It `0.1` mole `H_3 PO_x` is completely neutralised by `5.6g KOH` then select the true statement.

Options

A. x=3 and given acid is dibasic
B. x=4 and given acid has no P-H linkage
C. x=2 and given acid does not form acid salt
D. all of the above

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of KOH
To find the number of moles of KOH, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Given:
- Mass of KOH = 5.6 g
- Molar mass of KOH = 56 g/mol

Calculating the moles of KOH:

\[ \text{Number of moles of KOH} = \frac{5.6 \, \text{g}}{56 \, \text{g/mol}} = 0.1 \, \text{moles} \]

### Step 2: Determine the relationship between moles of acid and base
We know that the neutralization reaction between an acid and a base follows the stoichiometry of the reaction. We have 0.1 moles of \( H_3PO_x \) and 0.1 moles of KOH.

### Step 3: Analyze the possible values of x
We need to determine the possible values of x in \( H_3PO_x \) that can react with KOH in a 1:1 ratio.

1. **For \( H_3PO_4 \)**:
- Reaction: \( H_3PO_4 + 3 KOH \rightarrow K_3PO_4 + 3 H_2O \)
- Here, 1 mole of \( H_3PO_4 \) requires 3 moles of KOH. Therefore, 0.1 moles of \( H_3PO_4 \) would require 0.3 moles of KOH. This does not match our condition.

2. **For \( H_3PO_3 \)**:
- Reaction: \( H_3PO_3 + 2 KOH \rightarrow K_2HPO_3 + 2 H_2O \)
- Here, 1 mole of \( H_3PO_3 \) requires 2 moles of KOH. Therefore, 0.1 moles of \( H_3PO_3 \) would require 0.2 moles of KOH. This does not match our condition.

3. **For \( H_3PO_2 \)**:
- Reaction: \( H_3PO_2 + KOH \rightarrow KH_2PO_2 + H_2O \)
- Here, 1 mole of \( H_3PO_2 \) requires 1 mole of KOH. Therefore, 0.1 moles of \( H_3PO_2 \) would require 0.1 moles of KOH. This matches our condition.

### Conclusion
The only value of x that satisfies the condition of complete neutralization with 5.6 g of KOH is when \( x = 2 \). Thus, the true statement is that \( H_3PO_x \) is \( H_3PO_2 \).

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