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 is the pH of the solution at the equivalence point?

Answer 1

D 5ml of 0.250M NH₂ 0.1 MHC kn 1.75 X10-5 NH₃ + Hd → NH 4 the number of moles of NH₃ = Molarity & Volume is 0.25 M X 0.005 L - 4 0.00125 ml Acc to equation moles of NH₃ = moles of Hd. -0.00125 and Hell Vuch a moles - 0.00125ml Molarity 01 mol/L 1 0 :0125 L . - 12. 5 ml total volume = 5ml + 12.5ml en - 17.5 ml INH+ = 0.0012 5 moles a 0.0714M 17.5*10-31 NHT & HO - NH₂ at H₂O 0.0714 -a 0.0714-2 al se koka kw = 5.7 X 10-10 - 195x10-5 5.7 x 100 = [NH₃ ] [H₂ ot1 = x2 [NH + ] - 0.0714-X x = 5.7 x 10-10x0.0714 outthon (xcL0107142 21- 6.38 X10-6 - [H2011 pH = -log [H₃O+] = -log (6.38 x 10-6) = 5.19