ws6s08key

# ws6s08key - CH 302 Worksheet 6 Key 1 You have a 750 mL...

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Unformatted text preview: CH 302 Spring 20 08 Worksheet 6 Key 1. You have a 750 mL solution of 0.1 M methylamine. You can’t find the K b for methylamine but notice that the K a for its conjugate acid is 1 x 10-9 . What is the pH of the methylamine solution? Answer: K w = K a K b = 1 x 10-14 so K b = 1 x 10-5 [OH- ] = (K b C b ) 1/2 = [(0.1)(10-5 )] 1/2 = 10-3 pOH = 3 pH = 14-pOH = 11 2. You decide to titrate it against 1 M hydrochloric acid. When you’ve added 25 mL of the HCl to the solution, what is the pH? Answer: You have 0.075 mol ammonia and 0.025 mol HCl. Neutralize: You end up with 0.05 mol ammonia and 0.025 mol ammonium ion (NH 4 + ). This is a buffer. [OH- ] = K b (C b /C a ) = 10-5 (0.05/0.025) = 2 x 10-5 pOH = -log(2 x 10-5 ) = 4.7 pH = 14-4.7 = 9.3 3. You continue the titration. What is the pH when you’ve added 75 mL HCl total? What is this point called? Answer: You have 0.075 mol of each. Neutralize: You end up with 0.075 mol of ammonium ion. This is a weak acid. [H + ] = (K a C a ) 1/2 Remember Ka = Kw/Kb = 10-14 /10-5 = 10-9 Also remember that the total volume is 75 mL + 750 mL = 775 mL [H + ] = [(10-9 )(0.075 mol/0.775 L)] 1/2 = 9.8 x 10-6 M pH = -log(9.8 x 10-6 ) = 5.0 4. You keep going until you’ve added 100 mL HCl. What is this final pH?...
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ws6s08key - CH 302 Worksheet 6 Key 1 You have a 750 mL...

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