Animated Solution for Chemistry - Ionic Equilibrium: The $p{K}_a$ of a weak acid, HA is 4.80. The $p{K}_b$ of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be

Imagine you are standing in a chemistry lab, holding a beaker of pure water. You drop a pinch of a special salt, denoted as $BA$, into it. This isn't your everyday table salt. This salt is the offspring of a weak acid ($HA$) and a weak base ($BOH$).

When this salt dissolves, it completely shatters into its constituent ions: $B^{+}$ and $A^{-}$. But the story doesn't end there. Because both of these ions come from weak parents, they are highly reactive with water. They both want to steal pieces of the water molecule to reform their parent compounds. This dual tug-of-war is what we call salt hydrolysis.

To find the pH of such a complex battlefield, we don't need to calculate the individual concentrations of every species. Chemistry gives us a beautifully elegant and symmetric formula that cuts right to the chase:

Notice something fascinating about this equation? There is no concentration term ($C$) in it! For a salt of a weak acid and a weak base, the final pH depends only on the relative strengths of the acid and the base, not on how much salt you dissolved.

Let's bring in the numbers from our problem. We are given the acid dissociation constant in its logarithmic form, $p{K}_a=4.80$, and the base dissociation constant, $p{K}_b=4.78$.

We substitute these directly into our master equation:

First, we evaluate the battle inside the parentheses. The difference is $4.80-4.78=0.02$. Because the $p{K}_a$ is slightly larger than the $p{K}_b$, it means the acid is slightly weaker than the base.

Next, we take half of that tiny difference:

Finally, we add this to the neutral baseline of 7:

And there we have it! The final pH is 7.01. The math perfectly mirrors the physical reality: because the base was just a tiny bit stronger than the acid, the final solution ends up being just a tiny bit basic.