Find the the valency factor for following acids
`(i) CH_(3)CO OH " " (ii) NaH_(2)PO_(4) " " (iii) H_(3)BO_(3)`

Answer

To find the valency factor for the given acids, we need to determine how many hydrogen ions (H⁺) can be released by each acid when dissolved in water. The valency factor is essentially the number of moles of H⁺ ions produced per mole of acid.

Let's analyze each acid step by step:

### Step 1: Analyze CH₃COOH (Acetic Acid)
1. **Structure**: Acetic acid consists of a carboxylic group (-COOH) attached to a methyl group (CH₃).
2. **Ionization**: When acetic acid ionizes, it releases one hydrogen ion (H⁺):
\[ \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}^+ \]
3. **Valency Factor**: Since it releases 1 H⁺ ion, the valency factor for acetic acid is **1**.

### Step 2: Analyze NaH₂PO₄ (Sodium Dihydrogen Phosphate)
1. **Structure**: Sodium dihydrogen phosphate can be represented as H₂PO₄⁻.
2. **Ionization**: When NaH₂PO₄ ionizes, it can release two hydrogen ions (H⁺):
\[ \text{NaH}_2\text{PO}_4 \rightarrow \text{Na}^+ + \text{H}_2\text{PO}_4^- \rightarrow \text{Na}^+ + \text{H}^+ + \text{HPO}_4^{2-} \]
3. **Valency Factor**: Since it can release 2 H⁺ ions, the valency factor for NaH₂PO₄ is **2**.

### Step 3: Analyze H₃BO₃ (Boric Acid)
1. **Structure**: Boric acid has the formula H₃BO₃.
2. **Ionization**: When boric acid ionizes, it can release three hydrogen ions (H⁺):
\[ \text{H}_3\text{BO}_3 \rightarrow \text{B(OH)}_4^- + \text{H}^+ \]
3. **Valency Factor**: Since it can release 3 H⁺ ions, the valency factor for H₃BO₃ is **3**.

### Summary of Valency Factors
- For CH₃COOH: Valency Factor = **1**
- For NaH₂PO₄: Valency Factor = **2**
- For H₃BO₃: Valency Factor = **3**