Question

1) A 27.7 mL sample of 0.242 M diethylamine, (C₂H₅)₂NH, is titrated with 0.252 M perchloric acid. After adding 40.2 mL of perchloric acid, the pH is?

2) A 24.8 mL sample of 0.369 M ammonia, NH₃, is titrated with 0.253 M hydrobromic acid. At the equivalence point, the pH is?

Answer 1

(alts) N H Molarity = 0.242m It clos: Molarity zo. 2529 volume = 27.7m1= 0.02772 I volume a 40.2=0.04022 ma 7 no-ot moles= 6.252 X6-0402 =ooololoy lese o.242X0-0277 no ot moles = 0.00 67034 (Ets) NH+ Halos » (sets), NH₂ cloy 0.oo 67034 0.0201 304 -0.0062034 -0.00 6 2034 o 10.00.3427 Total volume = 0.0277 + 0.0402=0-06292 0-003427 -0.05047M 0-0629 [Help] =24+] = 7 0- pH = -log [H+] = -log (0.050 47) =1-297

-2 NH = molarity = 0.3699 k of NH : 1-88105 1 pkb log kb _-10g (18x105) volume=24.8m1 = 0.82482 okb=4.745 prw_14 no. of moles=o.369 X0.0248 -0.0091512 at cecivalence point moles of moles of -0.0091512 NH = HBM no. of moles of HBA Molarity of HBr - volume of HB9 (2) 0.00 91512 0.253= volume of HB7 volume of HBr = 0.036172 Nh+ HBT - NH Br weak base strong acid salt Io.00 91512 0.0091512 clo-00915121-0.0045121 to 081512 O 0260915121 -0.150094 = [c] 0.0091512 0.06097 v Total volume =0.0248+ 0.03617= 0.000972 INH, B13 - 2014 for this salt pH formula pH + [pkw - pkb Loge] pH = [14-4-745-log (0.15 609)] pH = 5.0393

pH formula's for salt's weak acid & strong base -salft to pH = ± [ptw toka + log c] strongacid & weakbase salf toto pH = ₃ [ pk w _pkbe_loge