test3solution_pdf

# 1 55 2 67 3 45 correct 4 23 5 21 explanation both 001

• Notes
• 10

This preview shows 4 out of 7 pages.

1. 5.5 2. 6.7 3. 4.5 correct 4. 2.3 5. 2.1 Explanation: Both 0.01 moles of H 3 A and OH are added to the solution, neutralizing to give 0.01 moles of H 2 A : H 3 A + OH -→ H 2 O + H 2 A This is an amphoteric acid form, giving: pH = pK a1 + pK a2 2 = 2 . 3 + 6 . 7 2 = 4 . 5 015 3.2points Fe(OH) 3 (s) is very insoluble in water ( K sp = 1 . 6 × 10 39 ; however, Fe 3+ forms a strong complex with EDTA (FeEDTA , K f = 1 . 3 × 10 25 ). For a solution which is at equi- librium and contains Fe(OH) 3 (s) precipitate, which of the following occurs if EDTA is added to the solution? 1. More Fe(OH) 3 (s) precipitates because [Fe 3+ ] decreases. 2. Nothing happens because K sp is much smaller than K f . 3. No more Fe(OH) 3 (s) dissolves, but Fe 3+ complexes with EDTA. 4. More Fe(OH) 3 (s) dissolves. correct Explanation: 016 3.2points Please note that the question is asking for the pKb! The curve for the titration of hydroxy- lamine base (NH 2 OH) with HCl(aq) acid is given below.

Subscribe to view the full document.

Version 221 – Exam 3 – holcombe – (51395) 5 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of acid (mL) pH Estimate the p K b of hydroxylamine base. C a = 0 . 58 M, C b = 0 . 4408 M, and the volume of NH 2 OH is 100 mL. 1. 76 2. 5 . 1 correct 3. 4 . 8 4. 8 . 9 5. 38 Explanation: 0 2 4 6 8 10 12 14 0 20 40 60 80 100 120 140 Volume of acid (mL) pH (76,4 . 8) (38,8 . 9) The equivalence point of this titration is when the curve is at an inflection point; i.e. , at a volume of 76 mL . The pH at the equivalence point of this titration is 4 . 8 . The p K a for the conjugate acid of hydrox- ylamine can be found at one-half the volume of the equivalence point; i.e. , at 38 mL. The p K a is 8 . 9 from looking at the graph. p K a + p K b = p K w p K b = p K w - p K b = 14 - 8 . 9 = 5 . 1 017 3.2points In the reaction H 2 O + NH 3 OH + NH 4 + , the substances acting as Bronsted-Lowry acids are 1. H 2 O and OH . 2. H 2 O and NH 4 + . correct 3. NH 3 and OH . 4. NH 3 and NH 4 + . 5. NH 3 and H 2 O. Explanation: 018 3.2points K sp for ZnS is 1 . 1 × 10 21 . At what S 2 concentration will ZnS precipitate for a 0.20 M solution of Zn(NO 3 ) 2 ? Zn(NO 3 ) 2 is a very soluble salt. 1. 5 . 5 × 10 20 M 2. 5 . 5 × 10 21 M correct 3. 2 . 2 × 10 20 M 4. 3 . 3 × 10 11 M 5. 2 . 4 × 10 10 M Explanation:
Version 221 – Exam 3 – holcombe – (51395) 6 K sp for ZnS = 1.1 × 10 21 ZnS Zn 2+ + S 2 K sp = [Zn 2+ ] [S 2 ] Since Zn(NO 3 ) 2 is very soluble, there will be an extra 0 . 20 M Zn 2+ in solution. 1 . 1 × 10 21 = ( x + 0 . 20) ( x ) We can disregard x in x + 0 . 20 because the K sp is very small, which means ZnS doesn’t dissociate much. So the equation now be- comes 1 . 1 × 10 21 = (0 . 20) ( x ) x = 5 . 5 × 10 21 M AlternateSolution: K sp ZnS = [Zn 2+ ] [S 2 ] = 1 . 1 × 10 21 Zn 2+ + H 2 ZnS Initial 0.2 - - Change x x - Final 0 . 2 + x x - K sp = (0 . 2 + x ) ( x ) = 1 . 1 × 10 21 x will be very small compared to 0.2, so 0 . 2 + x 0 . 2, and K sp = 0 . 2 x = 1 . 1 × 10 21 x = 1 . 1 × 10 21 0 . 2 = 5 . 5 × 10 21 019 3.2points Of the following, a solution with which pH would be the most basic solution?

Subscribe to view the full document.

You've reached the end of this preview.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern