# If the pH is 14.75, what is [OH^-]?

In aqueous solution under standard conditions, $p H + p O H = 14$
Given the above, $p O H$ $=$ $- 0.75$.
Thus $\left[H {O}^{-}\right]$ $=$ ${10}^{- \left(- 0.75\right)}$ $=$ ${10}^{+ 0.75}$ $=$ $5.62 \cdot m o l \cdot {L}^{-} 1$.