Question

What is the pH change when 19.5 mL of 0.108 M NaOH Is added to 71.0 mL of a buffer solution consisting of 0.127 M NH3 and 0.161 M NH4? ( for ammonium ion is 5.6*10^-10.) pH change =

Answer 1

ka for NHy + = 5.68 10-10 Kb for NH3 = 10-14 kw 10-14 10-14 = 1.79 x 10-5 ka 5.68 10-10 Befor adding NaOH [NH3 ] = 0.127M [NH4+] = 0.161M Using Henderson Hanalbalch equation pkb= -log ky = -log (1.79810-5) pOH = pkb + 109 [NHy t] (NH₃] pkb = 4.75 pH= 14-port - 14-pH = 4.75 +109 10.161m) por = 14-pH 10.127M pH = 9.15 ÞH before addition - 9.15 After adding NaOH. mmol = concentration (ro) x volume (ML) mmol of NooH = 0.108M X 19.5 mL = 2.106 mmol mmol of NH3 = 0.127M X 7.0ML = 9.017 mmol mmol of NH4+ = 0.161 M X 70mL = 11.431 mmol 710ml + 19.5 mL = 90.5 mL Now, final volume of solution = [NaOH] = 2.106 mmol 90.5 mL 0.02327M (NH3 ] = mmol 9.0178 mmol M final volume 90. 5ml [NHyt] = mmol ting I volume = = 11.431 mmol 90.5ml 0.12631M NaOH, strong base, crociates NaOH Cagl Natlaql After dinociation 0.02327 completely in aqwow solution + OH- (aq) = [OH-] = 0.023 27 0.02327

ICE Table + + H₂O(1) concentration (MI NHy + (aq) 0.12631 - 0.02327 OH lag) 0.02327 - 0.02327 NH3199) 0.09964 + 0.02327 0.12291 0.10304 {NHy+] = 0.10304 M [NH3) = 0.12291M using Henderson Hassalbalch equation pOH = pp Ko + log (NH4+) (NH3] 14-pH = 4.75 + 10.10304 M 0.12291 M pH = 9,32 pH after addition = 9.32 change in ÞH = ÞH after addition - pH before addition = 9.32 - 9.15 = 0.17 Aniwer