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Horvath Practice Problems 73

# Horvath Practice Problems 73 - 1 L = 0 1 Y mol NaOH(17.6...

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Before the equivalence point NaOH is the limiting reagent. If X L of NaOH is added then 0 . 100 M NaOH 1 L × X L = 0 . 1 X mol NaOH (17.3) have reacted with the HCl. Thus 0 . 0025 0 . 1 X moles of HCl remains. Then, £ H + ¤ = 0 . 0025 0 . 1 X 0 . 025 + X (17.4) pH = log μ 0 . 0025 0 . 1 X 0 . 025 + X At the equivalence point the HCl and NaOH exactly neutralize one another and the pH = 7 . 00 . This occurs at 0 . 0025 mol HCl × 1 mol NaOH 1 mol HCl × 1 L NaOH 0 . 100 mol NaOH = 0 . 025 L NaOH (17.5) After the equivalence point HCl is the limiting reactant. If Y L of NaOH is added then Y L NaOH × 0 . 100 mol NaOH
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Unformatted text preview: 1 L = 0 . 1 Y mol NaOH (17.6) have been added. Of these moles 0.0025 mol react with the HCl leaving . 1 Y − . 0025 mol NaOH unreacted. From this £ OH − ¤ = . 1 Y − . 0025 . 025 + Y (17.7) pOH = − log μ . 1 Y − . 0025 . 025 + Y ¶ Thus, pH = 14 − ∙ − log μ . 1 Y − . 0025 . 025 + Y ¶¸ (17.8) The titration curve is shown in the f gure below 72 TITRATIONS PART I...
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