# How does a strong acid differ from a weak acid?

Oct 10, 2016

Consider the equilirium:

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

#### Explanation:

Of course, the acidity of an acid is dependent on the solvent system. Most of the time (in fact almost all of the time), we deal with water, $O {H}_{2}$, as the solvent.

For strong acids, the given equilibrium,

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

lies strongly to the right. And thus for $1 \cdot m o l \cdot {L}^{-} 1$ concentrations of acids such as $H C l$, $H B r$, $H N {O}_{3}$, $\left[{H}_{3} {O}^{+}\right]$, the concentration it expresses in solution, $\cong$ $1 \cdot m o l \cdot {L}^{-} 1$. What would be $\left[{H}_{3} {O}^{+}\right]$ for a $1 \cdot m o l \cdot {L}^{-} 1$ solution of ${H}_{2} S {O}_{4}$, a strong acid?

On the other hand, for weak acids, the given equilibrium lies to the left, towards the side of the undissociated acid. The acid is solvated by water molecules, but formation of ${H}_{3} {O}^{+}$, $\text{acidium ion}$, is impeded.