3 5 4 11 5 3 explanation nacl is a neutral salt it

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3. 5. 4. 11. 5. 3. Explanation: NaCl is a neutral salt. It does not alter pH. 009 3.3points If you have four weak bases and four K b val- ues and don’t know which matches with which but do know that hydrazine is the strongest, which K b below belongs to hydrazine? 1. 1 . 7 × 10 - 6 correct 2. 1 . 2 × 10 - 7 3. 5 × 10 - 12 4. 7 . 8 × 10 - 9 Explanation: Base strength is directly proportional to K b . 010 3.3points 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 = 9.26 correct 2. pH = 7.12 3. pH = 3.63 4. pH = 4.74 5. pH = 10.37 6. pH = 9.48 7. pH = 8.93 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 . 26 011 3.3points The pH of 0 . 015 M HNO 2 (nitrous acid) aque- ous solution was measured to be 2 . 63. What is the value of p K a of nitrous acid? 1. 3.36 correct 2. 0.97 3. 2.90 4. 2.63 5. 4.24 6. 3.59 Explanation: M = 0 . 015 M pH = 2 . 63 Analyzing the reaction with molarities, HNO 2 + H 2 O H 3 O + + NO - 2 0 . 015 - 0 0 - x - x x 0 . 015 - x - x x [H 3 O + ] = [NO - 2 ] = 10 - pH = 10 - 2 . 63 = 0 . 00234423 mol / L .
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casey (rmc2555) – Homework 8 – holcombe – (51395) 4 The K a is K a = [H 3 O + ][NO - 2 ] [HNO 2 ] = (0 . 00234423) 2 0 . 015 - 0 . 00234423 = 0 . 000434222 and the p K a is p K a = - log(0 . 000434222) = 3 . 36229 . 012 3.3points A solution of 0.2 M boric acid is prepared as an eye wash. What is the approximate pH of this solution? For boric acid K a = 7 . 2 × 10 - 10 . 1. pH = 6 2. pH = 7 3. pH = 5 correct 4. pH = 3 5. pH = 4 Explanation: 013 3.3points 0.50 moles of HCN are added to a liter of water. What is the pH? ( K a of HCN is 4 . 0 × 10 - 10 ) 1. 4.85 correct 2. 5.35 3. 4.35 4. 9.40 5. 4.69 Explanation: HCN is not a strong acid so [H + ] will not be 0.5 M. To figure it out, we must look at the K a . HCN -→ H + + CN - Initial 0.5 0 0 Change - x + x + x Equili- 0 . 5 - x (but x x x brium is negligible) x is negligible compared to 0.5 in this situa- tion because the K a is so small (which means the reaction isn’t going to go very much. We leave in the other two x ’s in because they are not negligible compared to zero: K a = [CN - ][H + ] [HCN] 4 × 10 - 10 = x 2 0 . 5 x = 1 . 4 × 10 - 5 = [H + ] pH = - log ( 1 . 4 × 10 - 5 ) = 4 . 85 014 3.3points The term “ K a for the ammonium ion” de- scribes the equilibrium constant for which of the following reactions? 1. The term is misleading, because the am- monium ion is not an acid. 2. NH + 4 + OH - NH 3 + H 2 O 3. NH 3 + H 3 O + NH + 4 + H 2 O 4. NH 4 Cl(solid) + H 2 O NH + 4 + Cl - 5. NH 3 + H 2 O NH 4 + + OH - 6. NH + 4 + H 2 O NH 3 + H 3 O + correct Explanation: 015 3.3points What would be the pH of a solution of hypo- bromous acid (HOBr) prepared by dissolving 9.7 grams of the acid in 20 mL of pure wa- ter (H 2 O)? The Ka of hypobromous acid is 2 × 10 - 9 1. 1 2. 13
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casey (rmc2555) – Homework 8 – holcombe – (51395) 5 3. 4 correct 4. 10 5. 6 Explanation: 9 . 7 g HOBr × 1 mol 97 g = 0 . 1 mol HOBr 0 . 1 mol HOBr 0 . 02 L H 2 O = 5 M HOBr [H + ] = (K a · C a ) 1 / 2 = (2 × 10 - 9 · 5) 1 / 2 = (10 - 8 ) 1 / 2 = 10 - 4 pH = - log[H + ] = - log(10 - 4 ) = 4 016 3.3points Without doing any calculations,
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