Animated Solution for Chemistry - Ionic Equilibrium: 100 mL of 0.1M HCl is taken is a beaker and to it 100 mL of 0.1 M NaOH is added in steps of 2 mL and the pH continuously measured. Which of the following graphs correctly depicts the change in pH?

The process of titration is like a delicate dance between two fierce rivals: the acidic protons ($H^{+}$) and the basic hydroxides ($O{H}^{-}$). When they meet, they neutralize each other to form water. But how does this battle unfold on a pH graph? Let's dive into the fascinating world of strong acid-strong base titrations!

The Initial State

Imagine you have a beaker filled with $100\text{ mL}$ of $0.1\text{ M HCl}$. Hydrochloric acid is a strong acid, meaning it completely dissociates in water. Before we even add a single drop of our base, the concentration of $H^{+}$ ions is exactly $0.1\text{ M}$.

Because pH is defined as the negative logarithm of the hydrogen ion concentration ($pH=-\log [H^{+}]$), we can easily calculate our starting point: This tells us that our graph must start at a very low, highly acidic value.

Pre-Equivalence Region

Now, we start adding $0.1\text{ M NaOH}$ drop by drop. The $O{H}^{-}$ ions from the base immediately seek out the $H^{+}$ ions, neutralizing them to form neutral water molecules ($H_{2}O$). As the amount of $H^{+}$ decreases, the solution becomes less acidic, and the pH begins to rise.

But here is the catch: the pH rises very slowly at first. Why? Because the pH scale is logarithmic! When you have a large concentration of $H^{+}$ ions, removing a few of them doesn't significantly change the overall order of magnitude. The graph gently curves upwards, biding its time.

The Equivalence Point

The real magic happens when we add exactly $100\text{ mL}$ of NaOH. At this precise moment, the moles of base added perfectly equal the initial moles of acid. This is known as the equivalence point.

Every single $H^{+}$ ion has been neutralized by an $O{H}^{-}$ ion. What is left in the beaker? Just water and $NaCl$ (table salt). Since $NaCl$ is a neutral salt that does not undergo hydrolysis, the solution is perfectly neutral. The pH instantly shoots up to exactly $7$.

Because the concentration of $H^{+}$ is so incredibly tiny near this point, even a fraction of a drop of NaOH causes a massive swing in pH. This is why the graph shows a sharp, almost vertical line crossing $pH=7$.

Post-Equivalence Region

What happens if we keep adding NaOH after the equivalence point? Now, there is no acid left to fight back. The excess $O{H}^{-}$ ions flood the solution, making it highly basic.

The pH continues to rise, but just like in the beginning, the rate of change slows down. The graph flattens out at a high pH value, approaching $13$ (since the concentration of excess $O{H}^{-}$ approaches $0.1\text{ M}$).

When we put it all together, the complete titration curve forms a beautiful, symmetrical "S" shape. It starts low, leaps vertically at the equivalence point, and flattens out high.

Looking at our options: - Graph (a) perfectly captures this S-shaped curve. - Graph (b) is a straight line, completely ignoring the logarithmic nature of pH. - Graph (c) flattens out at $pH=7$, which would only happen if we stopped titrating or used a very weak base. - Graph (d) shows the pH decreasing, which is the exact opposite of what happens when you add a base!

Therefore, Graph (a) is the undisputed winner.