A 1L solution contains 0.2M `NH_(4)OH` and 0.2M `NH_(4)Cl.` If 1.0 mL of 0.001 M HCl is added to it what will be the `[OH^(-)]` of the resulting solution `(K_(b)=2xx10^(-5))`

Options

A. `2xx10^(-5)`
B. `5xx10^(-10)`
C. `2xx10^(-3)`
D. None of these

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the components of the solution
We have a solution containing:
- **0.2 M NH₄OH** (a weak base)
- **0.2 M NH₄Cl** (the salt of the weak base, which acts as a source of NH₄⁺ ions)

### Step 2: Understand the effect of adding HCl
When we add **1.0 mL of 0.001 M HCl** to the solution, we are introducing H⁺ ions. The H⁺ ions will react with the NH₄OH (the weak base) to form NH₄⁺ and water. This will decrease the concentration of NH₄OH and increase the concentration of NH₄⁺.

### Step 3: Calculate the moles of HCl added
To find the moles of HCl added:
- Volume of HCl = 1.0 mL = 0.001 L
- Concentration of HCl = 0.001 M

Moles of HCl = Volume (L) × Concentration (M) = 0.001 L × 0.001 M = **1.0 × 10⁻⁶ moles**

### Step 4: Calculate the initial moles of NH₄OH and NH₄Cl
Since the solution is 1 L:
- Moles of NH₄OH = 0.2 M × 1 L = **0.2 moles**
- Moles of NH₄Cl = 0.2 M × 1 L = **0.2 moles**

### Step 5: Determine the new concentrations after HCl addition
When HCl is added, it reacts with NH₄OH:
- Moles of NH₄OH after reaction = 0.2 moles - 1.0 × 10⁻⁶ moles ≈ **0.2 moles** (since the change is negligible)
- Moles of NH₄Cl after reaction = 0.2 moles + 1.0 × 10⁻⁶ moles ≈ **0.200001 moles**

The total volume of the solution after adding HCl is approximately 1.001 L.

New concentrations:
- [NH₄OH] = 0.2 moles / 1.001 L ≈ **0.1998 M**
- [NH₄Cl] = 0.200001 moles / 1.001 L ≈ **0.1998 M**

### Step 6: Use the Henderson-Hasselbalch equation
For a basic buffer, the Henderson-Hasselbalch equation can be used:
\[ \text{pOH} = \text{pK}_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]

Where:
- \( K_b = 2 \times 10^{-5} \)
- \( \text{pK}_b = -\log(K_b) = -\log(2 \times 10^{-5}) \approx 4.70 \)

Now, substituting the concentrations:
\[ \text{pOH} = 4.70 + \log\left(\frac{0.200001}{0.1998}\right) \]

Since the concentrations of salt and base are very close, the log term will be very small.

### Step 7: Calculate the concentration of OH⁻
Using the relationship:
\[ [OH^-] = K_b \cdot \frac{[\text{Salt}]}{[\text{Base}]} \]

Substituting the values:
\[ [OH^-] = 2 \times 10^{-5} \cdot \frac{0.200001}{0.1998} \approx 2 \times 10^{-5} \]

Thus, the concentration of OH⁻ in the resulting solution is approximately **2 × 10⁻⁵ M**.

### Final Answer
The concentration of \([OH^-]\) in the resulting solution is **2 × 10⁻⁵ M**.

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