Zumdahl_chapt_15

Zumdahl_chapt_15 - Intro to Applications of Acid-Base...

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Unformatted text preview: Intro to Applications of Acid-Base Equilibria 0.500 L of 1.00 M NaOH plus 0.500 L of 2.00 M formic acid HCOOH 0.500 L of 1.00 M NaCOOH plus 0.500 L of 1.00 M formic acid HCOOH 0.500 L of 1.00 M HCl plus 0.500 L of 2.00 M sodium formate NaCOOH Three Problems Four Answers One Solution Intro to Applications of Acid-Base Equilibria 0.500 L of 1.00 M NaOH plus 0.500 L of 2.00 M formic acid HCOOH 0.500 L of 1.00 M NaCOOH plus 0.500 L of 1.00 M formic acid HCOOH 0.500 L of 1.00 M HCl plus 0.500 L of 2.00 M sodium formate NaCOOH Three Problems Four Answers One Solution Intro to Applications of Acid-Base Equilibria 0.500 L of 1.00 M NaOH plus 0.500 L of 2.00 M formic acid HCOOH 0.500 L of 1.00 M NaCOOH plus 0.500 L of 1.00 M formic acid HCOOH 0.500 L of 1.00 M HCl plus 0.500 L of 2.00 M sodium formate NaCOOH Three Problems Four Answers One Solution Intro to Applications of Acid-Base Equilibria 0.500 L of 1.00 M NaOH plus 0.500 L of 2.00 M formic acid HCOOH 0.500 L of 1.00 M NaCOOH plus 0.500 L of 1.00 M formic acid HCOOH 0.500 L of 1.00 M HCl plus 0.500 L of 2.00 M sodium formate NaCOOH Three Problems Four Answers One Solution Intro to Applications of Acid-Base Equilibria 0.500 L of 1.00 M NaOH plus 0.500 L of 2.00 M formic acid HCOOH 0.500 L of 1.00 M NaCOOH plus 0.500 L of 1.00 M formic acid HCOOH 0.500 L of 1.00 M HCl plus 0.500 L of 2.00 M sodium formate NaCOOH Answer #3 Treat as Acid Intro to Applications of Acid-Base Equilibria 0.500 L of 1.00 M NaOH plus 0.500 L of 2.00 M formic acid HCOOH 0.500 L of 1.00 M NaCOOH plus 0.500 L of 1.00 M formic acid HCOOH 0.500 L of 1.00 M HCl plus 0.500 L of 2.00 M sodium formate NaCOOH Answer #4 Treat as Base Intro to Applications of Acid-Base Equilibria 0.500 L of 1.00 M NaOH plus 0.500 L of 2.00 M formic acid HCOOH 0.500 L of 1.00 M NaCOOH plus 0.500 L of 1.00 M formic acid HCOOH 0.500 L of 1.00 M HCl plus 0.500 L of 2.00 M sodium formate NaCOOH Answer #1 [H + ]=1.8 x 10-4 pH = 3.75 Answer #2 [H + ]=1.8 x 10-4 pH = 3.75 Answer #3 pH = 3.75 Answer #4 pH = 3.75 Intro to Applications of Acid-Base Equilibria 0.500 L of 1.00 M NaOH plus 0.500 L of 2.00 M formic acid HCOOH 0.500 L of 1.00 M NaCOOH plus 0.500 L of 1.00 M formic acid HCOOH 0.500 L of 1.00 M HCl plus 0.500 L of 2.00 M sodium formate NaCOOH ] O [H ] O [H-] O [H 3 3 HCOOH 3 COOH-+ + + × + = C C K a Intro to Applications of Acid-Base Equilibria What concentration of HCl is required to have a pH of 3.75?...
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Zumdahl_chapt_15 - Intro to Applications of Acid-Base...

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