Final answer:
To determine the pH after the addition of KOH to HNO₃, the moles of both reactants were calculated, and the excess of H⁺ was determined. The concentration of H⁺ in the final solution was then calculated, leading to a pH calculation of approximately 0.541.
Step-by-step explanation:
To calculate the pH after 18.38 ml of 0.650 M KOH is added to 60.0 ml of 0.575 M HNO₃, we need to determine the moles of KOH and HNO₃, see if there's an excess of OH⁻ or H⁺, and then calculate the pH.
Firstly, calculate the moles of KOH added: moles of KOH = volume (L) × concentration (M) = 0.01838 L × 0.650 M = 0.011947 moles of KOH.
Next, calculate the moles of HNO₃: moles of HNO₃ = 0.0600 L × 0.575 M = 0.0345 moles of HNO₃.
KOH completely reacts with HNO₃ since they are a strong base and a strong acid, respectively. The reaction goes to completion, KOH + HNO₃ → KNO₃ + H₂O, so calculate the remaining moles after the reaction.
Moles remaining = initial moles of HNO₃ - moles of KOH = 0.0345 moles - 0.011947 moles = 0.022553 moles of HNO₃.
Since there are still moles of HNO₃ remaining, the solution is acidic. Now, calculate the concentration of the H⁺ ions in the new total volume (0.060 L + 0.01838 L): [H⁺] = 0.022553 moles / (0.07838 L) ≈ 0.2878 M.
Finally, the pH is given by -log([H⁺]): pH = -log(0.2878) ≈ 0.541.