Question

Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid: (a) 8.00 mI (b) 29.30 mL pH = (c) 39.00 mL

Answer 1

Formula used, calculation and answer shown in the following image

PRTE Balanced Neutrilization reaction is KOH IHB - Hoa K Br 8:00 ml Noot mole - molarity x volume of solution in liter Maple at 30 ml mole of KOH = 0.030 X 2: 1000 = 0:0030 mele 01030 liter & ml=B008 mole of HBr = 0.0.0.8 xD: 1000 = 0'000R roole BE Roh and Hog geact equimolas proportion therefore 0:0008 mole H By react with 0:006 Imole Koh. unreacted koh remain in Solution La 0.0030 - 0008 = 0.0022 mole Volume at Solution = 30+8 = 38 ml = 8:038 liter molarity of t = No. of molel volume in likes molasity of Kot = 0.0022 0.038=0.0579 M ROH is shond base dissociate completly therefore ThoHE LOG]=00stein pot = -106 2013 -log[o-os 7g) = 1024 - 14-1024 5 12.76 OH= 14-DOH pH = \2. 4. b) 29. 30 ml No. of mole = molarity x volume of solution in the 3001 = 0.030 liter 39.30 ME 010293 likes Imole of kot 50000 X 0.030 E 0.0030 mole mole of ABS = 011000x 01029320100293 mole Kot of HBr reach equimolany therefore 10 10 0243 mole HBr Jeact with 0:00243 mole KOH mus Kot semain after reaction 0.0030-0:00243 = 0.000007. mole

volume of solution - 30+ 29120 = 89.30 at malou = 0.0311 of male volume of some molatot KAH = 000 HoiasaD-018 OHSS hand base dissociate complety nexttore DROHT=TOH pOH = 204 TOT 0.001er 104 10601R pa 2045 la- por H. 2093 1007 PH 11.07. c) 39 m N ot molemolarity xuolume of solution in it mole of KOH = 1000 X 0.030 = 00030 mole of H a looox 0.039 = 0.0034 KOHof- HXCALE Equimolax retou mole Kot react seith 0:0030 mole bag HBO remain after tubilizalon -0.0039-00030 -0.009 mole volume of solution=304 39=6am acoba liter malarity of moleluolumbo Solo Han Thites molasty of HBG=B009/01069 0.13 M |H Bg is strong and dissociate completa the recore Dhort [H20504319 Re d ag et367 log 2043)= 0.89 PH= 0.89.