Exam 3-solutions - Version 447 – Exam 3 – sutcliffe...

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Unformatted text preview: Version 447 – Exam 3 – sutcliffe – (51045) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points For the weak diprotic acid system, which of the following statements is NOT true? 1. A 2 − is a stronger base than HA − 2. H 2 A dissociates completely correct 3. K a2 < K a1 4. HA − is amphiprotic 5. HA − and A 2 − form a basic buffer Explanation: A weak acid only slightly dissociates. 002 10.0 points What is the pH of a 0 . 3 M solution of am- monium nitrate (NH 4 NO 3 )? K b for ammonia (NH 3 ) is 1 . 8 × 10 − 5 . 1. 4.89 correct 2. 4.74 3. 6.35 4. 2.63 5. 7.00 6. 4.65 7. 11.4 8. 5.02 Explanation: M NH 4 NO 3 = 0 . 3 M K b = 1 . 8 × 10 − 5 It’s a salt of a weak base (BHX). This means you need a K a for the weak acid BH + . Use K a = K w K b and you’ll get the K a = 5 . 55556 × 10 − 10 . You CAN use the approximation for the equilibrium which means that [H + ] = radicalbig K a · C BH + = 1 . 29099 × 10 − 5 M pH =- log(1 . 29099 × 10 − 5 ) = 4 . 88908 003 10.0 points A 100 ml sample of 0.100 M NH 3 solution is titrated to the equivalence point with 50 ml of 0.200 M HCl. What is the final [H 3 O + ]? The ionization constant of NH 3 is 1 . 8 × 10 − 5 . 1. 8 . 61 × 10 − 6 M 2. 3 . 70 × 10 − 11 M 3. 1 . 00 × 10 − 7 M 4. 6 . 09 × 10 − 6 M correct 5. 1 . 10 × 10 − 3 M Explanation: [NH 3 ] = 0.1 M [HCl] = 0.2 M Initially, for NH 3 , (0.1 M)(100 mL) = 10 mmol for HCl, (0.2 M)(50 mL) = 10 mmol Neutralization: NH 3 + HCl → NH 4 Cl ini 10 mmol 10 mmol Δ- 10 mmol- 10 mmol 10 mmol fin 0 mmol 0 mmol 10 mmol [NH + 4 ] = 10 mmol 150 mL = 0 . 067 M Equilibria re-established: K a NH + 4 ⇀ ↽ NH 3 + H + 0.067 M-- . 067- x x x K w = 10 − 14 K b = 1 . 8 × 10 − 5 K a = K w K b = 10 − 14 1 . 8 × 10 − 5 = 5 . 56 × 10 − 10 K a = x 2 . 067- x Version 447 – Exam 3 – sutcliffe – (51045) 2 Assumption: x = [H + ] = radicalbig K a × . 067 = radicalBig (5 . 56 × 10 − 10 )(0 . 067) = 6 . 09 × 10 − 6 004 10.0 points Determine if a precipitate will form when . 96 g Na 2 CO 3 is combined with 0 . 2 g BaBr 2 in a 10 L solution (For BaCO 3 , K sp = 2 . 8 × 10 − 9 ). 1. BaCO 3 does not precipitate 2. BaCO 3 precipitates correct Explanation: C Na 2 CO 3 = 0 . 96 g / 10 L C BaBr 2 = 0 . 2 g / 10 L K sp = 2 . 8 × 10 − 9 Na 2 CO 3 + BaBr 2 (aq)-→ 2 NaBr(aq) + BaCO 3 (s) BaCO 3 (s) ⇀ ↽ Ba 2+ (aq) + CO 2 − 3 (aq) K sp = [Ba 2+ ] [CO 2 − 3 ] = 2 . 8 × 10 − 9 parenleftbigg . 96 g Na 2 CO 3 10 L parenrightbigg parenleftBigg 1 mol CO 2 − 3 105 . 96 g Na 2 CO 3 parenrightBigg = 0 . 001812 mol / L CO 3 2 − parenleftbigg . 2 g BaBr 2 10 L parenrightbiggparenleftbigg 1 mol Ba 2+ 297 . 13 g BaBr 2 parenrightbigg = 6 . 73106 × 10 − 5 mol / L Ba 2+ [Ba 2+ ] [CO 2 − 3 ] = (0 . 001812 mol / L) × (6 . 73106 × 10 − 5 mol / L) = 1 . 21967 × 10 − 7 > K sp BaCO 3 precipitates....
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Exam 3-solutions - Version 447 – Exam 3 – sutcliffe...

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