{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw5 - amg2845 hw05 McCord(53580 This print-out should have...

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

View Full Document Right Arrow Icon
amg2845 – hw05 – McCord – (53580) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The term “ K a for the ammonium ion” de- scribes the equilibrium constant for which of the following reactions? 1. NH + 4 + H 2 O NH 3 + H 3 O + correct 2. NH + 4 + OH NH 3 + H 2 O 3. NH 3 + H 2 O NH 4 + + OH 4. NH 3 + H 3 O + NH + 4 + H 2 O 5. The term is misleading, because the am- monium ion is not an acid. 6. NH 4 Cl(solid) + H 2 O NH + 4 + Cl Explanation: 002 10.0 points Hydroxylamine is a weak molecular base with K b = 6 . 6 × 10 9 . What is the pH of a 0.0500 M solution of hydroxylamine? 1. pH = 8.93 2. pH = 4.74 3. pH = 3.63 4. pH = 9.26 correct 5. pH = 9.48 6. pH = 7.12 7. pH = 10.37 Explanation: Hydroxylamine is a weak base, so use the equation to calculate weak base [OH ] con- centration (note that this is the approximate equation. Why? Because K b is very small and the concentration is reasonable) : [OH ] = radicalbig K b C b = radicalBig (6 . 6 × 10 9 ) (0 . 0500) = 1 . 82 × 10 5 After finding [OH ], you can find pH using either method below: A) pOH = log ( 1 . 82 × 10 5 ) = 4 . 74 pH = 14 4 . 74 = 9 . 26 or B) [H + ] = K w [OH ] = 1 . 0 × 10 14 1 . 82 × 10 5 = 5 . 52 × 10 10 pH = log ( 5 . 52 × 10 10 ) = 9 . 26 003 (part 1 of 2) 10.0 points Calculate the pH of the solute in an aque- ous solution of 0 . 25 M C 5 H 5 N(aq) (pyridine) if the K b is 1 . 8 × 10 09 . Correct answer: 9 . 32661. Explanation: C pyridine = 0 . 25 M K b = 1 . 8 × 10 09 C 5 H 5 N + H 2 O C 6 H 5 NH + + OH 0 . 25 0 0 x + x + x 0 . 25 x x x K b = [C 6 H 5 NH + ][OH ] [C 5 H 5 N] 1 . 8 × 10 09 = x 2 0 . 25 x x 2 0 . 25 x = [OH ] = radicalBig 0 . 25(1 . 8 × 10 09 ) = 2 . 12132 × 10 05 mol / L . The pOH is pOH = log(2 . 12132 × 10 05 ) = 4 . 67339 , and the pH is pH = 14 4 . 67339 = 9 . 32661 .
Background image of page 1

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

View Full Document Right Arrow Icon
amg2845 – hw05 – McCord – (53580) 2 004 (part 2 of 2) 10.0 points What is the percentage protonation of the solute?
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.

{[ snackBarMessage ]}