Question #c8734

Apr 24, 2016

$\text{Option B}$

Explanation:

Water undergoes autoprotolysis under standard conditions. What does this mean? It means that the water molecule undergoes the following equilibrium where an $O - H$ bond is broken:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

This is an equilibrium reaction, which can certainly be measured, but it measures $\left[{H}_{3} {O}^{+}\right]$ and $\left[H {O}^{-}\right]$ from all sources.

i.e. ${K}_{w} = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{-} 14$

Now sulfurous acid contributes to the equilibrium:

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

Since we know that $\left[{H}_{3} {O}^{+}\right] = 1 \times {10}^{-} 3$, $\left[H {O}^{-}\right] = {K}_{w} / \left[\left[{H}_{3} {O}^{+}\right]\right]$ $=$ ${10}^{-} 11 \cdot \text{mol} \cdot {L}^{-} 1$