`pH` of an aqueous solution of `0.6M NH_(3)` and `0.4M NH_(4)Cl` is `9.4 (pK_(b) = 4.74)`. The new `pH` when `0.1M Ca(OH)_(2)` solution is added to it.

Options

A. `9.86`
B. `10.14`
C. `10.2`
D. `10.86`

Explanation

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

### Step 1: Understand the Initial Conditions
We have a buffer solution made of:
- 0.6 M NH₃ (weak base)
- 0.4 M NH₄Cl (salt of the weak base)

The initial pH of this buffer solution is given as 9.4, and we know the pKₐ value of the weak base (NH₃) is related to the pKₑ value (pKₑ = 14 - pKₐ).

### Step 2: Calculate the Initial pOH
Since we know the initial pH is 9.4, we can calculate the initial pOH using the formula:
\[ pOH = 14 - pH \]
\[ pOH = 14 - 9.4 = 4.6 \]

### Step 3: Determine the Effect of Adding Ca(OH)₂
When we add 0.1 M Ca(OH)₂ to the buffer solution, it dissociates into:
\[ Ca(OH)₂ \rightarrow Ca^{2+} + 2OH^{-} \]
This means that for every 1 mole of Ca(OH)₂, we get 2 moles of OH⁻ ions. Therefore, from 0.1 M Ca(OH)₂, we will get:
\[ [OH^-] = 0.1 \times 2 = 0.2 M \]

### Step 4: Update the Concentrations of NH₃ and NH₄Cl
According to the buffer action:
- The concentration of NH₃ (the weak base) will increase by the amount of OH⁻ added.
- The concentration of NH₄Cl (the salt) will decrease by the same amount.

Thus, we have:
- New concentration of NH₃ = 0.6 M + 0.2 M = 0.8 M
- New concentration of NH₄Cl = 0.4 M - 0.2 M = 0.2 M

### Step 5: Calculate the New pOH Using the Henderson-Hasselbalch Equation
For a basic buffer, the pOH can be calculated using the Henderson-Hasselbalch equation:
\[ pOH = pK_b + \log\left(\frac{[Salt]}{[Base]}\right) \]
Given that \(pK_b = 4.74\):
\[ pOH = 4.74 + \log\left(\frac{0.2}{0.8}\right) \]
Calculating the logarithm:
\[ \log\left(\frac{0.2}{0.8}\right) = \log(0.25) = -0.602 \]
Thus:
\[ pOH = 4.74 - 0.602 = 4.138 \]

### Step 6: Calculate the New pH
Finally, we can find the new pH using:
\[ pH = 14 - pOH \]
\[ pH = 14 - 4.138 = 9.862 \]

### Final Answer
The new pH of the solution after adding 0.1 M Ca(OH)₂ is approximately **9.86**.

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