Consider mixing 150 ml, 0.1M, HCI with 100 ml, 0.2M, KOH solution. Determine the pH of final solution.



Answer to a math question Consider mixing 150 ml, 0.1M, HCI with 100 ml, 0.2M, KOH solution. Determine the pH of final solution.

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To determine the pH of the final solution, we need to calculate the concentration of the resulting solution after mixing HCI and KOH.

Step 1: Calculate the number of moles of HCI and KOH:
- HCI: moles = volume (L) x concentration (M) = 0.150 L x 0.1 M = 0.015 mol
- KOH: moles = volume (L) x concentration (M) = 0.100 L x 0.2 M = 0.020 mol

Step 2: Determine the limiting reagent:
Since the stoichiometric ratio of HCI to KOH is 1:1, the limiting reagent is the one with the lesser number of moles, which in this case is HCI.

Step 3: Calculate the remaining moles of the excess reactant:
- Excess KOH: moles = moles - limiting reagent moles = 0.020 mol - 0.015 mol = 0.005 mol

Step 4: Calculate the total volume of the final solution:
- Total volume = volume of HCI + volume of KOH = 150 ml + 100 ml = 250 ml = 0.250 L

Step 5: Calculate the concentration of the final solution:
- Final concentration = total moles / total volume = (moles of HCI + moles of excess KOH) / total volume = (0.015 mol + 0.005 mol) / 0.250 L = 0.08 M

Step 6: Calculate the pOH of the final solution:
- pOH = -log[OH-]
- Since KOH is a strong base and fully dissociates, [OH-] = concentration of KOH = 0.2 M
- pOH = -log(0.2) = 0.70

Step 7: Calculate the pH of the final solution:
- pH + pOH = 14
- pH = 14 - pOH = 14 - 0.70 = 13.30

Answer: The pH of the final solution is 13.30.

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