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

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## 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.

Timmothy
4.8
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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