Question

What is the normality of HCl if 15 ml of 0.1N Na\(_2\)CO\(_3\) is consumed to titrate 10 ml of HCl?

• A. 1.5
• B. 1.0
• C. 0.5
• D. 0.15

Hint:

To quickly solve basic titration problems, multiply the volume and normality of the known reactant, then divide by the volume of the unknown: $(15 \times 0.1) / 10 = 1.5 / 10 = 0.15\text{ N}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Volumetric titration is governed by the law of equivalence.
According to this law, at the equivalence point of a titration, the number of equivalents of the acid is exactly equal to the number of equivalents of the base.
Key Formula or Approach:
The normality equation is:
\[ N_1 V_1 = N_2 V_2 \]
where:
\( N_1 \) and \( V_1 \) are the normality and volume of the base ($\text{Na}_2\text{CO}_3$)
\( N_2 \) and \( V_2 \) are the normality and volume of the acid ($\text{HCl}$)

Step 2: Detailed Explanation:
Let us perform the calculations:
Given:
Normality of \( \text{Na}_2\text{CO}_3 \) base (\( N_1 \)) = $0.1\text{ N}$
Volume of \( \text{Na}_2\text{CO}_3 \) base (\( V_1 \)) = $15\text{ mL}$
Volume of \( \text{HCl} \) acid (\( V_2 \)) = $10\text{ mL}$
We need to find the normality of \( \text{HCl} \) (\( N_2 \)):
Using the normality equation:
\[ 0.1 \times 15 = N_2 \times 10 \]
\[ 1.5 = 10 N_2 \]
\[ N_2 = \frac{1.5}{10} = 0.15\text{ N} \]

Step 3: Final Answer:
The normality of the HCl solution is 0.15 N, which corresponds to option (D).