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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- Spring '08
- pH, Trigraph, Hydroxide, hydroxide ions, free hydroxide ions