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82 x 10 3 m x chem 102 d miller solutions to review

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Unformatted text preview: M = x Chem 102 D. Miller Solutions to Review Problems for Acid/Base Chemistry 3. If 13.2 g NaC2H3O2 (FW = 82.0) are added to the 800 mL of solution in Problem 2, what is the resulting pH? The addition of C2H3O2- to a solution of HC2H3O2 creates a HC2H3O2 / C2H3O2buffer. initially, [HC2H3O2] = 0.195 M and mol C2H3O2- = 13.2 g/82.0 g/mol = 0.161 mol [C2H3O2-] = 0.161 mol/0.800 L = 0.201 M HC2H3O2 + H2O init 0.195 M equil ~0.195 X H3O+ + C2H3O2~10-7 M x x = 1.66 x 10-5 M = [H3O+] 0.201 M ~0.201 Y pH = 4.78 Chem 102 D. Miller Solutions to Review Problems for Acid/Base Chemistry 4. The resulting 800 mL of solution in Problem 3 is divided into two 400-mL samples. If 5.0 mL of 6.0 M HCl are added to one sample, and 5.0 mL of 6.0 M NaOH are added to the other, what is the resulting pH in each case? The added HCl is neutralized by the weak base and a new buffer is formed. mol HCl added = (6.0...
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This document was uploaded on 03/06/2014 for the course CHEM 102 at CSU Northridge.

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