CHEM 302 HW9 - jrr944 – Homework 9 – Lyon –(53565 1...

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Unformatted text preview: jrr944 – Homework 9 – Lyon – (53565) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This HW assignment is due Monday, March 31, by 11PM. 001 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. 1 . 00 × 10- 7 M 2. 8 . 61 × 10- 6 M 3. 1 . 10 × 10- 3 M 4. 6 . 09 × 10- 6 M correct 5. 3 . 70 × 10- 11 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 Assumption: x = [H + ] = radicalbig K a × . 067 = radicalBig (5 . 56 × 10- 10 )(0 . 067) = 6 . 09 × 10- 6 002 10.0 points A 50.00 mL sample of 0 . 1 M KOH is being titrated with 0 . 1 M HNO 3 . Calculate the pH of solution after 52.00 mL of HNO 3 is added. 1. 2.71 correct 2. 6.50 3. 3.01 4. 2.41 5. 7.00 Explanation: V KOH = 50.0 mL [KOH] = 0 . 1 M V HNO 3 = 52.0 mL [HNO 3 ] = 0 . 1 M Initially (ini): n KOH = (0.05 L)(0.1 M) = 0.005 n HNO 3 = (0.052 L)(0.1 M) = 0.0052 KOH + HNO 3 → K + + NO- 3 + H 2 O ini, mol . 005 . 0052 Δ, mol − . 005 − . 005 . 005 0 . 005 fin, mol . 005 . 005 0 . 005 K + and NO- 3 are spectator ions. 0 . 005 M HNO 3 (a strong acid) remains in 102.0 mL of solution. HNO 3 + H 2 O → H 3 O + + NO- 3 [H 3 O + ] = . 005 mol . 102 L = 49 . 0196 × 10- 3 M pH = − log(49 . 0196 × 10- 3 M) = 1 . 30963 003 10.0 points What is the pH at the stoichiometric point for the titration of 0.26 M CH 3 NH 2 (aq) with 0.26 M HClO 4 (aq)? For CH 3 NH 2 , K b = 3 . 6 × 10- 4 . jrr944 – Homework 9 – Lyon – (53565) 2 1. 5.72 correct 2. 2.01 3. 5.57 4. 2.16 5. 7.00 Explanation: 004 10.0 points What is the concentration of acetate ions at the stoichiometric point in the titration of 0.018 M CH 3 COOH(aq) with 0.036 M NaOH(aq)? For acetic acid, K a = 1 . 8 × 10- 5 ....
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This homework help was uploaded on 04/09/2008 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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CHEM 302 HW9 - jrr944 – Homework 9 – Lyon –(53565 1...

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