Chemistry HW 8 - ekk86 hw08 Holcombe(53570 This print-out...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ekk86 – hw08 – Holcombe – (53570) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points At the stoichiometric point in the titra- tion oF 0.130 M HCOOH(aq) with 0.130 M KOH(aq), 1. the pH is less than 7. 2. the pH is 7.0. 3. the pH is greater than 7. correct 4. [HCO - 2 ] = 0.130 M. 5. [HCOOH] = 0.0650 M. Explanation: 002 10.0 points What is the pH at the halF-stoichiometric point For the titration oF 0.22 M HNO 2 (aq) with 0.01 M KOH(aq)? ±or HNO 2 , K a = 4 . 3 × 10 - 4 . 1. 3.37 correct 2. 7.00 3. 2.31 4. 2.16 5. 2.01 Explanation: 003 10.0 points ±or the titration oF 50.0 mL oF 0.020 M aque- ous salicylic acid with 0.020 M KOH(aq), cal- culate the pH aFter the addition oF 55.0 mL oF KOH(aq). ±or salycylic acid, p K a = 2.97. 1. 7.00 2. 12.30 3. 10.98 correct 4. 11.26 5. 12.02 Explanation: 004 10.0 points Consider the titration oF 50.0 mL oF 0.0200 M HClO(aq) with 0.100 M NaOH(aq). What is the Formula oF the main species in the solution aFter the addition oF 10.0 mL oF base? 1. ClO - correct 2. NaOH 3. ClO 2 4. ClOH 5. HClO 2 Explanation: 005 (part 1 oF 2) 10.0 points 0 10 20 30 40 50 60 0 1 2 3 4 5 6 7 8 9 10 11 12 Titration Curve mL oF NaOH pH What is the pH at the equivalence point oF this titration? 1. 4 . 83 2. 2 . 92
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
ekk86 – hw08 – Holcombe – (53570) 2 3. 10 . 39 4. 3 . 88 5. 8 . 79 correct 6. 5 . 62 7. 6 . 81 Explanation: The infection points are shown below. 0 10 20 30 40 50 60 0 1 2 3 4 5 6 7 8 9 10 11 12 Titration Curve mL oF NaOH pH (17 . 5, 4 . 83) (35, 8 . 79) 006 (part 2 oF 2) 10.0 points What is the p K a oF this acid? 1. 10 . 39 2. 3 . 88 3. 8 . 79 4. 6 . 81 5. 2 . 92 6. 5 . 62 7. 4 . 83 correct Explanation: 007 10.0 points You have a solution that is bu±ered at pH = 2.0 using H 3 PO 4 and H 2 PO - 4 (p K a1 =2 . 12; p K a2 =7 . 21; p K a3 = 12 . 68). You decide to titrate this bu±er with a strong base. 15.0 mL are needed to reach the ²rst equivalence point. What is the total volume oF base that will have been added when the second equivalence point is reached? 1. < 30 mL 2. A second equivalence point in the titra- tion will never be observed. 3. 30 mL 4. > 30 mL correct Explanation: 008 10.0 points 50.0 mL oF 0.0018 M aniline (a weak base) is titrated with 0.0048 M HNO 3 . How many mL oF the acid are required to reach the equiva- lence point? 1. 133 mL 2. 18.8 mL correct 3. Need to know the K b oF aniline. 4. Bad titration since HNO 3 is not a strong acid. 5.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/01/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

Page1 / 7

Chemistry HW 8 - ekk86 hw08 Holcombe(53570 This print-out...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online