172% of titration

Question

Titrate a 0.05 M solution of 87 cm3 of monovalent strong base with 0.1 M HCl. What is the pH of the solution at 118% titration?

Answer 1

Let XOH be the monovalent strong base.

XOH + HCl == XCl + H2O

87 cm3 x 1 L / 1000 cm3 x 0.05 mol/L = 0.00435 moles XOH

At 100% titration, moles HCl = 0.00435 mols

Volume of HCl: (0.1 mol/L)(x L) = 0.00435 mols and x = 0.0435 L = 43.5 ml

An additional 18% will be 43.5 x 1.18 = 51.3 mls HCl

Total moles HCl added = 0.0513 L x 0.1 mol/L = 0.00513 mols HCl

Moles xs HCl = 0.00513 mols - 0.00435 mols = 0.00078 mols

Total volume = 87 ml + 51.3 ml = 138 mls

[H⁺] = 0.00078 mol / 0.138 L = 0.00565 M

pH = -log 0.00565

pH = 2.25