0.01 mole `H_(3)PO_(x)` is completely neutralised by 0.56 gram of KOH hence :

Options

A. x=3 and given acid is dibasic
B. x=2 and given acid is monobasic
C. x=3 and given acid is monobasic
D. x=4 and given acid forms three series of salt

Answer

To solve the problem step by step, we need to determine the value of \( x \) in the compound \( H_3PO_x \) based on the neutralization reaction with KOH.

### Step 1: Understand the Reaction
The neutralization reaction can be represented as:
\[ H_3PO_x + KOH \rightarrow \text{Salt} + H_2O \]
In this reaction, the number of equivalents of the acid must equal the number of equivalents of the base.

### Step 2: Calculate the Number of Moles of KOH
Given that the mass of KOH is 0.56 grams, we need to calculate the number of moles of KOH using its molar mass.

- Molar mass of KOH:
- Potassium (K) = 39 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol
- Total = 39 + 16 + 1 = 56 g/mol

Now, calculate the number of moles of KOH:
\[ \text{Number of moles of KOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.56 \text{ g}}{56 \text{ g/mol}} = 0.01 \text{ moles} \]

### Step 3: Determine the Gram Equivalents of KOH
The n-factor for KOH is 1 (since it can donate one hydroxide ion). Therefore, the gram equivalents of KOH can be calculated as:
\[ \text{Gram equivalents of KOH} = \text{Number of moles} \times \text{n-factor} = 0.01 \times 1 = 0.01 \text{ equivalents} \]

### Step 4: Calculate the Gram Equivalents of \( H_3PO_x \)
Since the acid is completely neutralized by KOH, the gram equivalents of \( H_3PO_x \) must also equal 0.01 equivalents.

### Step 5: Relate the Gram Equivalents to the n-factor of \( H_3PO_x \)
The formula for gram equivalents is:
\[ \text{Gram equivalents} = \text{Number of moles} \times \text{n-factor} \]
For \( H_3PO_x \):
\[ 0.01 = 0.01 \times n \]
From this, we can see that \( n \) (the n-factor for \( H_3PO_x \)) must be 1.

### Step 6: Determine the Value of \( x \)
The n-factor of \( H_3PO_x \) indicates how many protons (H\(^+\)) can be replaced. Since \( H_3PO_x \) has 3 hydrogen atoms and the n-factor is 1, it means only one hydrogen is replaceable. Therefore, the structure of the acid is \( H_3PO_2 \) (where \( x = 2 \)).

### Step 7: Determine the Basicity
Since there is only one replaceable hydrogen ion, \( H_3PO_2 \) is classified as a monobasic acid.

### Conclusion
Thus, the value of \( x \) is 2, and the acid \( H_3PO_2 \) is monobasic.