Question

A 900.0 mL sample of 0.18 M HClO₄ is titrated with 0.27 M LiOH at 25 degree C. Determine the pH of the solution after the addition of 600.0 mL of LiOH.

Answer 1

Given:
M(HClO4) = 0.18 M
V(HClO4) = 900 mL
M(LiOH) = 0.27 M
V(LiOH) = 600 mL


mol(HClO4) = M(HClO4) * V(HClO4)
mol(HClO4) = 0.18 M * 900 mL = 162 mmol

mol(LiOH) = M(LiOH) * V(LiOH)
mol(LiOH) = 0.27 M * 600 mL = 162 mmol


We have:
mol(HClO4) = 1.62*10^2 mmol
mol(LiOH) = 1.62*10^2 mmol
1.62*10^2 mmol of both will react to form neutral solution
hence pH of solution will be 7
Answer: 7.00