Partial Question 6 3/5 pts e Given: HOCI has a Ka = 2.9 x 10- and a pka = 7.54 You are titrating 20.0 mL of a 0.200 M HO

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Partial Question 6 3/5 pts e Given: HOCI has a Ka = 2.9 x 10- and a pka = 7.54 You are titrating 20.0 mL of a 0.200 M HO

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Partial Question 6 3 5 Pts E Given Hoci Has A Ka 2 9 X 10 And A Pka 7 54 You Are Titrating 20 0 Ml Of A 0 200 M Ho 1
Partial Question 6 3 5 Pts E Given Hoci Has A Ka 2 9 X 10 And A Pka 7 54 You Are Titrating 20 0 Ml Of A 0 200 M Ho 1 (50.1 KiB) Viewed 43 times
Partial Question 6 3 5 Pts E Given Hoci Has A Ka 2 9 X 10 And A Pka 7 54 You Are Titrating 20 0 Ml Of A 0 200 M Ho 2
Partial Question 6 3 5 Pts E Given Hoci Has A Ka 2 9 X 10 And A Pka 7 54 You Are Titrating 20 0 Ml Of A 0 200 M Ho 2 (49.65 KiB) Viewed 43 times
Partial Question 6 3/5 pts e Given: HOCI has a Ka = 2.9 x 10- and a pka = 7.54 You are titrating 20.0 mL of a 0.200 M HOCI with a 0.200 M NaOH. The equivalence point volume is therefore 20.0 mL of NaOH. You stop the titration early when you have added 15.0 mL of NaOH to check the pH and see if it matches the theoretical value. Theoretical pH Analysis (in buffer range) . Before titration, you have 4.00 millimoles of HOCI The 15.0 mL of NaOH adds Select) millimoles of OH. • Since the OH" reacts almost completely, you should make Select millimoles of OCH This would leave 2.00 millimoles of unreacted HOCI. • Since this is now a buffer solution, the pH would equal Select . he Answer 1: 4.00 Answer 2: 3.00 Answer 3: 1.00 Answer 4: 200 Answer 5: 8.02
Incorrect Question 7 0/5 pts Using the question above, you overshoot the endpoint by adding an extra 10.0 mL of NaOH (you added 30.0 mL instead of 20.0 mL). The total volume is now 50.0 ml. What is the pH at this point? (Hint: pH past equivalence point is governed by the excess OH-in the total volume) (Hint: [OH-] = mmol of excess OH- divided by 50.0 mL)
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