# What is pH for a 1.95*g mass of pure sulfuric acid added to enough water to give a final volume of 1*dm^3?

Mar 25, 2017

$p H = 1.40 \ldots \ldots . .$

#### Explanation:

We assume complete dissociation of the sulfuric acid:

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

And so $\left[{H}_{2} S {O}_{4}\right]$ $\equiv$ $\frac{\frac{1.95 \cdot g}{98.08 \cdot g \cdot m o {l}^{-} 1}}{1 \cdot {\mathrm{dm}}^{3}}$ $=$ $0.0199 \cdot m o l \cdot {L}^{-} 1$. However, this is a FORMAL concentration and corresponds to $\left[S {O}_{4}^{2 -}\right]$. $\left[{H}_{3} {O}^{+}\right] = 0.0398 \cdot m o l \cdot {L}^{-} 1$, i.e. TWICE the formal concentration.

Now since, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, pH=-log_10(0.0398)=??