Question

I need help to answer the bold questions please. More so help on 100% of HCl volume added and 105% of HCl volume added, but if 20%, 50%, and 95% can be shown so i can compare my answers that would be great! Thank you

pH Titration Curve itrated with a 0.50 M HCl solution. A 40.0 mL sample of a 0.30 M solution of NH. on when the following percentages of the HCl solution rear Calculate the pH of the solution when the following ner reach the equivalence point have been added. olution before the titration begins (0% of the Plot the data you determined for the pH at various points in the titration What is the pH of the ammonia solution before the titration required HCI added)? • 2% of the total volume of HCI? • 5% of the total volume of HCI? o 20% of the total volume of HCI? • 35% of the total volume of HCI? o 50% of the total volume of HCI? • 65% of the total volume of HCI? • 90% of the total volume of HCI? o 95% of the total volume of HCI? . 98% of the total volume of HCI? o 100% of the total volume of HCI? . 101% of the total volume of HCI? o 105% of the total volume of HCI? • 120% of the total volume of HCI?

Kb of ammonia = 1.8 x 10^-5

Answer 1

weak base vs strong acid(S.A)

no of mol of NH3 = 40*0.3 = 12 mmol

no of mol of HCl required to reach equivalence point = 12 mmol

a) before addition of S.A

pkb of NH3 = -logkb = -log(1.8*10^-5) = 4.74

C = concentration of NH3 = 0.1 M

pOH = 1/2(pkb-logC)

      = 1/2(4.74-log0.3)

      = 2.63

pH = 14-2.63 = 11.37

b) afetr 20% of the total volume of HCl

no of mol of HCl added = 12*20/100 = 2.4 mmol = No of mol of NH4+

no of mol of NH3 = 40*0.3 = 12 mmol

pOH = pkb+log(NH4+/NH3)

     = 4.74 + log(2.4/(12-2.4))

     = 4.14

pH = 14-4.14 = 9.86

c) afetr 50% of the total volume of HCl

no of mol of HCl added = 12*50/100 = 6 mmol = No of mol of NH4+

no of mol of NH3 = 40*0.3 = 12 mmol

pOH = pkb+log(NH4+/NH3)

     = 4.74 + log(6/(12-6))

     = 4.74

pH = 14-4.74 = 9.26

d) afetr 95% of the total volume of HCl

no of mol of HCl added = 12*95/100 = 11.4 mmol = No of mol of NH4+

no of mol of NH3 = 40*0.3 = 12 mmol

pOH = pkb+log(NH4+/NH3)

     = 4.74 + log(11.4/(12-11.4))

     = 6.02

pH = 14-6.02 = 7.98

c) afetr 100% of the total volume of HCl

no of mol of HCl added = 12 mmol = No of mol of NH4cl

volume of HCl added = 12/0.5 = 24 ml

no of mol of NH3 = 40*0.3 = 12 mmol

concentration of salt(NH4Cl) = n/V

                               = 12/(40+24)

                               = 0.1875 M
                              

pH = 7-1/2(pkb+logC)

    = 7-1/2(4.74+log0.1875)

    = 5

   

d) afetr 100% of the total volume of HCl

no of mol of HCl added = 12*105/100 = 12.6 mmol  

volume of HCl added = 12.6/0.5 = 25.2 ml

no of mol of NH3 = 40*0.3 = 12 mmol

concentration of excess HCl = (12.6-12)/(25.2+40)

                              = 0.0092 M

pH = -log(H3O+)

     = -log0.0092

     = 2.04