C) The weak base hydroxylamine, HONH 2 , is titrated with HCl. What is the pH af

Question

C)   The weak base hydroxylamine, HONH2, is titrated with HCl. What is the pH after 9.05 mL of 0.20 M HCl is added to 30.0 mL of 0.20 M HONH2? Kb of HONH2 = 1.1 x 10-8 (Give your answer to two decimal places.)

Quizzes-Inorganic & Qu × Take Test: Q7-Titrations × / Chegg Study l Guided Sol × Chegg Study | Guided Sox C https://www.chegg.com/homework help/questions and answers/using-titration curve shown-volume acid added-ph-pka-0-ml-b 25-mc-50-r : : Apps My Lab / Mastering WebCOM" 2.0 N Today's Front Pages N Today's Front Pages Liaison International. Chegg BoOKS STUDY MORE uided 11× BOOKS STUDY MORE Find books, solutions, tutors and more Chemistry tutors who can help right now Question A) Using the titration curve shown below, at what volume of acid added will pHpKa? Inna K. The University of No. Priti S Utkal University, Orissa 667 12 Joanna B Georgia Tech 10 934 See more tutors 0 20 40 60 80 100 120 140 vol titrant added (mL) 0 mL 25 mL 50 mL 100 mL 1 points B) Using the titration curve shown below, at what volume of acid added will the pH depend only on the amount of excess strong acid added? 12:56 PM 10/19/2015

Explanation / Answer

Solution :-

Part A ) pH= pka at the half equivalence point of the titration in the given titration plot the 100 ml acid is added to react the equivalence point of the titration therefore to reach the half equivalence point volume of acid needed = 100 ml / 2 = 50 ml

Therefore pH= pka at the 50 ml acid added .

Part B)

The 100 ml of acid added to the titration reaches the equivalence point therefore after the 100 ml of acid the pH of the solution only depends on the volume of the acid.

Therefore the pH depends only on volume of acid added when 120 ml strong acid is added

So the answer is 120 ml strong acid

Part C) Lets first calculate the moles of the HONH2 and HCl

Moles = molarity * volume in liter

Moles of HONH2 = 0.20 mol per L * 0.030 L = 0.006 mol HONH2

Moles of HCl =0.20 mol per L * 0.00905 L = 0.00181 mol HCl

After the reaction HCl react with the HONH2

Therefore new moles of HONH2 = 0.006 – 0.00181 = 0.00419 mol

Moles of conjugate acid H2ONH2+ = 0.00181 mol

New molarities at the total volume 30.0 ml + 9.05 ml = 39.05 ml = 0.03905 L are as follows

[HONH2]= 0.00419 mol / 0.03905 L = 0.1073 M

[H2ONH2+] = 0.00181 mol / 0.03905 L = 0.04635 M

Now lets calculate the pOH using the Henderson equation

pOH = pkb + log ([acid]/[base])

pkb = -log kb

       = -log 1.1*10^-8

      = 7.96

pH= 7.96 + log [0.04635/0.1073]

pH= 7.60

therefore pH= 7.60

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