# What is [HO^-] of a solution for which pH=3.10?

Mar 8, 2017

$\left[H {O}^{-}\right] = 1.27 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$.............

#### Explanation:

We use the relationship $14 = p H + p O H$

Now $p H = - {\log}_{10} \left(7.90 \times {10}^{-} 4\right) = - \left(- 3.10\right) = 3.10$

And $p O H = 14 - 3.10 = 10.80$

And we can now take antilogarithms,

$\left[H {O}^{-}\right] = {10}^{- 10.8} = 1.27 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$