A solution of 0.1 M weak base (B) is titrated with 0.1 M of a strong acid (HA). The variation of pH

Q: Why can't we just use the equivalence point to find the pKb?

At the equivalence point, you only have the salt $\text{BH}^+\text{A}^-$. While you *can* calculate $\text{p}K_b$ using the salt hydrolysis formula ($\text{pH} = 7 - \frac{1}{2}(\text{p}K_b + \log C)$), reading the exact pH at the steep equivalence point from the graph is highly inaccurate. The half-equivalence point lies on a flatter region, making it much more reliable to read.

Q: Why is the pH not 7 at the equivalence point?

We are titrating a weak base with a strong acid. The resulting salt ($\text{BH}^+\text{A}^-$) undergoes cationic hydrolysis. The $\text{BH}^+$ ion reacts with water to release $\text{H}^+$ ions, making the solution acidic at the equivalence point ($\text{pH} < 7$).

Q: How do we know the temperature is 25°C to use pH + pOH = 14?

In JEE problems, unless a specific temperature is explicitly mentioned, it is standard convention to assume room temperature ($25^\circ\text{C}$ or $298\text{ K}$), where $K_w = 10^{-14}$ and $\text{pH} + \text{pOH} = 14$.

Q: What if the concentrations of the acid and base were different?

The equivalence volume would change based on the stoichiometry ($M_1V_1 = M_2V_2$). However, the core concept of the half-equivalence point remains exactly the same: at half the equivalence volume, $\text{pOH} = \text{p}K_b$ for a weak base.

Q: Why does the curve flatten out after the equivalence point?

After the equivalence point, the weak base is completely neutralized. Any further addition of the strong acid (HA) simply adds excess $\text{H}^+$ ions to the solution. The pH is now entirely controlled by the concentration of this excess strong acid, which causes the pH to drop and flatten out at a low value.

Question

A solution of 0.1 M weak base (B) is titrated with 0.1 M of a strong acid (HA). The variation of pH of the solution with the volume of HA added is shown in the figure below. What is the $	ext{p}K_b$ of the base? The neutralization reaction is given by $	ext{B} + 	ext{HA} ightarrow 	ext{BH}^+ + 	ext{A}^-$.

Sigma Percentile · Accepted Answer

### Decoding the Titration Curve

Imagine you are in a chemistry lab, carefully adding drops of a strong acid into a beaker containing a weak base.

As each drop falls, the $	ext{pH}$ of the solution slowly decreases. This entire journey is beautifully captured in the titration curve provided in the problem.

Our goal is to find the $	ext{p}K_b$ of this weak base. While it might seem like we need complex equilibrium calculations, this graph holds a secret shortcut.

### The Equivalence Point: The Vertical Drop

Look closely at the graph. Do you see that sharp, almost vertical drop in the curve?

This dramatic plunge represents the **equivalence point**. It is the exact moment when the amount of strong acid added perfectly matches the amount of weak base initially present.

By tracing this vertical drop down to the x-axis, we can see it occurs exactly at $6 	ext{ mL}$.

This tells us that $V_{	ext{eq}} = 6 	ext{ mL}$. It takes exactly $6 	ext{ mL}$ of the strong acid to completely neutralize our weak base.

### The Magic of the Half-Equivalence Point

Now, here is where the magic happens. If complete neutralization requires $6 	ext{ mL}$, what happens when we have only added half of that amount?

At exactly $3 	ext{ mL}$, we reach what is known as the **half-equivalence point**.

At this specific volume, exactly half of our weak base ($	ext{B}$) has reacted with the acid to form its conjugate acid ($	ext{BH}^+$).

Because exactly half has reacted, the concentration of the remaining weak base is perfectly equal to the concentration of the newly formed conjugate acid.

Mathematically, $[	ext{B}] = [	ext{BH}^+]$. We have essentially created a perfect basic buffer right in our beaker!

### The Buffer Equation and the Final Strike

Let's bring in the Henderson-Hasselbalch equation for a basic buffer:

$$ 	ext{pOH} = 	ext{p}K_b + \log \frac{[	ext{BH}^+]}{[	ext{B}]} $$

Since we established that $[	ext{B}] = [	ext{BH}^+]$ at the half-equivalence point, the ratio becomes $1$.

And since $\log(1) = 0$, the entire log term vanishes! This leaves us with a beautifully simple relationship:

$$ 	ext{pOH} = 	ext{p}K_b $$

This is the ultimate shortcut. To find the $	ext{p}K_b$, all we need is the $	ext{pOH}$ at the half-equivalence point.

Let's go back to our graph. At $V = 3 	ext{ mL}$, if we trace up to the curve and over to the y-axis, we read a $	ext{pH}$ of exactly $11$.

Assuming standard room temperature, we know that:

$$ 	ext{pH} + 	ext{pOH} = 14 $$

Substituting our $	ext{pH}$ value:

$$ 11 + 	ext{pOH} = 14 \implies 	ext{pOH} = 3 $$

And since $	ext{pOH} = 	ext{p}K_b$ at this magical point, we have our final answer:

$$ 	ext{p}K_b = 3 $$

By simply understanding the geometry of the titration curve and the power of the half-equivalence point, we bypassed pages of algebra and arrived straight at the solution!

Enter Numerical Value:

Visualized Solution

Analyzing the Titration Curve

Equivalence Point

Half-Equivalence Point

Buffer Equation

Relationship at Half-Equivalence

Extracting Data from Graph

Final Calculation

The Sigma Insight: pH, Buffer and Indicator

Decoding the Titration Curve

The Equivalence Point

The Magic of the Half-Equivalence Point

The Buffer Equation and the Final Strike

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