Question #69eae

May 17, 2017

$\text{Dissociate.............}$

Explanation:

The strength of an acid relates to the extent which the following equilibrium is completed:

$H X \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {X}^{-}$

For strong acids, ${X}^{-} = H S {O}_{4}^{-} , \text{halide} \left(X \ne F\right) , C l {O}_{4}^{-}$, the equilibrium lies STRONGLY to the right. For weaker acids, $H F , {H}_{2} S {O}_{3} , H {O}_{2} C - C {H}_{3}$, the equilibrium lies to the left.

And this we can use the equilibrium constant, ${K}_{a}$, as a determinant:

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{X}^{-}\right]}{\left[H X\right]}$

The given acids, have ${K}_{a}$ values so large in water that the dissociation is considered to be quantitative.