Question

Determine the hydronium ion concentration during the titration of 6.72 mL of 0.700 M ethylamine (C2H5NH2. Kb = 4.30x10-4) by 19.4 ml of 0.700 M HBr.

Answer 1

* Concenklation af GMs ruhig = 0.700m volume of Cq HG nutog 6.72 mL 7 no. of milimales of Cp #is It? = [conc.] x [volume comes = no. of milimales of Ca Hg Nha 0.700 X 6.72 =) no. of milimales of Calls NH₂ = 4.704 mmal conc. of HBr = o.Toom Volume of her 19.4mL No. of miljmoles of HBr = [conc.] * [volume cmus] of milimales of HBr = 0.100 X 19.4 다. ब mo = 13.58mmal Now the reaction between CQH, NHQ (Base) and HBr (acid) Ca Hs Whop + HBr CH; NH Be Initial mmol 4704 13.58 final mmol 13.58-4.104 45704 As we are left with HBO in Solution ) 80; no. of milimales of HBr left= 13.58–4.704

no. of moles of Hbo left www 8.876 mmol Now HBois Shung quid; So it will dissociate as = Hoot + B8 HBr + H₂O Initial 8.876 excess מוניטור excels final mmol 8.876 8.816 11 Thus ; mo. of milimales of He of 8.876 mmol Voleme of solution 6,72 +19.45 = 26.12 mL Conc. of H3o+= no. of milimales volume (me) 8.876 26.12 =) Cone of H₂ot = 0.3398 m Thug Concesition of Hydoonium ion 0.3398 m Ane.