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 \( \boxed{0.13 \, \text{N}} \).

Answer 2

To determine the normality of the resulting solution with respect to NO₃⁻, we need to consider the contributions from both solutions.

Step 1: Calculate moles of NO₃⁻ from each solution.

Ca(NO₃)₂ Solution:

The concentration of Ca(NO₃)₂ is \( M \times 10^{-1} \). For Ca(NO₃)₂, each formula unit provides 2 NO₃⁻ ions. Thus, the effective concentration of NO₃⁻ is \( 2 \times M \times 10^{-1} \). Given the volume is 100 mL (or 0.1 L), the moles of NO₃⁻ are:

\( \text{Moles of NO₃⁻ from Ca(NO₃)₂} = 2 \times M \times 10^{-1} \times 0.1 = 0.02M \) moles

KNO₃ Solution:

The concentration of KNO₃ is \( M \times 10^{-1} \), and it provides 1 NO₃⁻ ion per formula unit. The effective concentration of NO₃⁻ is \( M \times 10^{-1} \). Given the volume is 200 mL (or 0.2 L), the moles of NO₃⁻ are:

\( \text{Moles of NO₃⁻ from KNO₃} = M \times 10^{-1} \times 0.2 = 0.02M \) moles

Step 2: Total moles of NO₃⁻.

Total moles of NO₃⁻ from both solutions:

\( 0.02M + 0.02M = 0.04M \) moles

Step 3: Calculate the normality of the resulting solution.

The total volume of the resulting solution is:

\( 100 \text{ mL} + 200 \text{ mL} = 300 \text{ mL} = 0.3 \text{ L} \)

Normality (N) is defined as moles of solute per liter of solution. Thus, the normality with respect to NO₃⁻ is:

\( \text{Normality (N)} = \frac{0.04M}{0.3} = 0.13M \)

Hence, the normality of the resulting solution with respect to NO₃⁻ is 0.13 N.