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hw8solution_pdf

hw8solution_pdf - So(dds785 Homework 8 Sutclie(52440 This...

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So (dds785) – Homework 8 – Sutclife – (52440) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. 001 10.0 points A 130 mL portion o± 0 . 4 M acetic acid is be- ing titrated with 0 . 4 M NaOH solution. What is the pH o± the solution a±ter 110 mL o± the NaOH solution has been added? The ioniza- tion constant o± acetic acid is 1 . 8 × 10 - 5 . 1. pH = 5 . 49 correct 2. pH = 3 . 99 3. pH = 6 . 22 4. pH = 4 . 44 5. pH = 5 . 71 Explanation: V CH 3 COOH = 130 mL [CH 3 COOH] = 0 . 4 M V NaOH = 110 mL [NaOH] = 0 . 4 M K a = 1 . 8 × 10 - 5 ²or CH 3 COOH, (0 . 4 M)(0 . 13 L) = 0 . 052 mol ²or NaOH, (0 . 4 M)(0 . 11 L) = 0 . 044 mol CH 3 COOH + NaOH NaCH 3 COO + H 2 O 0 . 052 mol 0 . 044 mol 0 mol - 0 . 044 mol - 0 . 044 mol +0 . 044 mol 0 . 008 mol 0 mol 0 . 044 mol CH 3 COOH CH 3 COO - + H + 0 . 008 mol 0 . 044 mol 0 . 24 L 0 . 24 L 0 . 0333333 M 0 . 183333 M x Thus K a = b CH 3 COO - Bb H + B [CH 3 COOH] 1 . 8 × 10 - 5 = 0 . 183333 x 0 . 0333333 x = b H + B = K a [CH 3 COOH] [CH 3 COO - ] = ( 1 . 8 × 10 - 5 ) (0 . 0333333) 0 . 183333 = 3 . 27273 × 10 - 6 Thus pH = - log b H + B = 5 . 48509 002 10.0 points What is the Fnal volume when 100 mL o± 0.200 M acetic acid solution is titrated to the equivalence point with 0.0400 M Ba(OH) 2 ? 1. 110 mL 2. 140 mL 3. 350 mL correct 4. 600 mL 5. 250 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 0 0 - 20 - 20 20 20 0 0 20 20 100 mL CH 3 COOH(aq) × 0 . 2 mol CH 3 COOH 1000 mL × 1 mol Ba(OH) 2 2 mol CH 3 COOH × 1000 mL Ba(OH) 2 0 . 04 mol Ba(OH) 2 = 250 mL Ba(OH) 2 (aq) Total volume = 350 mL 003 10.0 points A 100 ml sample o± 0 . 200 M NH 3 solution is titrated to the equivalence point with 50 mL o± 0 . 400 M HCl. What is the Fnal [H 3 O + ]? The ionization constant o± NH 3 is 1 . 8 × 10 - 5 . 1. 3 . 70 × 10 - 11 M
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So (dds785) – Homework 8 – Sutclife – (52440) 2 2. 8 . 61 × 10 - 6 M correct 3. 1 . 05 × 10 - 5 M 4. 1 . 00 × 10 - 7 M 5. 6 . 09 × 10 - 6 M 6. 1 . 55 × 10 - 3 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 ] = 0 . 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 + = r K a [NH + 4 ] = r (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 pH b b I II The pOH at point I is equal to the and the pH at point II is pH 7. 1. pH oF the base, less than 2. p K b oF the base, greater than 3. p K b oF the base, less than correct 4. p K b oF the base, equal to 5. pH oF the base, greater than Explanation: 005 10.0 points Calculate the pH at the equivalence point For the titration oF a solution containing 1250.0 mg oF hilariamine (MW = 115 . 25 g/mol, K b = 7 . 1 × 10 - 4 ) with 0.1000 M HCl solution.
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