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 find the pH of the solution after adding 50 mL of 0.1 M NaOH to 50 mL of 0.1 M acetic acid (CH₃COOH), we can follow these steps:

### Step 1: Calculate the number of moles of NaOH and CH₃COOH
- **For NaOH:**
\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume (in L)} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{mol} \, (5 \, \text{mmol}) \]

- **For CH₃COOH:**
\[ \text{Moles of CH₃COOH} = \text{Molarity} \times \text{Volume (in L)} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{mol} \, (5 \, \text{mmol}) \]

### Step 2: Determine the reaction
The reaction between NaOH (strong base) and CH₃COOH (weak acid) is:
\[ \text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]
Since both reactants are present in equal amounts (5 mmol each), they will completely neutralize each other.

### Step 3: Calculate the moles of products formed
After the reaction:
- Moles of CH₃COOH = 0
- Moles of NaOH = 0
- Moles of CH₃COONa = 5 mmol

### Step 4: Calculate the concentration of CH₃COONa
The total volume of the solution after mixing is:
\[ 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]
The concentration of CH₃COONa is:
\[ \text{Concentration of CH}_3\text{COONa} = \frac{\text{Moles}}{\text{Volume}} = \frac{0.005 \, \text{mol}}{0.1 \, \text{L}} = 0.05 \, \text{M} \]

### Step 5: Calculate the pH using the formula for the salt of a weak acid and strong base
Since CH₃COONa is a salt of a weak acid (CH₃COOH) and a strong base (NaOH), we use the formula:
\[ \text{pH} = \frac{1}{2} \left( pK_w + pK_a + \log c \right) \]
Where:
- \( pK_w = 14 \)
- \( pK_a = 4.7447 \)
- \( c = 0.05 \, \text{M} \)

Substituting the values:
\[ \text{pH} = \frac{1}{2} \left( 14 + 4.7447 + \log(0.05) \right) \]
Calculating \( \log(0.05) \):
\[ \log(0.05) = -1.3010 \]
Now substituting this back into the pH equation:
\[ \text{pH} = \frac{1}{2} \left( 14 + 4.7447 - 1.3010 \right) \]
Calculating the inside:
\[ = \frac{1}{2} \left( 17.4437 \right) = 8.72185 \]

### Final Answer
The pH of the solution is approximately **8.7218**.

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