What is the pH of a solution made by mixing 20.00 mL of 0.100 M HCl with 40.00 mL of 0.100 M KOH? Assume that the volumes of the solutions are additive.Group of answer choices0.4812.5213.521.48

To solve this problem, we need to first determine the moles of HCl and KOH in the solution.

1. Calculate the moles of HCl: Moles = Molarity * Volume(L) Moles of HCl = 0.100 M * 20.00 mL / 1000 = 0.002 moles

2. Calculate the moles of KOH: Moles of KOH = 0.100 M * 40.00 mL / 1000 = 0.004 moles

3. Determine the excess reactant: Since KOH is in excess (0.004 moles - 0.002 moles = 0.002 moles), the solution will be basic.

4. Calculate the concentration of OH- ions: [OH-] = moles of excess KOH / total volume(L) [OH-] = 0.002 moles / (20.00 mL + 40.00 mL) / 1000 = 0.033 M

5. Calculate the pOH of the solution: pOH = -log[OH-] pOH = -log(0.033) = 1.48

6. Calculate the pH of the solution: pH = 14 - pOH pH = 14 - 1.48 = 12.52

So, the pH of the solution is 12.52.