Question

Be sure to answer all parts. Find the pH of the equivalence point and the volume (mL) of 0.200 M HCl needed to reach the equivalence point in the titration of 65.5 mL of 0.234 M NH3.\

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Answer 1

NH₂ : Molarity = 0.234M volume = 65.5ml = 0.86552 Man pkb of NH = 4.745 pkw = 14 no-ot moles= 0.234 Ko 0655 | = 0.015327 at equivalence point moles of moles of Nity = Hal 0.015327 molarity of Hol moles of te volume of Hal 0.015327 0.2= volume of Hel volume of Hala o 0 76635L = 76.635ml اوک NH + Hol NH, CI weakbase strongacid I 10.015327 0.015327 </-0.015327 -0.015327 +0.015327 E o 1 o 10.015327 Total volume = 0.0655+ 0.076635= 0.1421352 -0.107834 M2 [6] ENHA] - 20-015327 0.142135 For this salt pH: formula pH = + 2 pk wapkblog c] = + [14-4-745-log(0-107834)] pH = soll