# Predicting acid-base equilibria

I was looking at acid-base equilibria of weak acids and weak bases. To determine the equilibrium constant of a reaction of those, I found a formula on the libretexts page:

Where ΔpKa is the difference of the product acid's pKa minus the reactant acid's pKa. This formula makes sense to me, but how can one arrive at this formula, starting with the general definitions of pKa and law of mass action?

An example:

Now if we add the two reactions together (using NaOH, hydronium and hydroxide cancel):

But how can we get the equilibrium constant for this reaction, using only the K values listed above, and thus arriving at the supposed result of K=0.01 (pKa of water is 14)? I am kind of stuck, any help is appreciated.

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Consider a generic acid dissociation reaction of the form:

$$\ce{HE <=> H^+ + E^-}$$

Let:

$$A$$ represent $$HE$$

$$B$$ represent $$H^+$$

$$C$$ represent $$E^-$$

So that the reaction becomes:

$$\ce{A <=> B + C}$$

We're mixing 2 different weak acids together, so each acid will have a different $$K_a$$ value. Assuming (1) represents the stronger acid (higher $$K_a$$), and (2) represents the weaker acid (lower $$K_a$$), we have:

$$\ce{A_1 <=> B_1 + C_1, K_a_1}$$

$$\ce{A_2 <=> B_2 + C_2, K_a_2}$$

Now, because (1) is stronger than (2), it will work as an acid and force (2) to work as a base.

This means we need to flip reaction (2) so that the products are the reactants and vice versa.

When we flip a reaction, the new $$K$$ value is the inverse of the old $$K$$ value, and when we add two reactions together, the resulting $$K$$ value is the product of each constant.

So, we get the following system:

$$\ce{A_1 <=> B_1 + C_1, K_a_1}$$

$$\ce{B_2 + C_2 <=> A_2, \frac{1}{K_a_2}}$$

$$------------------$$

$$\ce{A_1 + C_2 <=> A_2 + C_1 + B_1 - B_2, \frac{K_a_1}{K_a_2}}$$

Where: $$K_{eq}=\frac{K_{a1}}{K_{a2}}$$

If we take the base 10 logarithm on both sides:

$$log(K_{eq}) = log\left(\frac{K_{a1}}{K_{a2}}\right) = log(K_{a1})-log(K_{a2})=pK_{a2} - pK_{a1}$$

We then define:

$$\Delta P_{ka}=pK_{a2} - pK_{a1}$$

So then:

$$log(K_{eq})=\Delta P_{ka}$$

Finally:

$$K_{eq}=10^{\Delta P_{ka}}$$