6 - Elbel Brittany Homework 6 Due noon Inst J A Holcombe...

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Elbel, Brittany – Homework 6 – Due: Feb 27 2006, noon – Inst: J A Holcombe 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points According to the Bronsted-Lowry concept o± acids and bases, which o± the ±ollowing state- ments about a base is NOT true? 1. A base must contain a hydroxide group. correct 2. A base will share one o± its electron pairs to bind H + . 3. I± a base is strong, then its conjugate acid will be relatively weaker. 4. A base reacts with an acid to ±orm a salt. Explanation: According to the Bronsted-Lowry defni- tion o± bases, a base must be a proton accep- tor. It does not need to contain a hydroxide group. 002 (part 1 o± 1) 10 points Which is NOT a conjugate acid-base pair? 1. H 3 SO + 4 : H 2 SO 4 2. H 2 O : OH - 3. HCl : Cl - 4. H 2 SO 4 : SO 2 - 4 correct 5. H 2 : H - Explanation: Except ±or H 2 SO 4 and SO 2 - 4 , the members o± all o± the pairs di²er by one proton. 003 (part 1 o± 1) 10 points Consider the ±ollowing reaction: AX 4 , a molecule that contains only single bonds and obeys the octet rule, reacts with Y 1 - , an an- ion that has eight valence electrons. The only product is AX 4 Y 1 - , in which all atoms are bonded to A by single bonds. In this reaction AX 4 is acting as 1. a Lewis acid. correct 2. a Bronsted-Lowry base. 3. a Bronsted-Lowry acid. 4. an Arrhenius base. 5. a Lewis base. Explanation: 004 (part 1 o± 1) 10 points Calculate the resulting pH i± 400 mL o± 0.50 M HCl is mixed with 300 mL o± 0.080 M NaOH. 1. 0.4 2. 2005 3. 7.0 4. 0.6 correct 5. 1.3 6. 4.5 Explanation: V HCl = 400 mL [HCl] = 0.50 M V NaOH = 300 mL [NaOH] = 0.080 M To determine the pH o± the fnal mixture, we need to determine how much H + or OH - is le±t over a±ter the reaction. Remember that ±or complete neutralization we need H + and OH - in equal molar amounts. First calculate how many moles o± H + and OH - we have. ? mol H + = 0 . 400 L soln × 0 . 50 mol HCl 1 L soln × 1 mol H + 1 mol HCl = 0 . 20 mol H +
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Elbel, Brittany – Homework 6 – Due: Feb 27 2006, noon – Inst: J A Holcombe 2 ? mol OH - = 0 . 300 L soln × 0 . 080 mol NaOH 1 L soln × 1 mol OH - 1 mol NaOH = 0 . 024 mol OH - According to the equation H + + OH - H 2 O , we need 1 mole of H + for every 1 mol of OH - , but we have more H + than OH - . This means the OH - will be consumed and some H + will remain unreacted. Calculate how many mol H + will be unre- acted. ? mol H + = 0 . 20 mol H + - 0 . 024 mol H + = 0 . 176 mol H + Calculate the molarity of H + in the ±nal solu- tion and the pH. ? M H + = 0 . 176 mol H + 0 . 700 L = 0 . 251 M H + pH = - log[0 . 251] = 0 . 6 005 (part 1 of 1) 10 points How many grams of NaOH are needed to give a pH of 11.5 in a 14.5 L tank of water? 1.
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6 - Elbel Brittany Homework 6 Due noon Inst J A Holcombe...

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