Animated Solution for Chemistry - Ionic Equilibrium: pH of $0.005$ M calcium acetate ($p{K}_a$ of $C{H}_{3}COOH=4.74$) is

The problem of finding the pH of a salt solution is a classic in ionic equilibrium, but this specific question holds a hidden trap that catches many students off guard. Let's break down the chemistry and the math step-by-step to uncover the flawless answer.

Our journey begins with identifying the nature of the given salt, calcium acetate, mathematically written as $(C{H}_{3}COO{)}_{2}Ca$.

A salt is the product of a neutralization reaction between an acid and a base. In this case, the parent acid is acetic acid ($C{H}_{3}COOH$), which is a well-known weak acid. The parent base is calcium hydroxide ($Ca(OH{)}_2$), which is a strong base.

Because the strong base dominates the character of the resulting solution, we can immediately predict that the final solution will be basic, meaning its pH must be greater than 7.

When calcium acetate is dissolved in water, it undergoes complete dissociation. This is where we must be incredibly careful with our stoichiometry.

Notice the coefficient of $2$ in front of the acetate ion. One mole of calcium acetate yields two moles of acetate ions.

The problem states that the concentration of the calcium acetate solution is $0.005\text{ M}$. Therefore, the effective concentration of the acetate ions, which we will call $C$, is double that amount:

Many students mistakenly use $0.005\text{ M}$ as the concentration, which leads to an incorrect pH. Always account for the valency of the metal cation!

Once in solution, the acetate ion undergoes anionic hydrolysis. It reacts with water to reform a small amount of weak acetic acid, releasing hydroxide ions ($O{H}^{-}$) in the process:

This release of $O{H}^{-}$ is the chemical reason the solution becomes alkaline. To find the exact pH of a salt formed from a weak acid and a strong base, we use the standard hydrolysis formula:

We are given the $p{K}_a$ of acetic acid as $4.74$, and we just calculated our active concentration $C$ as $0.01\text{ M}$. Let's substitute these values into our master equation:

Now, we execute the mathematics. First, let's handle the logarithmic term. The value $0.01$ can be written as $10^{-2}$.

Substituting this back into our equation, we get:

The final pH is exactly 8.37.

A Note on the Reference Material: If you look closely at the printed solution in some textbooks, you might notice a bizarre mathematical typo. The book correctly sets up the expression $7+2.37+\frac{\log 0.01}{2}$, but then inexplicably evaluates the log term as $-2.30$ instead of $-1$, leading to an incorrect final answer of $7.04$. Trust the rigorous math—the flawless and correct answer is indeed 8.37.