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Exam3_Solutions

# Exam3_Solutions - Version 351 Exam 3 Sutclie(52410 This...

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Version 351 – Exam 3 – Sutcliffe – (52410) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. NOTE: The values of K a for acetic acid and K b for ammonia may be found in a different question; they are used in several questions but not quoted in all of them. 001 10.0 points What is the pH of a 0 . 2 M solution of Ca(OH) 2 ? Assume complete dissociation of the base. 1. 0.4 2. 10.8 3. 7 4. 13.6 correct Explanation: [Ca(OH) 2 ] = 0.2 M Ca(OH) 2 is a strong base which completely dissociates. Ca(OH) 2 −→ Ca 2+ + 2 OH [OH ] = 2 [Ca(OH) 2 ] = 2 (0 . 2 M) = 0 . 4 M pH = 14 pOH = 14 log 0 . 4 = 13 . 6 002 10.0 points Consider the ionization constants cyanic acid (HOCN) : K a = 3 . 5 × 10 4 ; ammonia (NH 3 ) : K b = 1 . 8 × 10 5 . A solution of ammonium cyanate (NH 4 OCN) is 1. basic, because the cation and the anion hydrolyze to the same extent. 2. neutral, because the cation hydrolyzes to a greater extent than the anion. 3. neutral, because NH 4 OCN is a weak base/weak acid salt. 4. neutral, because the cation and the anion hydrolyze to the same extent. 5. basic, because the cation hydrolyzes to a greater extent than the anion. 6. neutral, because the anion hydrolyzes to a greater extent than the cation. 7. acidic, because the anion hydrolyzes to a greater extent than the cation. 8. basic, because the anion hydrolyzes to a greater extent than the cation. 9. acidic, because the cation hydrolyzes to a greater extent than the anion. correct 10. acidic, because the cation and the anion hydrolyze to the same extent. Explanation: Compare NH + 4 + H 2 O NH 3 + H 3 O + K a = 1 K b = 1 1 . 8 × 10 5 = 55555 . 6 OCN + H 2 O HOCN + OH K b = 1 K a = 1 3 . 5 × 10 4 = 2857 . 14 The producton of H 3 O + occurs to the greater extent, making the solution acidic. 003 10.0 points Calculate the concentration of H 2 SO 3 present in 0 . 168 M Na 2 SO 3 (aq). 1. 4 . 16667 × 10 8 M 2. 6 . 66667 × 10 13 M correct 3. 1 . 33333 × 10 12 M 4. 3 . 33333 × 10 13 M 5. 8 . 33333 × 10 8 M 6. None of these Explanation:

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Version 351 – Exam 3 – Sutcliffe – (52410) 2 K a1 = 0 . 015 K a2 = 1 . 2 × 10 7 K w = 1 × 10 14 The equilibrium reactions of interest are now the base forms of the carbonic acid equi- libria, so K b values should be calculated for the following changes: SO 2 3 (aq) + H 2 O( ) OH (aq) + HSO 3 (aq) K b1 = K w K a2 = 1 × 10 14 1 . 2 × 10 7 = 8 . 33333 × 10 8 . HSO 3 (aq) + H 2 O( ) OH (aq) + H 2 SO 3 (aq) K b2 = K w K a1 = 1 × 10 14 0 . 015 = 6 . 66667 × 10 13 . Because the second hydrolysis constant is much smaller than the first, we can assume that the first step dominates: SO 2 3 + H 2 O OH + HSO 3 0 . 168 0 0 x + x + x 0 . 168 x x x Assuming that x 0 . 168 M, K b1 = [OH ][HSO 3 ] [SO 2 3 ] 8 . 33333 × 10 8 = x 2 0 . 168 x x 2 0 . 168 x 2 = (0 . 168) (8 . 33333 × 10 8 ) x = 0 . 000118322 M .
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