Animated Solution for Chemistry - Ionic Equilibrium: The pH of a $0.02\text{ M }{\text{NH}}_4\text{Cl}$ solution will be [Given $K_b({\text{NH}}_4\text{OH})=10^{-5}$ and $\log 2=0.301$]

Imagine you are standing in a laboratory, holding a beaker of pure water. When you dissolve a salt like ammonium chloride (${\text{NH}}_4\text{Cl}$) into it, a fascinating microscopic dance begins. The salt completely dissociates into ammonium (${\text{NH}}_4^{+}$) and chloride (${\text{Cl}}^{-}$) ions.

Now, not all ions are created equal. The chloride ion is the conjugate base of a very strong acid ($\text{HCl}$), which makes it an incredibly weak base. It acts merely as a spectator, floating aimlessly in the water. However, the ammonium ion is the conjugate acid of a weak base (${\text{NH}}_4\text{OH}$). Because it comes from a weak parent, it has a strong desire to react with water. This reaction, where the cation of the salt reacts with water to produce extra hydrogen ions (${\text{H}}^{+}$), is known as cationic hydrolysis.

Because the hydrolysis of the ammonium ion releases extra ${\text{H}}^{+}$ ions into the solution, the delicate balance of pure water is disrupted. The solution becomes acidic.

Before we even touch a calculator, our chemical intuition tells us a crucial fact: the pH of this solution must be less than 7. This is a powerful sanity check that you should always keep in your back pocket. Whenever you have a salt formed from a strong acid and a weak base, the strong acid's character dominates the final solution.

To find the exact pH, we don't need to derive the equilibrium expressions from scratch every time. We have a beautifully elegant master formula for the pH of a salt of a strong acid and a weak base:

Let's gather our tools to use this formula. We are given the base dissociation constant, $K_b$, as $10^{-5}$. The term $p{K}_b$ is simply the negative logarithm of $K_b$:

Next, we need the concentration of the salt, $C$. We are given $C=0.02\text{ M}$. To make our upcoming logarithmic calculations smoother, it is always a smart move to convert decimals into scientific notation. So, we write $C=2\times 10^{-2}\text{ M}$.

Now, let's carefully substitute these values into our master formula. This is where many students make silly algebraic mistakes, so we will take it step-by-step.

We need to expand the logarithm term using the fundamental property $\log (a\times b)=\log a+\log b$:

We are given that $\log 2=0.301$, and we know that $\log (10^{-2})=-2$. Substituting these back into our equation:

Let's simplify the terms inside the bracket. Five minus two gives us three, and adding zero point three zero one gives $3.301$.

Half of $3.301$ is exactly $1.6505$.

And there we have it! The final calculated pH is $5.35$. Notice how this perfectly aligns with our initial theoretical prediction: the value is indeed less than 7, confirming that our strong acid-weak base salt has created an acidic environment. Always trust the math, but verify it with your physical intuition!