# Why does pKa affect equilibrium?

Jul 18, 2018

Because $p {K}_{a}$ is a MEASUREMENT of the equilibrium?

#### Explanation:

For the generalized reaction.....

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

Now there is a rate forward....

$\text{Rate forward} = {k}_{f} \left[A\right] \left[B\right]$, and a $\text{rate backward} = {k}_{r} \left[C\right] \left[D\right]$.

And equilibrium does NOT specify cessation of chemical change; it does specify EQUALITY of FORWARD or REVERSE rates.

And so.......

$\text{Rate forward"=k_f[A][B]-="rate backward} = {k}_{r} \left[C\right] \left[D\right]$.

And thus the quotient $\frac{\left[C\right] \left[D\right]}{\left[A\right] \left[B\right]} = {k}_{r} / {k}_{f} = {K}_{\text{eq}}$

And ${K}_{\text{eq}}$ is otherwise known as the thermodynamic equilibrium constant.

$p {K}_{a}$ or $p {K}_{\text{eq}}$ is simply the logarithm of ${K}_{a}$, ${K}_{\text{eq}}$..

$p {K}_{a} = - {\log}_{10} {K}_{a}$