Question

100 mL of \( M \times 10^{-1} \) Ca(NO₃)₂ and 200 mL of \( M \times 10^{-1} \) KNO₃ solutions are mixed. What is the normality of the resulted solution with respect to NO₃⁻?

• A. 0.1 N
• B. 0.2 N
• C. 0.13 N
• D. 0.066 N

Hint:

To find the normality of a mixed solution, calculate the normality of each solution separately, find the total equivalents, and then divide by the total volume.

Answer 1

The Correct Option is C

We are given the following solutions:

• 100 mL of \( M \times 10^{-1} \) Ca(NO₃)₂, which provides two moles of NO₃⁻ ions per formula unit.
• 200 mL of \( M \times 10^{-1} \) KNO₃, which provides one mole of NO₃⁻ ion per formula unit.

Step 1:
For \( \text{Ca(NO₃)₂} \):
- The normality of \( \text{Ca(NO₃)₂} \) solution is \( N_1 = M \times 10^{-1} \times 2 = 0.2 \, N \).
- Volume of \( \text{Ca(NO₃)₂} \) solution = 100 mL = 0.1 L.
- Moles of NO₃⁻ from \( \text{Ca(NO₃)₂} \) = \( N_1 \times V = 0.2 \times 0.1 = 0.02 \, \text{equivalents} \).
Step 2:
For \( \text{KNO₃} \):
- The normality of \( \text{KNO₃} \) solution is \( N_2 = M \times 10^{-1} \times 1 = 0.1 \, N \).
- Volume of \( \text{KNO₃} \) solution = 200 mL = 0.2 L.
- Moles of NO₃⁻ from \( \text{KNO₃} \) = \( N_2 \times V = 0.1 \times 0.2 = 0.02 \, \text{equivalents} \).
Step 3:
The total volume of the mixture is \( 0.1 + 0.2 = 0.3 \, L \).
The total moles of NO₃⁻ in the mixture = \( 0.02 + 0.02 = 0.04 \, \text{equivalents} \).
The normality of the resulting solution with respect to NO₃⁻ is: \[ N_{\text{total}} = \frac{0.04 \, \text{equivalents}}{0.3 \, \text{L}} = 0.13 \, \text{N}. \] Thus, the normality of the resulting solution is \( \boxed{0.13 \, \text{N}} \).