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Unformatted text preview: = [HFor]/[For ] = 2.2 volume For = 20/2,2 = 9.0 mL 4. OH (aq) + HFor (aq) For (aq) + H 2 O The added hydroxide ions react with HFor and are used up. No free hydroxide ions are present so that the pH does not change appreciably....
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This lab report was uploaded on 04/08/2008 for the course CHEM 128 taught by Professor All during the Spring '08 term at UConn.
- Spring '08