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hw6solution_pdf

hw6solution_pdf - So(dds785 Homework 6 Sutclie(52440 This...

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So (dds785) – Homework 6 – Sutcliffe – (52440) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points If the value of K b for pyridine is 1 . 8 × 10 - 9 , calculate the equilibrium constant for C 5 H 5 NH + (aq) + H 2 O( ) C 5 H 5 N(aq) + H 3 O + (aq) . 1. 1 . 8 × 10 - 16 2. 1 . 8 × 10 - 9 3. - 1 . 8 × 10 - 9 4. 5 . 6 × 10 8 5. 5 . 6 × 10 - 6 correct Explanation: 002 10.0 points Calculate the resulting pH obtained by mixing 16.5 g HCl and 22.6 g NaCl in water to create 482 L solution. 1. 1.13 2. 3.92 3. 2.15 4. 4.31 5. 5.67 6. 3.03 correct Explanation: m HCl = 16.5 g m NaCl = 22.6 g V = 482 L The resulting pH is obtained from the [H + ]. ? mol H + = 16 . 5 g HCl × 1 mol HCl 36 . 45 g HCl × 1 mol H + 1 mol HCl × 1 482 L = 9 . 39 × 10 - 4 M H + pH = - log[H + ] = - log[9 . 39 × 10 - 4 ] = 3 . 03 003 10.0 points I found some ammonium sulfate sitting in a can. I mixed it with some water I had sitting in a pan. It really mixed up nicely, I thought that I would say the solution was ? on this lovely day. 1. still neutral 2. now acidic correct 3. now just basic Explanation: Ammonium sulfate is the salt of a strong acid, so it is an acidic salt. 004 10.0 points Which solution has the lowest pH? 1. 0.1 M KHCOO, K a HCOOH = 1 . 8 × 10 - 4 2. 0.1 M KCl, K a HCl = very large correct 3. 0.1 M KClO, K a HClO = 3 . 5 × 10 - 8 4. 0.1 M KC 2 H 3 O 2 , K a HC 2 H 3 O 2 = 1 . 8 × 10 - 5 5. 0.1 M KNO 2 , K a HNO 2 = 4 . 5 × 10 - 4 Explanation: The larger the K a , the stronger the acid. The stronger the acid, the lower the pH. Since the 0.1 M KCl has the highest K a , it has the lowest pH. 005 10.0 points What is [OH - ] in a solution made by dissolving 0.100 mole of sodium acetate (NaCH 3 COO) in enough water to make one liter of solution? K a for CH 3 COOH is 1 . 80 × 10 - 5 .

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So (dds785) – Homework 6 – Sutcliffe – (52440) 2 1. 1 . 34 × 10 - 9 2. 7 . 46 × 10 - 6 correct 3. 9.25 4. 5 . 56 × 10 - 10 5. 5.13 6. 8.87 7. 1 . 34 × 10 - 3 8. 1 . 80 × 10 - 6 9. 5.74 10. 2.87 Explanation: 006 10.0 points What is the pH of a 0 . 37 M solution of anilin- ium nitrate (C 6 H 5 NH 3 NO 3 )? K b for aniline is 4 . 2 × 10 - 10 . Your answer must be within ± 0.4% Correct answer: 2 . 52752. Explanation: M C 6 H 5 NH 3 NO 3 = 0 . 37 M K b = 4 . 2 × 10 - 10 It’s a salt of a weak base (BHX). This means you need a K a for the weak acid BH + : K a = K w K b = 1 . 0 × 10 - 14 4 . 2 × 10 - 10 = 2 . 38095 × 10 - 5 You CAN use the approximation for the
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hw6solution_pdf - So(dds785 Homework 6 Sutclie(52440 This...

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