this equation!

In your case, you have

$\left[{\mathrm{OH}}^{-}\right]={10}^{-4.6}$

$\left[{\mathrm{OH}}^{-}\right]=2.5\times {10}^{-5}\mathrm{M}$

Hence, option A is correct.