# What is the ph of 0.45 M of H_2SO_4?

Jun 23, 2016

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $- {\log}_{10} \left[0.90\right]$ $\cong 0.05$

#### Explanation:

Sulfuric acid is reasonably treated as a strong diprotic acid that gives stoichiometric $\left[{H}_{3} {O}^{+}\right]$ and $\left[S {O}_{4}^{2 -}\right]$ on dissolution.

${H}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right) \rightarrow 2 {H}_{3} {O}^{+} + S {O}_{4}^{2 -}$

Thus $\left[{H}_{3} {O}^{+}\right] = 2 \times 0.45 \cdot m o l \cdot {L}^{-} 1$ to a first approx.

And $p H = - {\log}_{10} \left[0.90\right]$

For a better approximation, we need $p K {a}_{2}$ for bisulfate anion, $H S {O}_{4}^{-}$, though this is still pretty low.