Part A Express your answer using two decimal places 285 Subm

Consider the titration of a 25.0 sample of 0.110 with 0.120. Determine each of the following.

Part A the initial Express your answer using two decimal places.    = 2.85 SubmitMy AnswersGive Up Correct Part B the volume of added base required to reach the equivalence point = SubmitMy AnswersGive Up Try Again; 2 attempts remaining Part C the at 4.00 of added base Express your answer using two decimal places. = SubmitMy AnswersGive Up Try Again; 2 attempts remaining Part D the at one-half of the equivalence point Express your answer using two decimal places. = SubmitMy AnswersGive Up Try Again; 5 attempts remaining Part E the at the equivalence point Express your answer using two decimal places. = SubmitMy AnswersGive Up Part F This question will be shown after you complete previous question(s). Provide FeedbackContinue

Solution

THIS SOLVED WILL BE HELPFUL FOR YOU Ka of acetic acid = 1.74*10^-5 = [H+][OAc-]/[HOAc] HOAc = H+ + OAc- If x moles dissociate: 0.11 - x = x + x [H+][OAc-]/[HOAc] = x*x/(0.11-x) = 1.74*10^-5 Assume x <<< 0.11, then 0.11-x ~ 0.11 x^2 = 0.11*1.74*10^-5 x = [H+] = 1.38*10^-3 pH = 2.86 Check assumption: Is 0.00138 << 0.11 Yes --------------------------- After 5 ml of added base Moles base added = 0.13*0.005 = 0.00065 This equals moles of acid reacted Moles acid before reaction = 0.025*0.11 = 0.00275 Moles acid remaining = 0.00275 - 0.00065 = 0.0021 Volume = 25 + 5 = 30mL = 0.03L Concentration acid remaining = 0.0021/0.03 = 0.07M If x moles dissociate: 0.07 - x = x + x [H+][OAc-]/[HOAc] = x*x/(0.07-x) = 1.74*10^-5 Assume x <<< 0.07, then 0.07-x ~ 0.07 x^2 = 0.07*1.74*10^-5 x = [H+] = 6.343*10^-4 pH = 3.20 Check assumption: Is 0.00063 << 0.07 Yes --------------- At one-half the equivalence point we have a buffer solution with equal concentrations of acid and salt. pH = pKa = 4.76 ----------------- At equivalence point we have a solution of sodium acetate. Volume NaOH required = 0.11*25/0.13 = 21.15mL Total volume = 46.15 mL Moles sodium acetate = initial moles acid = 0.00275 concentration = 0.00275/0.04615 = 0.0596 M The acetate ion will reassociate with water to form OH- OAc- + H2O = HOAc + OH- Let x moles react: 0.0596 - x = x + x Kb = Kw/Ka = 10^-14/1.74*10^-5 = 5.75*10^-10 [HOAc][OH-]/[OAc-] = 5.75*10^-10 Let [HOAc] = [OH-] = x x*x/(0.0596 - x) = 5.75*10^-10 Assume x <<< 0.0596 then (0.0596 - x) ~ 0.0596 x^2 = 0.0596*5.75*10^-10 x = 5.854*10^-6 = [OH-] Check assumption: Is 5.854*10^-6 <<< 0.0596 ? Yes pOH = 6.23 pH = 14 - 6.23 = 7.67