Question

An analytical chemist is titrating 249.1 ml of a 1.200 M solution of hydrazoic acid (HN) with a 0.4500 M solution of KOH. The pk of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 481.8 ml of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places. pH = 0

Answer 1

Molarity of Hal = 1.2009 Volume in L = 1. 200mole x 249.14 I 1000 1 Mhz HNspot koteas) - KN3+H₂O volume of th₃ = 249.1 mL & Polarity of HN no. of moles of Haz - Molarity & Volume in L = 1.200 = 0.299 moles no. of moles of koH= Molarity of koHx Volume of kofta no. of roles of KOH= 0.450X481-8x mole x k = 0.217 moles 1000 L pka of HN3 = 4.72. Tolal volume = 249.1 ml + 481.8ml = 730.9L Total volume. = 730.9 L = 0.73092 1000 looo HN3 t koH KN₃ + H₂O I 0.299 0.217 C -0.217 -0.217 +0.217 E 0.082 - 0.217 As pH = pkat log (Balt : (salt] = [kN ₂] = 0.217 m 0.7309 pH = 4.72 + log 0.297 [acid] ? facid] = [Hag] = 01082 of TpH= 5.14] (HN3] = 0.1129 0.7309 [KN3] = 0.297M 0,112