Question

what is the concentration of potassium ions in a solution obtained by mixing 180 ml of 0.100m potassium hydroxide (aq) with 28- ml of 0.150 m potassium sulfate (aq)

Answer 1

To find the concentration of potassium ions after mixing potassium hydroxide with potassium sulfate, you must calculate moles from each and divide by total volume, yielding a concentration of 0.127 M.

Step-by-step explanation:

To calculate the concentration of potassium ions in the solution after mixing 180 ml of 0.100M potassium hydroxide (KOH) with 28 ml of 0.150M potassium sulfate (K2SO4), follow these steps:

1. Calculate the moles of potassium ions contributed by KOH:
   Moles of K+ from KOH = Volume (L) × Molarity (M) = 0.18 L × 0.100 M = 0.018 moles of K+.
2. Calculate the moles of potassium ions contributed by K2SO4: Moles of K+ from K2SO4 = 2 × (Volume (L) × Molarity (M)) = 2 × (0.028 L × 0.150 M) = 0.0084 moles of K+.
3. Add the moles of K+ from each solution to get the total moles in the mixture: Total moles of K+ = 0.018 moles + 0.0084 moles = 0.0264 moles of K+.
4. Calculate the total volume of the mixture in liters: Total volume = 0.18 L + 0.028 L = 0.208 L.
5. Divide the total moles of K+ by the total volume to get the concentration of potassium ions:
   [K+] = 0.0264 moles / 0.208 L = 0.127 M.

Therefore, the concentration of potassium ions in the mixed solution is 0.127 M.