Question

Find the pH of the equivalence point and the volume (mL) of 0.175 M HCl needed to reach the equivalence point in the titration of 65.5 mL of 0.234 M NH3.

Answer 1

H: Molasity -o2.34 M NH1810 volume 6s.smlso.06s52 no-omales 0234x0-0655pb 245 O.015327 at eguivalence potnt moles tsN Hel 0.0153 27 moles o Molat hy dr moles ot tel Hel velume CL) O0153 27 175 volume dc volume d tel= 0.CA75P32= P7-583 m salt wakbase OI327 1oiS327 to-O15327 cto-0153 22 -015327 0I5327 o.0655+ 0-0 P7533o. I5303 Tfal volume O OIS327 O1001222 M £c Inzel O153083 Po Hhs t pe dsalf p fotmula kb pH 4-4745-1o3 (0.1o0121) 5-127 pt

ph. foamula's fon salt's weakacrd+ strong base> salf+ ,0 PHLpha og salf to Strong asid+weak base PEwpb10g] ph.