# What is the pH of a solution of "sulfuric acid" of 2xx10^-2*mol*L^-1 concentration?

Aug 9, 2017

$p H = 1.70 \ldots \ldots \ldots . .$

#### Explanation:

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

Now for ${H}_{2} S {O}_{4}$ with a concentration of ${10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$, clearly, $\left[{H}_{3} {O}^{+}\right] = 2 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$.

That is to a first approx. ${H}_{2} S {O}_{4}$ is considered to be a diacid in aqueous solution and gives 2 equiv of ${H}_{3} {O}^{+}$. (In the absence of a value for $p {K}_{a 2}$ I am perfectly justified in making this assumption!)

And so $p H = - {\log}_{10} \left(2 \times {10}^{-} 2\right) = - {\log}_{10} \left(0.02\right) = 1.70$