The normality of 2 M `H_(3)BO_(3)` is

Options

A. 6
B. 4
C. 2
D. 1

Explanation

To find the normality of 2 M H₃BO₃, we can follow these steps:

### Step 1: Understand the relationship between normality, molarity, and the n factor.
Normality (N) is defined as:
\[ \text{Normality (N)} = \text{Molarity (M)} \times \text{n factor} \]
Where:
- Molarity (M) is the concentration of the solution in moles per liter.
- n factor is the number of equivalents of the solute that react or are produced in a reaction.

### Step 2: Identify the molarity of the solution.
From the question, we know that the molarity (M) of H₃BO₃ is given as 2 M.

### Step 3: Determine the n factor for H₃BO₃.
The n factor for an acid is the number of protons (H⁺ ions) it can donate. H₃BO₃ (boric acid) can donate 1 proton (H⁺) when it dissociates in water:
\[ \text{H}_3\text{BO}_3 \rightarrow \text{B(OH)}_4^- + \text{H}^+ \]
Thus, the n factor for H₃BO₃ is 1.

### Step 4: Calculate the normality.
Now that we have both the molarity and the n factor, we can calculate the normality:
\[ \text{Normality (N)} = \text{Molarity (M)} \times \text{n factor} \]
\[ \text{Normality (N)} = 2 \, \text{M} \times 1 = 2 \, \text{N} \]

### Conclusion
The normality of 2 M H₃BO₃ is 2 N.

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