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Title: Titration and calculating the pH after adding certain mL of HCl?? Post by: fodog414 on Mar 28, 2015 Consider the titration of 100.0 mL of 0.500 M CH3NH2 by 0.250 M HCl.
(Kb for CH3NH2 = 4.4×10-4) Part 1 Calculate the pH after 0.0 mL of HCl added. pH = Part 2 Calculate the pH after 25.0 mL of HCl added. pH = Part 3 Calculate the pH after 80.0 mL of HCl added. pH = Part 4 Calculate the pH at the equivalence point. pH = Part 5 Calculate the pH after 300.0 mL of HCl added. pH = Part 6 At what volume of HCl added, does the pH = 10.64? We never went over this, but if someone could explain a few or if they would like all of them it should be straight forward with a thorough explanation. Its due tomorrow :-\so any help is truly appreciated. Thank you! Title: Re: Titration and calculating the pH after adding certain mL of HCl?? Post by: psyche360 on Mar 28, 2015 Similar Question Consider the titration of 100.0 mL of 0.500 M CH3NH2 by 0.250 M HCl. (Kb for CH3NH2 = 4.4 x 10^-4) At what volume of HCl Added does the pH=10.64 [OH-] = sqrt(Kb*0.5) = 0.01483 M initial milli-moles of OH- = (0.01483*100) = 1.483 when pH = 10.64, pOH = 14-10.64 = 3.36 final [OH-] = 4.365x10^-4 M let V ml of 0.25 M HCl be added total volume of solution = (100+V) ml final milli-moles of OH- = (100+V)*(4.365x10^-4) change in milli-moles of OH- = 1.483-[(100+V)*(4.365x10^-4)] = [1.44-(4.365x10^-4)*V] this must be equal to milli-moles of H+ added which is equal to 0.25V hence, 0.25V = [1.44-(4.365x10^-4)*V] solving, V = 5.75 ml so, pH becomes 10.64 when 5.75 ml acid is added Similar Question Consider the titration of 100.0mL of 0.500 M NH₃ (Kb = 1.8 x 10^-5) with a 0.500 M HCl. After 50.0 mL HCl have been added, the [H₃0] of the solution is: Original moles of NH3 = 100/1000 x 0.500 = 0.05 mol Moles of HCl added = 50/1000 x 0.500 = 0.025 mol HCl + NH3 => NH4Cl After addition of HCl, Moles of NH3 left = 0.05 - 0.025 = 0.025 mol Moles of NH4Cl formed = 0.025 mol Kb(NH3) = 1.8 x 10-5 => Ka(NH4Cl) = Kw/Kb(NH3) = 10-14/1.8 x 10-5 = 5.556 x 10-10 pH = pKa + log ([NH3]/[NH4Cl]) = -log (5.556 x 10-10) + log [(0.025/0.150)/(0.025/0.150)] = 9.255 pH = -log [H3O+] = 9.255 [H3O+] = 10-9.255 = 5.556 x 10-10 M Title: Re: Titration and calculating the pH after adding certain mL of HCl?? Post by: fodog414 on Mar 28, 2015 Thank you so much for the quick reply!
So following your format for part 2 i got 10.49 part 3 10.46 but for part 5 i hit a problem i got -.025 finding what is left .05-.075? Should this be reversed? and you don't have to say but is part 1-4 correct? Title: Re: Titration and calculating the pH after adding certain mL of HCl?? Post by: psyche360 on Mar 29, 2015 Tbh, I didn't answer these myself, I was hoping you'd know what to do with them. As far as I know, they might even be wrong :(
Title: Re: Titration and calculating the pH after adding certain mL of HCl?? Post by: fodog414 on Mar 29, 2015 Yeah they were all incorrect :(
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