Exam 2-solutions-3 - Version 001 Exam 2 laude (51635) 1...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 001 Exam 2 laude (51635) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 6.0 points Consider the titration of equal volumes of 0.1 M HCl and 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH. Which of the following would be the same for both titrations? 1. Two of the other answers are correct. 2. the initial pH 3. the volume of NaOH added to reach the equivalence point correct 4. the pH at the half equivalence point Explanation: Both HCl and CH 2 C 3 O 2 H 2 are monoprotic acids and so if the volume concentrations are identical, the amounts of replaceable H + in both solutions are identical. 002 6.0 points Oxalic acid is a relatively strong diprotic acid with K 1 = 5 10- 2 and K 2 = 5 10- 5 . What is the [HC 2 O- 4 ] concentration in an oxalic acid solution that is 0.10 M? 1. 5 10- 5 2. 0.10 M 3. 0.025 M 4. 0.075 M 5. 0.05 M correct Explanation: K 1 = 5 10- 2 K 2 = 5.3 10- 5 In this case, since were considering a solu- tion of H 2 C 2 O 4 and HC 2 O- 4 , we should use the expression for K 1 here: K 1 = [H + ] [HC 2 O 4- ] [H 2 C 2 O 4 ] Substituting in values for the concentrations and K 1 , we have 5 10- 2 = ( x ) ( x ) (0 . 10- x ) As H 2 C 2 O 4 has a relatively large K 1 , we CANNOT assume that x is very small com- pared to 0.10 and consequently we must use the quadratic equation to solve for x : 5 10- 2 (0 . 10- x ) = x 2 . 005- 5 10- 2 x = x 2 x 2 + 5 10- 2 x- . 005 = 0 Using the quadratic equation, we discover that x is 0.05. Bearing in mind that we called [HC 2 O- 4 ] x in the equation, [HC 2 O- 4 ] is 0.05 M. 003 6.0 points For 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). For salycylic acid, p K a = 2.97. 1. 10.98 correct 2. 10.02 3. 12.02 4. 12.30 5. 7.00 Explanation: 004 6.0 points 100 mL of 0.50 M BaCl 2 is added to 100 mL of 0.50 M HCl. Which of the following equations is the correct charge balance for this system? 1. 2[Ba 2+ ] + [H + ] = [OH- ] + [Cl- ] correct 2. 2[Ba 2+ ] = [Cl- ] 3. [Ba 2+ ] = 2[OH- ] 4. [Ba 2+ ] + [H + ] = 3[Cl- ] 5. [Ba 2+ ] + [H + ] = [OH- ] + [Cl- ] 6. 2[Ba 2+ ] + [H + ] = 2[OH- ] + 2[Cl- ] Version 001 Exam 2 laude (51635) 2 7. 2[Ba 2+ ] + [H + ] = 2[OH- ] + [Cl-...
View Full Document

Page1 / 6

Exam 2-solutions-3 - Version 001 Exam 2 laude (51635) 1...

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