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`
When 100 mL of `0.1` M `NH_(4)OH` is added to 50 mL of `0.1M` HCl solution , the pH is

Options

A. `1.6021 `
B. ` 12.3979`
C. `4.7447`
D. `9.2553`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of NH₄OH and HCl
- **For NH₄OH:**
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume in liters} \]
\[ \text{Number of moles of NH₄OH} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{mol} \]

- **For HCl:**
\[ \text{Number of moles of HCl} = 0.1 \, \text{M} \times 0.05 \, \text{L} = 0.005 \, \text{mol} \]

### Step 2: Write the balanced chemical equation
The reaction between NH₄OH and HCl can be represented as:
\[ \text{NH}_4\text{OH} + \text{HCl} \rightarrow \text{NH}_4\text{Cl} + \text{H}_2\text{O} \]

### Step 3: Determine the limiting reactant and remaining moles
- Initially, we have:
- 0.01 mol of NH₄OH
- 0.005 mol of HCl

- Since HCl is the limiting reactant, it will completely react with NH₄OH:
\[ \text{Remaining moles of NH₄OH} = 0.01 - 0.005 = 0.005 \, \text{mol} \]
\[ \text{Moles of NH₄Cl formed} = 0.005 \, \text{mol} \]

### Step 4: Identify the buffer solution
After the reaction, we have:
- 0.005 mol of NH₄OH (weak base)
- 0.005 mol of NH₄Cl (salt)

This mixture forms a basic buffer solution.

### Step 5: Calculate the pOH of the buffer solution
Using the formula for pOH of a buffer:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Here, both the concentration of salt (NH₄Cl) and the base (NH₄OH) are equal, so:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{0.005}{0.005}\right) \]
Since \(\log(1) = 0\):
\[ \text{pOH} = 4.7447 + 0 = 4.7447 \]

### Step 6: Calculate the pH from pOH
Using the relation:
\[ \text{pH} + \text{pOH} = 14 \]
We can find pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 4.7447 = 9.2553 \]

### Final Answer
The pH of the solution is **9.2553**.

---