A weak base MOH was titrated against a strong acid. The `pH` at `1/4` th equivalence point was 9.3. What will be the `pH` at `3/4` th equivalence point in the same titration? `(log3=0.48)`

Explanation

To find the pH at the \( \frac{3}{4} \) equivalence point of the titration of a weak base \( \text{MOH} \) against a strong acid, we can follow these steps:

### Step 1: Understand the Equivalence Points
In a titration of a weak base with a strong acid, the equivalence points are the points at which the amount of acid added is stoichiometrically equivalent to the amount of base present. The \( \frac{1}{4} \) equivalence point means that one-fourth of the base has reacted with the acid, while at the \( \frac{3}{4} \) equivalence point, three-fourths of the base has reacted.

### Step 2: Calculate \( \text{pOH} \) at the \( \frac{1}{4} \) Equivalence Point
Given that the pH at the \( \frac{1}{4} \) equivalence point is 9.3, we can calculate the \( \text{pOH} \):
\[ \text{pOH} = 14 - \text{pH} = 14 - 9.3 = 4.7 \]

### Step 3: Use the Henderson-Hasselbalch Equation
At the \( \frac{1}{4} \) equivalence point, we can apply the Henderson-Hasselbalch equation for the weak base:
\[ \text{pOH} = \text{pK}_B + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
At the \( \frac{1}{4} \) equivalence point:
- Concentration of salt \( [\text{Salt}] = \frac{1}{4}X \)
- Concentration of base \( [\text{Base}] = 1 - \frac{1}{4}X = \frac{3}{4}X \)

Substituting these values into the equation:
\[ 4.7 = \text{pK}_B + \log\left(\frac{\frac{1}{4}X}{\frac{3}{4}X}\right) = \text{pK}_B + \log\left(\frac{1}{3}\right) \]
This simplifies to:
\[ 4.7 = \text{pK}_B - \log(3) \]
Given \( \log(3) = 0.48 \):
\[ 4.7 = \text{pK}_B - 0.48 \]
Thus, we can solve for \( \text{pK}_B \):
\[ \text{pK}_B = 4.7 + 0.48 = 5.18 \]

### Step 4: Calculate \( \text{pOH} \) at the \( \frac{3}{4} \) Equivalence Point
Now, we need to find \( \text{pOH} \) at the \( \frac{3}{4} \) equivalence point:
- Concentration of salt \( [\text{Salt}] = \frac{3}{4}X \)
- Concentration of base \( [\text{Base}] = 1 - \frac{3}{4}X = \frac{1}{4}X \)

Using the Henderson-Hasselbalch equation again:
\[ \text{pOH} = \text{pK}_B + \log\left(\frac{\frac{3}{4}X}{\frac{1}{4}X}\right) = \text{pK}_B + \log(3) \]
Substituting \( \text{pK}_B = 5.18 \) and \( \log(3) = 0.48 \):
\[ \text{pOH} = 5.18 + 0.48 = 5.66 \]

### Step 5: Calculate \( \text{pH} \) at the \( \frac{3}{4} \) Equivalence Point
Finally, we can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 5.66 = 8.34 \]

### Final Answer
The pH at the \( \frac{3}{4} \) equivalence point is **8.34**.

---