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 step by step, we will follow the process outlined in the video transcript.

### Step 1: Calculate the number of millimoles of NH₄OH and HCl

1. **Calculate the millimoles of NH₄OH:**
- Concentration of NH₄OH = 0.1 M
- Volume of NH₄OH = 100 mL = 0.1 L
- Millimoles of NH₄OH = Concentration × Volume = 0.1 mol/L × 0.1 L = 10 millimoles

2. **Calculate the millimoles of HCl:**
- Concentration of HCl = 0.1 M
- Volume of HCl = 50 mL = 0.05 L
- Millimoles of HCl = Concentration × Volume = 0.1 mol/L × 0.05 L = 5 millimoles

### Step 2: Determine the reaction between NH₄OH and HCl

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: Calculate the remaining moles after the reaction

- Initial moles of NH₄OH = 10 millimoles
- Initial moles of HCl = 5 millimoles

After the reaction:
- Moles of NH₄OH remaining = 10 - 5 = 5 millimoles
- Moles of NH₄Cl (salt formed) = 5 millimoles (since all HCl is consumed)

### Step 4: Determine the pOH of the resulting solution

Since we have a basic buffer solution (NH₄OH and NH₄Cl), we can use the formula for pOH:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]

Given:
- pK_b = 4.7447
- Concentration of salt (NH₄Cl) = 5 millimoles
- Concentration of base (NH₄OH) = 5 millimoles

Since the volumes are equal (both are in 100 mL total), the concentrations will cancel out:
\[ \text{pOH} = 4.7447 + \log\left(\frac{5}{5}\right) \]
\[ \text{pOH} = 4.7447 + \log(1) \]
\[ \text{pOH} = 4.7447 + 0 \]
\[ \text{pOH} = 4.7447 \]

### Step 5: Calculate the pH from pOH

Using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
We can find pH:
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 4.7447 \]
\[ \text{pH} = 9.2553 \]

### Final Answer

The pH of the resulting solution is approximately **9.2553**.