Question

5.00 mL of 0.250 M ammonia (NH₃) is titrated with 0.100 M hydrochloric acid (HCl). The Kb for ammonia is 1.75 x 10^-5. What volume of HCl is required to completely neutralize the NH₃?

Answer 1

- NH4 + HCl -> NH4Cl # 5 mL of_0:25 m_MHz __Molaxity = 0:25M Volume = 5 mL =5410-31_ 89 Moles of NH3 = Molarity x Volume (102) not F 0.25 x 5 x 10-3 mol =_ 1.25 x 10-3 mol - Nasz 1.25 mmol # Corrider if we add v mL of HCl Upto equevalent point for complete neutralisation Molexity = 0:1 M Volume = V mL nucl = 0.1 XV mpol so

a NH3 + HCl - NHy® + Cle • According to stoicheometry of weegsteong for complite neutralisation equal moly of NH₃ and Hel reequired so no norta = huet 1.95 mmusta 0:1 XV . SU, V = 12.5 ml