A 10.0-mL sample of \(0.250 \mathrm{MHNO}_{3}\) solution is titrated with $0.100 M$ KOH solution. Calculate the pH of the solution after the following volumes of base have been added: (a) \(20.0 \mathrm{~mL}\), (b) \(24.9 \mathrm{~mL}\) (c) \(25.0 \mathrm{~mL}\) (d) \(25.1 \mathrm{~mL}\), (e) \(30.0 \mathrm{~mL}\).

Before the equivalence point, we have an excess of acid in the solution. First, we will calculate the moles of remaining \(\mathrm{HNO}_{3}\) and the new total volume of the solution. Moles of \(\mathrm{HNO}_{3}\) = (0.010 L)(0.250 mol/L) = 0.00250 mol Moles of KOH = (0.020 L)(0.100 mol/L) = 0.00200 mol Since we have an excess of \(\mathrm{HNO}_{3}\), we will subtract the moles of KOH from the moles of \(\mathrm{HNO}_{3}\). Moles of remaining \(\mathrm{HNO}_{3}\) = 0.00250 mol - 0.00200 mol = 0.00050 mol Now, we calculate the new total volume of the solution (acid + base): Total Volume = 10.0 mL + 20.0 mL = 30.0 mL or 0.030 L Next, we calculate the concentration of remaining \(\mathrm{HNO}_{3}\): Concentration = 0.00050 mol / 0.030 L = 0.0167 M Since \(\mathrm{HNO}_{3}\) is a strong acid, it completely dissociates in water. The pH can be calculated using the remaining concentration of \(\mathrm{HNO}_{3}\): \( pH = -\log_{10} [H^{+}] \) \( pH = -\log_{10} [0.0167] \) pH = 1.78 Hence, the pH after adding 20.0 mL of KOH solution is 1.78.

First, we calculate the moles of added KOH: Moles of KOH = (0.0249 L)(0.100 mol/L) = 0.00249 mol This volume of KOH is just before the equivalence point, so we still have an excess of \(\mathrm{HNO}_{3}\) in the solution. We subtract the moles of KOH from the moles of \(\mathrm{HNO}_{3}\): Moles of remaining \(\mathrm{HNO}_{3}\) = 0.00250 mol - 0.00249 mol = 0.00001 mol Now, we calculate the new total volume of the solution: Total Volume = 10.0 mL + 24.9 mL = 34.9 mL or 0.0349 L Next, we calculate the concentration of remaining \(\mathrm{HNO}_{3}\): Concentration = 0.00001 mol / 0.0349 L = 0.00029 M Finally, we calculate the pH using the concentration of \(\mathrm{HNO}_{3}\): \( pH = -\log_{10} [0.00029] \) pH = 3.54 Hence, the pH after adding 24.9 mL of KOH solution is 3.54.

At the equivalence point, the moles of acid and base are equal. In this case, moles of KOH = 0.00250 mol, which means this is equal to the moles of \(\mathrm{HNO}_{3}\). Since nitric acid and potassium hydroxide are strong acid and strong base, they react to form a neutral solution. Total Volume = 10.0 mL + 25.0 mL = 35.0 mL or 0.035 L pH = 7 Hence, the pH at the equivalence point (after adding 25.0 mL of KOH) is 7.

Now, we are in the region past the equivalence point, where we have an excess of base in the solution. We calculate the moles of remaining KOH: Moles of added KOH = (0.0251 L)(0.100 mol/L) = 0.00251 mol Moles of remaining KOH = 0.00251 mol - 0.00250 mol = 0.00001 mol Next, we calculate the new total volume of the solution: Total Volume = 10.0 mL + 25.1 mL = 35.1 mL or 0.0351 L Then, we calculate the concentration of remaining KOH: Concentration = 0.00001 mol / 0.0351 L = 0.00029 M Now, we calculate the pOH using the concentration of remaining KOH: \( pOH = -\log_{10} [OH^{-}] \) \( pOH = -\log_{10} [0.00029] \) pOH = 3.54 Next, we calculate the pH using the relationship: pH = 14 - pOH pH = 14 - 3.54 pH = 10.46 Hence, the pH after adding 25.1 mL of KOH solution is 10.46.

We are still in the region past the equivalence point. We calculate the moles of remaining KOH: Moles of added KOH = (0.030 L)(0.100 mol/L) = 0.00300 mol Moles of remaining KOH = 0.00300 mol - 0.00250 mol = 0.00050 mol Next, we calculate the new total volume of the solution: Total Volume = 10.0 mL + 30.0 mL = 40.0 mL or 0.040 L Then, we calculate the concentration of remaining KOH: Concentration = 0.00050 mol / 0.040 L = 0.0125 M Now, we calculate the pOH using the concentration of remaining KOH: \( pOH = -\log_{10} [0.0125] \) pOH = 1.90 Next, we calculate the pH using the relationship: pH = 14 - pOH pH = 14 - 1.90 pH = 12.1 Hence, the pH after adding 30.0 mL of KOH solution is 12.1.