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solution for hw 3

# solution for hw 3 - Chem 211 Solutions to WebCT Homework...

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Unformatted text preview: Chem 211 Solutions to WebCT Homework 1.Weak Base Titration A 10.05 mL solution of 0.0550 M NaCN is diluted with 50.00 mL of water, then titrated with 0.0633 MHClO4. What is the pH, to two decimal places, at the equivalence point? Use the textbook appendix for Ka or Kb values. Titration reaction CN- + H+ HCN At the equivalence point, all we have in solution is HCN, i.e, a weak acid. So, we would use: [CN ][H ] x x Ka = = = [HCN] C HCN - x C HCN Ka for HCN from Appendix G = 6.2 10-10 What is CHCN? - + 2 2 Chem 211 Solutions to WebCT Homework Moles CN- initially = moles HCN at equivalence point = (10.05 mL)(0.0550 M) = 0.55275 mmol HCN What is the total volume at the equivalence point? VT = 10.05 mL + 50.00 mL+ Ve Equivalence point volume, Ve is: Ve = (10.05 mL)(0.0550 M ) = 8.732 mL 0.0633 M 0.55275 mmol = = 8.036 10 -3 M 10.05 + 50.00 + 8.732 mL Then , C HCN is : C HCN Then , x2 6.2 10 -10 = 8.036 10 -3 x = 2.232 10 -6 M = [H + ] pH = 5.65 ...
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solution for hw 3 - Chem 211 Solutions to WebCT Homework...

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