If the change in pH between 50% and 80% neutralisation of a weak acid `(P^(Ka) = 6)` by a strong base is given as logr, then find 'x'?

Answer

To solve the problem of finding the value of 'x' given the change in pH between 50% and 80% neutralization of a weak acid (with \( pK_a = 6 \)) by a strong base, we can follow these steps:

### Step 1: Understand the Neutralization Process
When a weak acid (HA) is neutralized by a strong base (OH⁻), it produces its conjugate base (A⁻) and water. The reaction can be represented as:
\[ \text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O} \]

### Step 2: Initial Concentration
Assume we start with 1 M concentration of the weak acid (HA). Therefore, before any neutralization, the concentrations are:
- [HA] = 1 M
- [A⁻] = 0 M

### Step 3: Calculate pH at 50% Neutralization
At 50% neutralization, half of the weak acid is converted to its conjugate base:
- [HA] = 0.5 M
- [A⁻] = 0.5 M

Using the Henderson-Hasselbalch equation:
\[ \text{pH} = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Substituting the values:
\[ \text{pH}_1 = 6 + \log\left(\frac{0.5}{0.5}\right) = 6 + \log(1) = 6 \]

### Step 4: Calculate pH at 80% Neutralization
At 80% neutralization, 80% of the weak acid is converted:
- [HA] = 0.2 M
- [A⁻] = 0.8 M

Using the Henderson-Hasselbalch equation again:
\[ \text{pH}_2 = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Substituting the values:
\[ \text{pH}_2 = 6 + \log\left(\frac{0.8}{0.2}\right) = 6 + \log(4) \]

### Step 5: Calculate the Change in pH
Now, we find the change in pH:
\[ \Delta \text{pH} = \text{pH}_2 - \text{pH}_1 = (6 + \log(4)) - 6 = \log(4) \]

### Step 6: Relate to Given Expression
According to the problem, the change in pH is given as \( \log R \). Therefore:
\[ \log R = \log(4) \]

### Step 7: Solve for x
Since \( R = 4 \), we can conclude that:
\[ x = 4 \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{4} \]

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