The formula weight of an acid is `82 amu .In titration 100 cm^(3)` of a solution of this acid containing `39.0 g` of the acid per litre were completely neutralised by `95.0 cm^(3)` of aqueous `NaOH` containing `40.0 g` of `NaOH` per litre. What is the basicity of the acid?

Options

A. 2
B. 1
C. 4
D. 3

Answer

To solve the problem step by step, we need to determine the basicity of the acid based on the given data. The basicity of an acid refers to the number of hydrogen ions (H⁺) it can donate in a reaction.

### Step 1: Calculate the Normality of the Acid
The normality (N) of the acid can be calculated using the formula:
\[ N = \frac{\text{Given weight}}{\text{Equivalent weight}} \times \frac{1}{\text{Volume in liters}} \]
Given:
- Given weight of the acid = 39 g/L
- Molecular weight of the acid = 82 g/mol

The equivalent weight of the acid can be expressed as:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{Basicity (n)}} \]
Thus, we can rewrite normality as:
\[ N = \frac{39}{\frac{82}{n}} = \frac{39n}{82} \]

### Step 2: Calculate the Normality of NaOH
Similarly, we calculate the normality of NaOH:
\[ N = \frac{\text{Given weight}}{\text{Equivalent weight}} \times \frac{1}{\text{Volume in liters}} \]
Given:
- Given weight of NaOH = 40 g/L
- Molecular weight of NaOH = 40 g/mol
- n factor for NaOH = 1 (since it can donate 1 OH⁻)

Thus, the normality of NaOH is:
\[ N_{NaOH} = \frac{40}{40} = 1 \text{ N} \]

### Step 3: Use the Neutralization Equation
For complete neutralization, the equivalents of acid must equal the equivalents of base:
\[ \text{Equivalents of acid} = \text{Equivalents of NaOH} \]
This can be expressed as:
\[ N_{acid} \times V_{acid} = N_{NaOH} \times V_{NaOH} \]
Where:
- \( V_{acid} = 100 \, \text{cm}^3 = 0.1 \, \text{L} \)
- \( V_{NaOH} = 95 \, \text{cm}^3 = 0.095 \, \text{L} \)

Substituting the known values:
\[ \frac{39n}{82} \times 0.1 = 1 \times 0.095 \]

### Step 4: Solve for n (Basicity)
Now we can solve for n:
\[ \frac{39n}{82} \times 0.1 = 0.095 \]
\[ 39n \times 0.1 = 0.095 \times 82 \]
\[ 39n = \frac{0.095 \times 82}{0.1} \]
\[ 39n = 0.779 \]
\[ n = \frac{0.779}{39} \approx 2 \]

### Conclusion
The basicity of the acid is 2, indicating that it is a dibasic acid.