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# 1 010 m nh 3 and 010 m nh 4 cl 2 010 m nh 3 and 020 m

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1. 0.10 M NH 3 and 0.10 M NH 4 Cl 2. 0.10 M NH 3 and 0.20 M NH 4 Cl correct 3. 0.20 M NH 3 and 0.050 M NH 4 Cl 4. 0.20 M NH 3 and 0.10 M NH 4 Cl 5. 0.050 M NH 3 and 0.20 M NH 4 Cl Explanation: pOH = 5.05 K b = 1 . 8 × 10 - 5 for NH 3 Applying the Henderson-Hasselbalch equa- tion, pOH = p K b + log [NH + 4 ] [NH 3 ] log [NH + 4 ] [NH 3 ] = pOH - p K b = pOH - [ - log ( K b )] = 5 . 05 + log ( 1 . 8 × 10 - 5 ) = 0 . 305273

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casey (rmc2555) – Homework 9 – holcombe – (51395) 5 [NH + 4 ] [NH 3 ] = 10 0 . 305273 = 2 . 01963 The only combination above which yields this ratio is 0.10 M NH 3 and 0.20 M NH 4 Cl. 014 3.4points A buffer was prepared by mixing 0.50 mole of HX acid and 0.50 mole of NaX to form an aqueous solution with a total volume of 1.00 liter. The pH of this buffer was 4 . 725. Then, to 400 mL of this buffer solution was added 25.0 mL of 2.0 M HCl. What is the pH of this new solution? Correct answer: 4 . 50315. Explanation: n HX = 0.50 mol n NaX = 0.50 mol V soln = 1.0 L p K a = 4 . 725 V HCL = 25.0 mL [HCl] = 2.0 M Notice that the p K a of this acid (HX) must be 4 . 725 by definition. The HCl converts X - into HX. 25 mL of 2.0 M is equal to 50 mmol of H + added. Take this amount AWAY from the X - and add it TO the HX: 200 mmol initial of HX + 50 = 250 mmol HX 200 mmol initial of X - ( - 50) = 150 mmol X - Now substitute into the Henderson- Hasselbalch equation: pH = 4 . 725 + log parenleftbigg 150 250 parenrightbigg = 4 . 50315 015 3.4points One liter of a buffer containing 2 M H 2 CO 3 and 1.5 M NaHCO 3 has 24 g of LiOH added to it. What is the new pH? Assume the K a1 of H 2 CO 3 is 2 . 5 × 10 - 4 . 1. 6 2. 8 3. 3 4. 4 correct 5. 11 6. 10 Explanation: (24 g LiOH) parenleftbigg 1 mol 24 g parenrightbigg = 1 mol LiOH [H + ] = K a parenleftbigg C a C b parenrightbigg = ( 2 . 5 × 10 - 4 ) parenleftbigg 1 M 2 . 5 M parenrightbigg = 10 - 4 pH = 4 016 3.4points Choose the effective pH range of an aniline- anilinium chloride buffer. The value of the K b for aniline is 4.3 × 10 - 10 . 1. 10.1 to 12.1 2. 8.4 to 10.4 3. 5.1 to 7.1 4. 1.1 to 3.1 5. 3.6 to 5.6 correct Explanation: 017 3.4points What is the buffer capacity of 50 mL of 0.3 M HNO 2 and 100 mL of 0.5 M NaNO 2 ? 1. 0.05 mol of OH - and 0.015 mol of H + 2. 0.3 mol of OH - and 0.5 mol of H + 3. 0.015 mol of OH - and 0.05 mol of H + correct 4. 0.5 mol of OH - and 0.3 mol of H + 5. 0.15 mol of OH - and 0.5 mol of H + Explanation: V 1 = 50 mL M 1 = 0 . 3 M
casey (rmc2555) – Homework 9 – holcombe – (51395) 6 V 2 = 100 mL M 2 = 0 . 5 M HA A - + H + 50 mL 100 mL 0.3 M 0.5 M 15 mmol 50 mmol Adding a strong base will introduce OH - ; the base will react with HA. There is only 0.015 mol of HA so only 0.015 mol of OH - can be added before the buffer capacity is exceeded. Adding a strong acid introduces H + ; the acid will react with A - . There is only 0.05 mol of A - so only 0.05 mol of H + can be added before exceeding buffer capacity. 018 3.4points A 97 mL solution of 0 . 1 M ammonium chloride is titrated with 67 mL of 0 . 2 M NaOH. What is the pH of the resulting solution? K b for ammonia is 1 . 8 × 10 - 5 . 1. 11 . 72 2. 12 . 35 correct 3. 2 . 24 4. 1 . 65 Explanation: V NH 4 Cl = 97 mL [NH 4 Cl] = 0 . 1 M V NaOH = 67 mL [NaOH] = 0 . 2 M NH 4 Cl + NaOH NaCl+NH 3 + H 2 O 97 ml 67 mL 0 . 1 M 0 . 2 M 9 . 7 mmol 13 . 4 mol - 9 . 7 mmol - 9 . 7 mmol 0 mmol 3 . 7 mmol [OH - ] = 3 . 7 mmol 164 mL = 0 . 022561 M pOH = - log[OH - ] = - log(0 . 022561) = 1 . 65 pH = 14 - pOH = 12 . 35 The contribution of pH from ammonium and ammonium chloride is minimal.

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