solution9_pdf

# solution9_pdf - toupal(rgt374 – Homework 9 – Sutcliffe...

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Unformatted text preview: toupal (rgt374) – Homework 9 – Sutcliffe – (52410) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A student titrated a sample containing 101 mL of 0 . 26 M NaOH with 0 . 7 M HCl. What is the pH of the solution after 23 mL of the hydrochloric acid has been added? 1. 13 . 92 2. 1 . 08 3. 12 . 92 correct 4. 7 5. . 08 Explanation: V NaOH = 101 mL [NaOH] = 0 . 26 M V HCl = 23 mL [HCl] = 0 . 7 M Initially n NaOH : (101 mL)(0 . 26 M) = 26 . 26 mmol n HCl : (23 mL)(0 . 7 M) = 16 . 1 mmol NaOH + HCl → NaCl +H 2 O ini 26 . 26 mmol 16 . 1 mmol Δ- 16 . 1 mmol- 16 . 1 mmol fin 10 . 16 mmol 0 mmol V new = 101 mL + 23 mL = 124 mL [NaOH] = 10 . 16 mmol 124 mL = 0 . 082 M = [OH- ] pOH =- log[OH- ] =- log(0 . 082) = 1 . 08 pH = 14- pOH = 14- 1 . 08 = 12 . 92 002 10.0 points One liter of an HCl solution (pH = 3.00) and one liter of a NaOH solution (pH = 10.00) are mixed. If we assume that the volumes are additive, what should be the pH of the final solution? 1. 8.53 2. 7.00 3. 3.35 correct 4. 6.50 5. 3.01 Explanation: pH of HCl = 3.0 V HCl = 1 L pH of NaOH = 10.0 V NaOH = 1 L [H 3 O + ] = 10- pH = 1 . × 10- 3 M [OH- ] = K w [H 3 O + ] = 1 . × 10- 14 1 . × 10- 10 = 1 . × 10- 4 M Initial condition (ini): n HCl = 1 . × . 001 = 0 . 001 mol n NaOH = 1 . × . 0001 = 0 . 0001 mol HCl + NaOH → Na + + Cl- + H 2 O ini, mol . 001 . 0001 Δ, mol- . 0001- . 0001 . 0001 0 . 0001 fin, mol 0 . 0009 . 0001 0 . 0001 HCl is a strong acid, and Na + and Cl- are spectator ions. Total Volume = 2 . 00 L [H 3 O + ] = . 0009 mol 2 . 0 L = 0 . 00045 M pH =- log(0 . 00045 M) = 3 . 34679 003 10.0 points A 100 mL portion of 0 . 300 M acetic acid is being titrated with 0.200 M NaOH solution. What is the [H + ] of the solution after 50.0 mL of the NaOH solution has been added? The ionization constant of acetic acid is 1 . 8 × 10- 5 . 1. 6 . 01 × 10- 4 2. 8 . 95 × 10- 6 toupal (rgt374) – Homework 9 – Sutcliffe – (52410) 2 3. 1 . 21 × 10- 5 4. 3 . 63 × 10- 5 correct 5. 9 . 94 × 10- 6 Explanation: V CH 3 COOH = 100 mL V NaOH = 50 mL [CH 3 COOH] = 0 . 300 M [NaOH] = 0 . 200 M K a = 1 . 8 × 10- 5 Initially, n CH 3 COOH = (100 mL)(0 . 3 M) = 30 mmol n NaOH = (50 mL)(0 . 2 M) = 10 mmol CH 3 COOH+NaOH → CH 3 COO- +Na + +H 2 O ini 30 10 Δ- 10- 10 10 10 fin 20 10 10 Na + is a spectator ion. CH 3 COOH and...
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solution9_pdf - toupal(rgt374 – Homework 9 – Sutcliffe...

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