Question

3. Determine the concentration of [H3O+] and the pH of a solution 0.025M of hydrozoic acid, HN3-. (PICTURE WITH ADDITIONAL INFO BELOW)

4. Determine the pH of a solution 0.125M of hydroxilamine, HONH2. (PICTURE WITH ADDITIONAL INFO BELOW)

3. Determine la concentración de [H3O+] y el pH de una solución 0.025M de ácido hidrazoico, HN3. HN, (g) + H20 (1) = H30+ + N3 K = 1.9 x 10-5 4. Determine el pH de una solución 0.125 M de hidroxilamina, HONH2 HONH2(g) + H20 (1) = HONH, (g) + OH-(g) Ky = 9.1 x 10-9

Answer 1

HN 3 (g) + Hgo (w - Mootto ka= 1.9x10s (given) table) MITTITTI 3 Hoof It Nă | HNo. concentration 0.0gsm concom al Coorn Initial change -30 to 18 8 equilibotim i 0.025-30 toca concentration of product foomed Concontration of product Reactant concontration Ka= karood oogs-se - At equilibtiung initial 0.025 = 1.318167 K 1.9x105 1.31x167 very longe them soo so we can - neglect ac from "0.095-8c" in equation ④

kaze Now PH 1.9x10-5xo.oas = 22 xa frax105x0.095 = 0.0006892 x = 6.88104 Encontration of H₂0l = 6.88104 for 0.025m Hydrozoic acid? We have concontration of Naota 6.88 104 we know, PH= -log[304] - - Log [6.8x107 = 3.167

HONHg thgO - HONH ton Ce table! = HONNA + oH mitial VIALTEZIILTI TIT change HONHg Onlgs - SC olas-x toc toc Equilibrium Oh formed soph will be in bassic range Poodud concentration Reactant concentration Knaca Kb - 0.125- I At equilibriom mitial oilgs L =.3736x10!! K 9:18109 As compare soo, looges value so. we discard the ac from olgs-xc and this become olas gra = 0.185 X Kb 185 x 9.1 X10 9 = 0.0000337 = 3.378105

Jca encontration of on= 3.374 105 PH" - Now As we know, [Por= -log[CH] POH = -Lag [3.374105] 4.4783 I PH+ Pola 14 PH + 4.4 = 14 PMC 14-4,4 -9.6. (Bassic)