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Unformatted text preview: Chem 211 Solutions to WebCT Homework 1. Weak Acids and bases The pH of 0.015 M BH+ is 5.50. Find pKb for the weak base, B. BH+ B + H+ [B][H + ] x2 x2 Ka = = = + [BH ] C BH + - x C BH + But, we know the pH and thus we know x x2 (3.162 10 -6 ) 2 = Ka = C BH + 0.015 K a = 6.667 10 -10 pK a = 9.17 6 pK b = 14 - pK a = 4.82 Chem 211 Solutions to WebCT Homework 2. Buffers A sufficient amount of NaOH was added to 100 mL of 0.13 M acetic acid (HOAc) to form a buffer of pH 5.25. To this buffer 1.1 mL of 0.10 M HCl was added. Calculate the new pH. Initial mmol HOAc = (100 mL)(0.13 M) = 13 mmol OH- will react with HOAc to form some OAc- HOAc + OH- OAc- + H2O
i c f 13 -x 13-x x -x 0 0 +x x So, for our initial buffer [OAc- ] pH = pK a + log [HOAc] (x) 5.25 = 4.756 + log (13 - x ) x = 9.8438 mmol Chem 211 Solutions to WebCT Homework 2. Buffers (cont) So in our initial buffer we have : 9.84 mmol OAc- and 13 - 9.84 = 3.16 mmol HOAc The added H+ will react with the OAc- OAc- + H+ HOAc
i c e 9.84 -0.11 9.73 0.11 -0.11 0 3.16 +0.11 3.27 [OAc- ] pH = pK a + log [HOAc] (9.73) pH = 4.756 + log (3.27) pH = 5.23 Note: Total volumes will cancel in concentration ratio. ...
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