If first dissociation of `X(OH)_(3)` is `100%` where as second dissociation is `50%` and third dissociation is negligible then the pH `4xx10^(-3) M X(OH)_(3)` is :

Options

A. `11.78`
B. `10.78`
C. `2.5`
D. `2.22`

Answer

To solve the problem, we need to find the pH of a solution of `X(OH)₃` with a concentration of `4 x 10^(-3) M`. The dissociation information provided is as follows:

1. The first dissociation of `X(OH)₃` is 100%.
2. The second dissociation is 50%.
3. The third dissociation is negligible.

### Step-by-Step Solution:

**Step 1: Determine the first dissociation.**
- The first dissociation of `X(OH)₃` can be represented as:
\[ X(OH)₃ \rightarrow X(OH)₂^+ + OH^- \]
- Since the first dissociation is 100%, all of `X(OH)₃` will dissociate. Therefore, if the initial concentration is `C = 4 x 10^{-3} M`, then:
\[ [X(OH)₂^+] = 4 x 10^{-3} M \quad \text{and} \quad [OH^-] = 4 x 10^{-3} M \]

**Step 2: Determine the second dissociation.**
- The second dissociation of `X(OH)₂^+` can be represented as:
\[ X(OH)₂^+ \rightarrow X(OH)^+ + OH^- \]
- The second dissociation is 50%, meaning half of the `X(OH)₂^+` will dissociate. The concentration of `X(OH)₂^+` after the first dissociation is `4 x 10^{-3} M`, so:
\[ \text{Dissociated amount} = 0.5 \times 4 x 10^{-3} M = 2 x 10^{-3} M \]
- Therefore, the concentration of `X(OH)^+` formed is also `2 x 10^{-3} M`, and the additional `OH^-` produced from this dissociation is:
\[ [OH^-] = 2 x 10^{-3} M \]

**Step 3: Calculate total OH⁻ concentration.**
- The total concentration of `OH^-` ions is the sum from both dissociations:
\[ [OH^-]_{\text{total}} = [OH^-]_{\text{from 1st}} + [OH^-]_{\text{from 2nd}} = 4 x 10^{-3} M + 2 x 10^{-3} M = 6 x 10^{-3} M \]

**Step 4: Calculate pOH.**
- The pOH can be calculated using the formula:
\[ pOH = -\log[OH^-] \]
- Substituting the value:
\[ pOH = -\log(6 x 10^{-3}) = -\log(6) - \log(10^{-3}) = -\log(6) + 3 \]
- Using the approximation \(\log(6) \approx 0.78\):
\[ pOH \approx 3 - 0.78 = 2.22 \]

**Step 5: Calculate pH.**
- The relationship between pH and pOH is given by:
\[ pH + pOH = 14 \]
- Therefore, we can find pH:
\[ pH = 14 - pOH = 14 - 2.22 = 11.78 \]

### Final Answer:
The pH of the `4 x 10^{-3} M` solution of `X(OH)₃` is approximately **11.78**.