A 200.0 mL sample of 0.180 M HClO₄ is titrated with 0.270 M KOH.Determine the pH of the solution after the addition of 150.0 mL of KOH. A) 12.1 B) 2.65 C) 11.35 D) 1.89 E) 13.06

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The pH after the addition of KOH can be calculated by first determining the moles of HClO and KOH, then finding the moles of excess reactant, and finally calculating the pH of the resulting solution. 1. Calculate moles of HClO:$$ \text{Moles of HClO} = 0.180 \, \text{M} \times 0.200 \, \text{L} = 0.036 \, \text{moles} $$2. Calculate moles of KOH:$$ \text{Moles of KOH} = 0.270 \, \text{M} \times 0.150 \, \text{L} = 0.0405 \, \text{moles} $$Since KOH is in excess (0.0405 moles of KOH vs. 0.036 moles of HClO), we calculate the moles of excess KOH:$$ \text{Excess moles of KOH} = 0.0405 \, \text{moles} - 0.036 \, \text{moles} = 0.0045 \, \text{moles} $$3. Calculate the concentration of OH- in the final solution:$$ \text{Volume of final solution} = 200.0 \, \text{mL} + 150.0 \, \text{mL} = 350.0 \, \text{mL} = 0.350 \, \text{L} $$$$ \text{Concentration of OH-} = \frac{0.0045 \, \text{moles}}{0.350 \, \text{L}} = 0.012857 \, \text{M} $$4. Calculate the pOH and then the pH:$$ \text{pOH} = -\log(0.012857) \approx 1.89 $$$$ \text{pH} = 14 - \text{pOH} = 14 - 1.89 = 12.1 $$Therefore, the pH of the solution after the addition of 150.0 mL of KOH is approximately 12.1.

A 2000 ML Sample of 0