When a 20 mL of 0.08 M weak base BOH is titrated with 0.08 HCl, the pH of the solution at the end point is 5. What will be the pOH if 10 mL of 0.04 M NaOH is added to the resulting solution ? [Given : log 2 = 0.30 and log 3 = 0.48]

Step-by-step solution

BOH + HCl ⟶ BCl + H 2 O Initial 1 . 6 mM 1 . 6 mM 0 0 0 0 1 . 6 mM 1 . 6 mM Total volume = 40 mL [BCl] = 1.6/40 = 0.04M p H = 1 2 p K w - p K b - log C 5 = 1 2 1 4 - p K b - log 0 . 0 4 pK b = 5.4 BCl + NaOH ⟶ BOH + NaCl 1 . 6 mM 0 . 4 mM 0 0 - 0 . 4 mM - 0 . 4 mM + 0 . 4 mM + 0 · 4 mM --------------- --------------- --------------- --------------- 1 · 2 mM 0 0 . 4 mM 0 . 4 mM p OH = p K b + log B + BOH = 5.4 + log 1.2 0.4 = 5 . 4 + log 3 = 5 . 4 + 0 . 4 8 = 5 . 8 8