When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations :
`pH = 1/2 [pK_(w) +pK_(a) + logc] ` (for salt of weak acid and strong base .)
`pH = 1/2 [pK_(w) - pK_(b) - logc] ` (for salt of weak base and strong acid ) .
`pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] ` (for weak acid and weak base ).
where 'c' represents the concentration of salt .
When a weak acid or a weak base not completely neutralized by strong base or strong acid
respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation :
`pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"])`
Answer the following questions using the following data :
`pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14`
50 mL `0.1 ` M NaOH is added to 50 mL of `0.1 ` M `CH_(3)COOH ` solution , the pH will be

Options

A. `4.7447`
B. `9.2553`
C. `8.7218`
D. `1.6021`

Answer

To solve the problem, we need to determine the pH of the solution when 50 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid (CH₃COOH). This is a classic case of a weak acid being partially neutralized by a strong base, resulting in the formation of a buffer solution.

### Step 1: Calculate the moles of acetic acid and NaOH

1. **Calculate moles of CH₃COOH:**
\[ \text{Moles of CH₃COOH} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \]

2. **Calculate moles of NaOH:**
\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \]

### Step 2: Determine the reaction between acetic acid and NaOH

The reaction between acetic acid (weak acid) and sodium hydroxide (strong base) is:
\[ \text{CH₃COOH} + \text{NaOH} \rightarrow \text{CH₃COONa} + \text{H₂O} \]

Since both reactants are present in equal amounts (0.005 moles), they will completely react with each other.

### Step 3: Calculate the concentrations of the resulting buffer

After the reaction, we will have:
- 0 moles of CH₃COOH (since it is completely neutralized)
- 0.005 moles of CH₃COONa (the salt formed)

The total volume of the solution after mixing is:
\[ 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]

Now, calculate the concentration of the acetate ion (CH₃COO⁻):
\[ \text{Concentration of CH₃COO⁻} = \frac{0.005 \, \text{moles}}{0.1 \, \text{L}} = 0.05 \, \text{M} \]

### Step 4: Use the Henderson-Hasselbalch equation to find the pH

Since we have formed a buffer solution, we can use the Henderson-Hasselbalch equation:
\[ \text{pH} = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

In this case:
- \( pK_a = 4.7447 \)
- \([\text{Salt}] = 0.05 \, \text{M}\)
- \([\text{Acid}] = 0 \, \text{M}\) (since all acetic acid has been neutralized)

Since \([\text{Acid}] = 0\), we cannot directly use the equation. However, we can consider that the pH will be determined by the salt formed, which is the acetate ion.

### Step 5: Calculate the pH using the pK_b of the conjugate base

Since we have no acetic acid left, we can use the \( pK_b \) of the acetate ion to find the pOH and then convert it to pH:
\[ pK_b = 4.7447 \quad \text{(given)} \]
Using the relation:
\[ pH + pOH = pK_w = 14 \]
We can find pOH:
\[ pOH = pK_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Since there is no base present, we can assume that the concentration of the base is negligible, leading to:
\[ pOH = 4.7447 \]
Now, calculate pH:
\[ pH = 14 - pOH = 14 - 4.7447 = 9.2553 \]

### Final Answer
The pH of the solution after adding 50 mL of 0.1 M NaOH to 50 mL of 0.1 M acetic acid is approximately **9.26**.