Final answer:
To find the pH after titrating a solution of HClO4 with LiOH, calculate the moles of both, determine the concentration of OH- in the new solution volume, calculate pOH, and then use pOH to find the pH.
Step-by-step explanation:
To determine the pH of the solution after adding 50.0 mL of 0.54 M LiOH to a 150.0 mL sample of 0.18 M HClO4, first calculate the moles of HClO4 and LiOH present initially: Moles of HClO4 = 0.18 M × 0.150 L = 0.027 mol, Moles of LiOH = 0.54 M × 0.050 L = 0.027 mol. Since the stoichiometry of the acid-base reaction is 1:1, all the HClO4 will be neutralized by the LiOH. As the volume of the solution has changed due to the addition of LiOH, we need to calculate the total volume of the solution to find the concentration of excess OH-: Total volume = 150.0 mL (original) + 50.0 mL (added) = 200.0 mL = 0.200 L
Since the moles of HClO4 and LiOH are equal, there are no excess H+ ions, and we only have the concentration of OH- in solution, which will be the same 0.027 mol in the new volume: Concentration of OH- = Moles of OH- / Total volume = 0.027 mol / 0.200 L = 0.135 M
The pOH of the solution can then be calculated: pOH = -log[OH-] = -log(0.135)
And the pH is calculated using the relationship pH + pOH = 14: pH = 14 - pOH
The precise pH value can be determined from this calculation. Since both HClO4 and LiOH are strong electrolytes (completely ionize in solution), the calculated pH would reflect the true pH of the solution.