Consider the titration of 30.0 mL of 0.251 M weak base B (Kb = 1.3 x 10⁻¹⁰) with 0.150 M HI. What would be the pH of the solution after the addition of 100.0 mL of HI?

Answer 1

B + HI ==> HB⁺ + I^-

mols B = 30.0 ml x 1 L / 1000 ml x 0.251 mol/L = 0.00753 mols

mols HI =100.0 ml x 1 L / 1000 ml x 0.150 mol/L = 0.0150 mols HI

All of the 0.00753 mols of B will be converted to 0.00753 mols of HB⁺

And there will be excess HI in the amount of 0.0150 - 0.00753 = 0.00747 mols

Final volume = 30 ml + 100 ml = 130 mls = 0.130 L

Final [HI] = 0.00747 mol / 0.130 L = 0.0575 M

Final [HB⁺] = 0.00753 mol / 0.130 L = 0.0579 M

The pH will be most significantly determined by the [HI] since it is a strong acid. The effect on pH of the HB⁺ will be insignificant.

pH = -log [H⁺] = -log 0.0575 M

pH = 1.24