`60ml` of `0.1M NH_(4)OH` is mixed with `40 ml` of `0.1M HCl`. Which of the following statement(s) is/are INCORRECT about the resulting solution/mixture? (Given `pK_(b)(NH_(4)OH)=4.74,log2=0.3`)

Options

A. a) The `pH` of the resulting solution is 5.04
B. b) The resulting solution is a basic buffer.
C. c) The `pOH` of the resulting solution is 9.26
D. d) The resulting solution is acidic due to hydrolysis of `NH_(4)Cl`

Explanation

To solve the problem, we need to analyze the reaction between `NH₄OH` (ammonium hydroxide) and `HCl` (hydrochloric acid) and determine the properties of the resulting solution. Here’s the step-by-step solution:

### Step 1: Calculate the number of moles of `NH₄OH` and `HCl`
- **For `NH₄OH`:**
- Volume = `60 ml = 0.060 L`
- Concentration = `0.1 M`
- Moles of `NH₄OH` = Volume × Concentration = `0.060 L × 0.1 mol/L = 0.006 moles`

- **For `HCl`:**
- Volume = `40 ml = 0.040 L`
- Concentration = `0.1 M`
- Moles of `HCl` = Volume × Concentration = `0.040 L × 0.1 mol/L = 0.004 moles`

### Step 2: Determine the reaction
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: Identify the limiting reactant
- We have `0.006 moles` of `NH₄OH` and `0.004 moles` of `HCl`.
- Since `HCl` is the limiting reactant, it will completely react with `0.004 moles` of `NH₄OH`.

### Step 4: Calculate the remaining moles after the reaction
- Moles of `NH₄OH` remaining = Initial moles - Moles reacted = `0.006 - 0.004 = 0.002 moles`
- Moles of `HCl` remaining = `0` (since it is completely reacted)

### Step 5: Calculate the concentration of the resulting solution
- Total volume after mixing = `60 ml + 40 ml = 100 ml = 0.1 L`
- Concentration of `NH₄OH` remaining = \(\frac{0.002 \text{ moles}}{0.1 \text{ L}} = 0.02 \text{ M}\)
- Concentration of `NH₄Cl` formed = \(\frac{0.004 \text{ moles}}{0.1 \text{ L}} = 0.04 \text{ M}\)

### Step 6: Calculate the pOH and pH of the solution
Using the formula for pOH:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Weak Base}]}\right) \]
Where:
- \(\text{pK}_b = 4.74\)
- \([\text{Salt}] = 0.04 \text{ M}\) (concentration of `NH₄Cl`)
- \([\text{Weak Base}] = 0.02 \text{ M}\) (concentration of remaining `NH₄OH`)

Calculating pOH:
\[ \text{pOH} = 4.74 + \log\left(\frac{0.04}{0.02}\right) \]
\[ \text{pOH} = 4.74 + \log(2) \]
Given \(\log(2) = 0.3\):
\[ \text{pOH} = 4.74 + 0.3 = 5.04 \]

Now, calculate pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 5.04 = 8.96 \]

### Step 7: Analyze the statements
1. **pH of the resulting solution is 5.04** - **Incorrect** (pH is 8.96)
2. **The resulting solution is a basic buffer** - **Correct**
3. **pOH of the resulting solution is 9.26** - **Incorrect** (pOH is 5.04)
4. **The resulting solution is acidic due to hydrolysis of NH₄Cl** - **Incorrect** (the solution is basic)

### Conclusion
The incorrect statements are:
- pH of the resulting solution is 5.04
- pOH of the resulting solution is 9.26
- The resulting solution is acidic due to hydrolysis of NH₄Cl