# What is the pH of a 0.055*mol*L^-1 solution of sulfuric acid in water?

May 29, 2016

Assuming quantitative dissociation, $p H = 0.96$

#### Explanation:

We assume (reasonably!) that dissociation is quantitative:

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

Thus $\left[{H}_{3} {O}^{+}\right]$ $=$ $0.110 \cdot m o l \cdot {L}^{-} 1$.

Now $p H$ $=$ $- {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $- {\log}_{10} \left(0.110\right)$ $=$ $0.96$