007 32points which of the following is not a common

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007 3.2points Which of the following is NOT a common strong acid? 1. HNO 3 2. HCl 3. H 2 SO 3 correct 4. HBr 5. All of the others answers are common strong acids. Explanation: Sulfurous acid (H 2 SO 3 ) is not a common strong acid. It does not ionize completely in water unlike the other choices: nitric acid (HNO 3 ), hydrochloric acid (HCl), and hydro- bromic acid (HBr). 008 3.2points A buffer solution is prepared by dissolving 0.50 mol HC 2 H 3 O 2 and 1.00 mol NaC 2 H 3 O 2 in enough water to make 1.00 L solution. What is the pH of the solution? ( K a = 1 . 8 × 10 5 .) 1. 4.44 2. 5.04 correct 3. 4.74 4. 7.00
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Version 221 – Exam 3 – holcombe – (51395) 3 5. 5.24 Explanation: [C 2 H 3 O 2 ] = 1.00 M [HC 2 H 3 O 2 ] = 0.50 M K a = 1 . 8 × 10 5 pH = p K a + log parenleftbigg [CH 3 COO ] [CH 3 COOH] parenrightbigg = - log(1 . 8 × 10 5 ) + log parenleftbigg 1 . 0 0 . 5 parenrightbigg = 5 . 04576 009 3.2points A 40 mL sample of 0.25 M NaCHOO is titrated with 0.2 M HCl. What is the pH of the solution after 140 mL of HCl has been added? 1. 2.0 2. 0 3. 1.0 correct 4. 2.1 5. -1.0 6. 3.0 Explanation: After the CHOOH has been neutralized, 0.018 mol of HCl remains in a solution with a total volume of 180 mL. [H + ] = 0 . 018 mol 180 mL = 0 . 1 M pH = 1 010 3.2points Sodium fluoride is added to pure water and stirred to dissolve. Compared to pure water, the new solution is 1. neutral pH 2. basic pH correct 3. unchanged pH 4. More than one of these is true. 5. acidic pH Explanation: Sodium fluoride is a basic salt and therefore changes solutions to a more basic pH. 011 3.2points Would the value of K w change if the temper- ature were raised from 25 C to 50 C? 1. No 2. Yes correct Explanation: 012 3.2points Rank following salts from least to most solu- ble: BiI K sp = 7 . 7 × 10 19 Cd 3 (AsO 4 ) 2 K sp = 2 . 2 × 10 33 AlPO 4 K sp = 9 . 8 × 10 21 CaSO 4 K sp = 4 . 9 × 10 5 1. AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 < CaSO 4 correct 2. Cd 3 (AsO 4 ) 2 < CaSO 4 < AlPO 4 < BiI 3. CaSO 4 < AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 4. BiI < Cd 3 (AsO 4 ) 2 < CaSO 4 < AlPO 4 Explanation: Molar solubility can be approximated by taking the n th root of the K sp where n is the number of ions in the salt. Doing so results in approximate molar solubilities of 10 10 , 10 7 , 10 11 and 10 3 for bismuth iodide, cadmium arsenate, aluminum phosphate and calcium sulfate, respectively. Arranging these from least to greatest produces: AlPO 4 < BiI < Cd 3 (AsO 4 ) 2 < CaSO 4 . 013 3.2points A 200 mL sample of 0 . 5 M N 2 H 4 (hy- drazine) is titrated to its equivalence point
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Version 221 – Exam 3 – holcombe – (51395) 4 with 200 mL of 0 . 5 M HCl. If the K b of N 2 H 4 is 2 . 5 × 10 6 , what is the pH of the solution at the equivalence point? 1. 8.5 2. 5 3. 4 4. 7 5. 4.5 correct Explanation: Since all 0 . 10 mol of the hydrazine will be converted to its conjugate acid, hydrazin- nium, and the total volume at the equivalence point is 0 . 40 L: C a = ( 0 . 10 mol N 2 H 5 + ) / (0 . 40 L) = 0 . 25 M N 2 H 5 + K a = K w /K b = 10 14 / (2 . 5 × 10 6 ) = 4 × 10 9 [H + ] = ( K a C a ) 1 / 2 = (4 × 10 9 · 0 . 25) 1 / 2 = (10 9 ) 1 / 2 = 10 4 . 5 pH = 4 . 5 014 3.2points A student adds 100 mL of a 0.1M triprotic acid, H 3 A ( pK a 1 = 2 . 3 ,pK a 2 = 6 . 7 ,pK a 3 = 11 . 2), and 50 mL of 0.2M NaOH to 850 mL of water. What is the pH of the solution?
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