Exam 3-solutions - Version 447 Exam 3 sutcliffe (51045) 1...

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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 Its 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 youll 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 - 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 (51045) 1...

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