Question

Suppose 0.80 L of 3.0 M aqueous K2MnO4 is combined with 0.20 L of 2.0 M aqueous KMnO4 to make a solution. What is the concentration of potassium ions in the solution?

A) 2.56 M.
B) 3.40 M.
C) 2.80 M.
D) 2.20 M.

Answer 1

Suppose 0.80 L of 3.0 M aqueous K₂MnO₄ is combined with 0.20 L of 2.0 M aqueous KMnO₄, the concentration of potassium ions is C) 2.80 M.

Step-by-step explanation:

To find the concentration of potassium ions in the solution, we need to calculate the total moles of potassium ions present in the solution and then divide it by the total volume of the solution.

Given:

• Volume of 3.0 M K₂MnO₄ solution = 0.80 L
• Volume of 2.0 M KMnO₄ solution = 0.20 L

Step 1: Calculate the moles of potassium ions in each solution:

• Moles of K+ ions in 3.0 M K₂MnO₄ solution = Molarity × Volume of solution = 3.0 mol/L × 0.80 L = 2.4 mol
• Moles of K+ ions in 2.0 M KMnO₄ solution = Molarity × Volume of solution = 2.0 mol/L × 0.20 L = 0.4 mol

Step 2: Sum up the moles of potassium ions:

• Total moles of K+ ions = 2.4 mol + 0.4 mol = 2.8 mol

Step 3: Calculate the concentration of potassium ions:

• Concentration of K+ ions = Total moles of K+ ions / Total volume of solution = 2.8 mol / (0.80 L + 0.20 L) = 2.8 mol / 1 L = 2.8 M

Therefore, the concentration of potassium ions in the solution is 2.8 M. The correct answer is C) 2.80 M.