Part A Express your answer using two decimal places. 2.85 Submit My Answers Give

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Part A 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 Express your answer using two decimal places. SubmitMy AnswersGive Up Try Again; 2 attempts remaining Part D Express your answer using two decimal places. SubmitMy AnswersGive Up Try Again; 5 attempts remaining Part E Express your answer using two decimal places. SubmitMy AnswersGive Up Part F This question will be shown after you complete previous question(s). Part F This question will be shown after you complete previous question(s). Provide FeedbackContinue 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

Explanation / Answer

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

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