Question

1. 25.0 mL of a 0.100 M solution of NH, is titrated with 0.150M HCl. After 10.0 mL of the HCl has been added, the resultant solution is: A) Basic and before the equivalence point B) Basic and after the equivalence point C) Acidic and before the equivalence point D) Acidic and after the equivalence point E) Neutral and at the equivalence point

Answer 1

Answer to the problem Titration of NH₃ with Hel: Strength of NH3 = 0.100 M Volume of NHg = 25 ml Moles of Mtz = 25 x0.100 x 10 %< kx mot = 205x103 mor Strength of Hel = 0.150 M volume of Hel = 10 min in moles of Hel = 10x 0.150 x 103 xxx mor = 1.5*103 mot , Moles of Nitz in excess after titration = (215_115)x103 mot = 1:0 x103 mot i. The resultant solution is - Basie and before the equivalence point.