At 25^ C , 40.0 mL of a 0.1 M weak monoacidic base BOH is titrated against 0.1 M HCl . The pH of the solution (a) at the start of the titration (when no HCl has been added) and (b) when 20.0 mL of HCl is added, are respectively: (Given: K_b = 1.0 10⁻⁵ )

Step-by-step solution

For part (a), at the start of the titration, the solution contains only the weak base BOH . The concentration of BOH is C = 0.1 M. Using the formula for the hydroxide ion concentration of a weak base: [OH^-] = K_b C = 1.0 10⁻⁵ 0.1 = 10⁻⁶ = 10⁻³ M pOH = - (10⁻³) = 3.0 Since pH + pOH = 14 at 25^ C , pH = 14 - 3.0 = 11.0 For part (b), when 20.0 mL of 0.1 M HCl is added to 40.0 mL of 0.1 M BOH : Initial millimoles of BOH = 40.0 0.1 = 4.0 mmol Millimoles of HCl added = 20.0 0.1 = 2.0 mmol The added HCl neutralizes half