To tackle your question, let's break down the concepts of molarity and normality, especially in the context of hydroxide ions (OH-) and sodium ions (Na+). Understanding these terms will help us analyze the solution you've described and determine the correctness of the statements provided.
Understanding Molarity and Normality
Molarity (M) is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution. Normality (N), on the other hand, is a measure of concentration that accounts for the reactive capacity of a solute. For acids and bases, normality is often related to the number of protons (H+) or hydroxide ions (OH-) that can be donated or accepted.
What Does Molarity with Respect to OH- or Na+ Mean?
When we refer to molarity with respect to OH- ions, we're looking at how many moles of hydroxide ions are present in one liter of the solution. Similarly, for Na+ ions, it indicates the concentration of sodium ions in the solution. This is crucial for understanding how these ions will react in a chemical reaction, such as neutralization.
Calculating Molarity of OH- Ions
In your case, we have a solution of magnesium hydroxide, Mg(OH)2. Each formula unit of Mg(OH)2 produces two hydroxide ions (OH-) when it dissociates in water. Let's calculate the molarity of OH- ions in the solution:
- First, determine the number of moles of Mg(OH)2 in the M/5 solution:
- Volume of the solution = 100 mL = 0.1 L
- Molarity of Mg(OH)2 = 1/5 M = 0.2 M
- Moles of Mg(OH)2 = Molarity × Volume = 0.2 mol/L × 0.1 L = 0.02 moles
- Since each mole of Mg(OH)2 produces 2 moles of OH-, the moles of OH- = 0.02 moles × 2 = 0.04 moles
- Now, to find the molarity of OH- in the final diluted solution (1 L):
- Molarity of OH- = Moles of OH- / Volume of solution = 0.04 moles / 1 L = 0.04 M
Determining Concentration of Na+ Ions
Next, let's find the concentration of sodium ions (Na+) from the sodium carbonate (Na2CO3) added to the solution:
- Mass of Na2CO3 = 1.06 g
- Molar mass of Na2CO3 = 23 (Na) × 2 + 12 (C) + 16 (O) × 3 = 106 g/mol
- Moles of Na2CO3 = Mass / Molar mass = 1.06 g / 106 g/mol = 0.01 moles
- Each mole of Na2CO3 produces 2 moles of Na+, so moles of Na+ = 0.01 moles × 2 = 0.02 moles
- To find the molarity of Na+ in the final solution:
- Molarity of Na+ = Moles of Na+ / Volume of solution = 0.02 moles / 1 L = 0.02 M
Evaluating the Statements
Now, let's evaluate the statements based on our calculations:
- A. Normality of the solution is 0.08N: This requires further calculation based on the total acidic or basic capacity. We need to consider the contributions from H2SO4 and Mg(OH)2.
- B. Molarity of the solution with respect to OH- ions is 0.04M: This is correct based on our calculations.
- C. Concentration of the resulting solution with respect to Na+ ions is 0.02M: This is also correct.
- D. 40 mL of 0.1N HCl will completely neutralize 50 mL of the above solution: This requires a neutralization calculation, which we can explore further.
In summary, the molarity with respect to OH- and Na+ ions helps us understand the concentration of these ions in the solution, which is essential for predicting how they will behave in reactions. The calculations show that statements B and C are correct. If you need further clarification on any specific part, feel free to ask!
