100 mL solution of an acid (molar mass 82) containing 39 g acid per litre was completely neutralized by 95.0 mL of aqueous NaOH solution containing 40 g NaOH per litre. The basicity of acid is ……

Explanation

To solve the problem step by step, we need to determine the basicity of the acid using the given information.

### Step 1: Calculate the number of moles of NaOH used in the reaction.

Given:
- Concentration of NaOH = 40 g/L
- Volume of NaOH used = 95 mL = 0.095 L

First, we calculate the number of grams of NaOH in 95 mL:

\[ \text{Mass of NaOH} = \text{Concentration} \times \text{Volume} = 40 \, \text{g/L} \times 0.095 \, \text{L} = 3.8 \, \text{g} \]

Now, we calculate the number of moles of NaOH:

\[ \text{Molar mass of NaOH} = 23 \, (\text{Na}) + 16 \, (\text{O}) + 1 \, (\text{H}) = 40 \, \text{g/mol} \]

\[ \text{Number of moles of NaOH} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{3.8 \, \text{g}}{40 \, \text{g/mol}} = 0.095 \, \text{mol} \]

### Step 2: Calculate the number of moles of the acid.

Given:
- Concentration of the acid = 39 g/L
- Volume of the acid solution = 100 mL = 0.1 L

First, we calculate the number of grams of acid in 100 mL:

\[ \text{Mass of acid} = \text{Concentration} \times \text{Volume} = 39 \, \text{g/L} \times 0.1 \, \text{L} = 3.9 \, \text{g} \]

Now, we calculate the number of moles of the acid:

\[ \text{Molar mass of the acid} = 82 \, \text{g/mol} \]

\[ \text{Number of moles of acid} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{3.9 \, \text{g}}{82 \, \text{g/mol}} = 0.04756 \, \text{mol} \]

### Step 3: Relate the moles of acid to the moles of NaOH.

From the neutralization reaction, the moles of H⁺ from the acid will equal the moles of OH⁻ from NaOH:

\[ \text{Moles of H}^+ = \text{Moles of OH}^- \]

Let \( n \) be the basicity of the acid (the number of H⁺ ions that can be released). Therefore:

\[ n \times \text{moles of acid} = \text{moles of NaOH} \]

Substituting the values we calculated:

\[ n \times 0.04756 = 0.095 \]

### Step 4: Solve for the basicity \( n \).

\[ n = \frac{0.095}{0.04756} \approx 1.997 \]

Rounding this gives us:

\[ n \approx 2 \]

### Conclusion

The basicity of the acid is **2**.

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