# If [H_3O^+]=4.5xx10^-3*mol*L^-1, what are pH, and pOH, and [HO^-]?

May 19, 2017

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] \ldots \ldots \ldots \ldots \ldots$

#### Explanation:

$\left(i\right)$ $p H = - {\log}_{10} \left(4.5 \times {10}^{-} 3\right) = - \left(- 2.35\right) = 2.35$.

For $p O H$, we recall that $14 = p O H + p H$ for protonolysis in water...

$\left(i i\right)$ $p O H = - {\log}_{10} \left(3.67 \times {10}^{-} 5\right) = - \left(- 2.35\right) = 4.44$; and thus, from the expression above, $p H = 9.56$.

I will let you do the last problem.