# How do you calculate Ka?

Jul 17, 2017

In fact you measure ${K}_{a}$........

#### Explanation:

For the hypothetical reaction..........

$A + B r i g h t \le f t h a r p \infty n s C + D$

There is a $\text{rate forward,}$ $= {k}_{f} \left[A\right] \left[B\right]$......

And a $\text{rate backward,}$ $= {k}_{r} \left[C\right] \left[D\right]$.......

And importantly the condition of chemical equilibrium does not specify the cessation of chemical change, but equality of forward and reverse rates (says he, channelling a text book)........

And thus ${k}_{f} \left[A\right] \left[B\right] = {k}_{r} \left[C\right] \left[D\right]$......

And ${k}_{f} / {k}_{r} = \frac{\left[C\right] \left[D\right]}{\left[A\right] \left[B\right]} = {K}_{\text{eq}}$, the thermodynamic equilibrium constant.

For acid base behaviour in water, we write......

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

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

the $\left[{H}_{2} O\right]$ term may be removed, because it is so large it is effectively constant. Apologies if I have missed the point of your question.......but for strong acids,......

$H A = H X \left(X \ne F\right) , {H}_{2} S {O}_{4} , H C l {O}_{4}$,

.......the equilibrium lies to the right......And the extent of equilibrium, as always, is established by measurement.