PH at the equivalence point?

What is the pH at equivalence point , when 40mL of 1.00M HCl is titrated with 20mL of 0.50M HONH2?

  • Answer:

    Kb = 1.1 x 10^-8 moles HONH2 = 0.020 L x 0.50 M=0.010 moles HCl required to reach the equivalence point = 0.010 Volume HCl = 0.010 mol / 1,00 = 0.010 L total volume = 0.010 + 0.020 = 0.030 L HONH2 + H+= HONH3+ moles HONH3+ = 0.010 [HONH3+]= 0.010 / 0.030 L=0.33 M HONH3+ + H2O = HONH2 + H3O+ Ka = Kw/Kb = 1.0 x 10^-14 / 1.1 x 10^-8=9.21 x 10^-7 = x^2 / 0.33-x x = [H3O+]= 0.00055 M pH = 3.3 moles HCl = 0.040 L x 1.00 M= 0.040 moles H+ in excess = 0.040 - 0.010 = 0.030 total volume = 0.060 L [H+]= 0.030 / 0.060 =0.50 M pH = 0.30

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Other answers

Kb = 1.1 x 10^-8 moles HONH2 = 0.020 L x 0.50 M=0.010 moles HCl required to reach the equivalence point = 0.010 Volume HCl = 0.010 mol / 1,00 = 0.010 L total volume = 0.010 + 0.020 = 0.030 L HONH2 + H+= HONH3+ moles HONH3+ = 0.010 [HONH3+]= 0.010 / 0.030 L=0.33 M HONH3+ + H2O = HONH2 + H3O+ Ka = Kw/Kb = 1.0 x 10^-14 / 1.1 x 10^-8=9.21 x 10^-7 = x^2 / 0.33-x x = [H3O+]= 0.00055 M pH = 3.3 moles HCl = 0.040 L x 1.00 M= 0.040 moles H+ in excess = 0.040 - 0.010 = 0.030 total volume = 0.060 L [H+]= 0.030 / 0.060 =0.50 M pH = 0.30

Dr.A

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