hw8 - khan (nsk338) Homework 8 Sutcliffe (52440) 1 This...

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Unformatted text preview: khan (nsk338) Homework 8 Sutcliffe (52440) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A 130 mL portion of 0 . 4 M acetic acid is be- ing titrated with 0 . 4 M NaOH solution. What is the pH of the solution after 100 mL of the NaOH solution has been added? The ioniza- tion constant of acetic acid is 1 . 8 10- 5 . 1. pH = 4 . 93 2. pH = 4 . 54 3. pH = 6 . 32 4. pH = 5 . 27 correct 5. pH = 5 . 05 Explanation: V CH 3 COOH = 130 mL [CH 3 COOH] = 0 . 4 M V NaOH = 100 mL [NaOH] = 0 . 4 M K a = 1 . 8 10- 5 For CH 3 COOH, (0 . 4 M)(0 . 13 L) = 0 . 052 mol For NaOH, (0 . 4 M)(0 . 1 L) = 0 . 04 mol CH 3 COOH + NaOH NaCH 3 COO + H 2 O . 052 mol . 04 mol 0 mol- . 04 mol- . 04 mol +0 . 04 mol . 012 mol 0 mol . 04 mol CH 3 COOH CH 3 COO- + H + . 012 mol . 04 mol . 23 L . 23 L . 0521739 M . 173913 M x Thus K a = bracketleftbig CH 3 COO- bracketrightbigbracketleftbig H + bracketrightbig [CH 3 COOH] 1 . 8 10- 5 = . 173913 x . 0521739 x = bracketleftbig H + bracketrightbig = K a [CH 3 COOH] [CH 3 COO- ] = ( 1 . 8 10- 5 ) (0 . 0521739) . 173913 = 5 . 4 10- 6 Thus pH =- log bracketleftbig H + bracketrightbig = 5 . 26761 002 10.0 points What is the final volume when 100 mL of 0.200 M acetic acid solution is titrated to the equivalence point with 0.0400 M Ba(OH) 2 ? 1. 250 mL 2. 600 mL 3. 350 mL correct 4. 140 mL 5. 110 mL Explanation: [CH 3 COOH] = 0.2 M V CH 3 COOH = 100 mL [Ba(OH) 2 ] = 0.04 M Initially: n CH 3 COOH = (100 mL)(0 . 2 M) = 20 . 0 mmol Ba(OH) 2 +2 CH 3 COOH Ba 2+ +2 CH 3 COO- + 2 H 2 O 20 20- 20- 20 20 20 20 20 100 mL CH 3 COOH(aq) . 2 mol CH 3 COOH 1000 mL 1 mol Ba(OH) 2 2 mol CH 3 COOH 1000 mL Ba(OH) 2 . 04 mol Ba(OH) 2 = 250 mL Ba(OH) 2 (aq) Total volume = 350 mL 003 10.0 points A 100 ml sample of 0 . 200 M NH 3 solution is titrated to the equivalence point with 50 mL of 0 . 400 M HCl. What is the final [H 3 O + ]? The ionization constant of NH 3 is 1 . 8 10- 5 . 1. 1 . 00 10- 7 M 2. 1 . 55 10- 3 M khan (nsk338) Homework 8 Sutcliffe (52440) 2 3. 1 . 05 10- 5 M 4. 6 . 09 10- 6 M 5. 8 . 61 10- 6 M correct 6. 3 . 70 10- 11 M Explanation: K w = 1 10- 14 K b = 1 . 8 10- 5 NH 3 + H + NH + 4 At the equivalence point of a titration, all of the NH 3 has reacted to form NH + 4 . Ini- tially, there were 0.02 moles of NH 3 , so at the equivalence point there are 0.02 moles of NH + 4 . The pH of the solution depends only on the weak acid now, so use the weak acid equation. The total new volume is 150 mL, so [NH + 4 ] = . 02 mL 150 mL 1000 mL 1 L = 0 . 133333 M You also need the K a NH + 4 of K a NH + 4 = K w K b = 1 10- 14 1 . 8 10- 5 = 5 . 55556 10- 10 H + = radicalBig K a [NH + 4 ] = radicalBig (5 . 55556 10- 10 )(0 . 133333) = 8 . 60663 10- 6 M 004 10.0 points Consider the titration curve of a weak base with a strong acid Volume of acid added...
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hw8 - khan (nsk338) Homework 8 Sutcliffe (52440) 1 This...

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