Question

For the titration of 50. mL of 0.10 M ammonia with 0.10 M HCl, calculate the pH For ammonia, NH3, Kh = 1.8 x 105. (a) Before the addition of any HCl solution pH= The number of significant digits is set to 4; the tolerance is +/-2% (b) After 20. mL of the acid has been added pH = The number of siqnificant digits is set to 3; the tolerance is +/-2% (c) After half of the NH3 has been neutralized pH The number of significant digits is set to 3; the tolerance is +/-2% (d) At the equivalence point pH The number of significant digits is set to 3; the tolerance is +/-2%

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Answer 1

a) Before any HCl addition NH3 + H₂O - NHyt toh 1. 8 o -5 eq'm; C-cd ca ca cicone of Nth = 0.1M 2: degree of dissociation :ka=cd² (Ostwald dilutions = 1 kq - 18 xlo-s = 0.01342 Com-3 = CX = (0.1) X0.0/342) = 1.342 xlo 3 M POH = -log [OH-] = -log (1.342 X/03) por= 2.87

14-2.87 pH = 14-POH LpH = 11-13 After ZomL of 16 0.1 MHCI Zomyx 0.1 mol = 2mmol of Hal NH₂ + HCL NHutch. (Buse (Acid) (salt) 1 mol of NH₃ reacts with 1 mo of Ha to form 1 mol of salt ammol and will react with 2 mmol of Base to produce 2 mmol salt. [salt) = salt = 2mmol Volume (50ml+20mL) Base acid = 2 h mol 70 mL = 0.0286M [Base] =3 mmol = 0.042869 (50+20m2 - Out of 5 mmol, 2 momol base have reacted to form salt. pOH = pkb + log (Salt) = 4.745 + log/0.0286 ) 0.04266 pOH = 4.57 pH=9.43] (Base] =1342 c) At half NH3 neutralised half equivalent point pH = pka & pot = pkp at half eq.pt. POH= 4.74 1PH = 9.26]

d) At equivalence point, all NH3 has been reacted with acid to form salt. 5 mmol of NH3 neutralised by Samol acid to form 5 mmd salt Volume of arid moles molarity = 5 mmol = 50mL oil molet = 0.05M [salt] = 5 mmol (50+50)ght Alid Base : Salt is of weak Base NH3 Strong Avid Hel pH=7-4 [pkb tleg c] Cicone. Of helt = 0.05M 27 - + [4.745 +logo-05) PH= 5.28] Thus, at equivalence point pH was found to be 5.28