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TITRATION

TITRATION. Titration of a strong acid with a strong base ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINT

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TITRATION

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  1. TITRATION Titration of a strong acid with a strong base ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINT At the end point for the titration of a strong acid with a strong base, the moles of acid (H+) equals the moles of base (OH-) to produce the neutral species water (H2O). If the mole ratio in the balanced chemical equation is 1:1 then the following equation can be used. MOLES OF ACID = MOLES OF BASE nacid = nbase Since M=n/V MAVA = MBVB

  2. TITRATION MAVA = MBVB 1. Suppose 75.00 mL of hydrochloric acid was required to neutralize 22.50 mLof 0.52 M NaOH. What is the molarity of the acid? HCl + NaOH  H2O + NaCl Ma Va = Mb Vb rearranges to Ma = Mb Vb / Va so Ma = (0.52 M) (22.50 mL) / (75.00 mL) = 0.16 M Now you try: 2. If 37.12 mL of 0.843 M HNO3 neutralized 40.50 mL of KOH, what is the molarity of the base? Mb = 0.773 mol/L

  3. TITRATION Titration of a strong acid with a strong base ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINT At the end point for the titration of a strong acid with a strong base, the moles of acid (H+) equals the moles of base (OH-) to produce the neutral species water (H2O). If the mole ratio in the balanced chemical equation is NOT 1:1 then you must rely on the mole relationship and handle the problem like any other stoichiometry problem. MOLES OF ACID = MOLES OF BASE nacid = nbase

  4. TITRATION 1. If 37.12 mL of 0.543 M LiOH neutralized 40.50 mL of H2SO4, what is the molarity of the acid? 2 LiOH + H2SO4 Li2SO4 + 2 H2O First calculate the moles of base: 0.03712 L LiOH (0.543 mol/1 L) = 0.0202 mol LiOH Next calculate the moles of acid: 0.0202 mol LiOH (1 mol H2SO4 / 2 mol LiOH)= 0.0101 mol H2SO4 Last calculate the Molarity: Ma = n/V = 0.010 mol H2SO4 / 0.4050 L = 0.248 M 2. If 20.42 mL of Ba(OH)2 solution was used to titrate29.26 mL of 0.430 M HCl, what is the molarity of the barium hydroxide solution? Mb = 0.260 mol/L

  5. PRACTICE PROBLEMS 10.8 mL 1. How many milliliters of 1.25 M LiOH must be added to neutralize 34.7 mL of 0.389 M HNO3? 2. How many mL of 0.998 M H2SO4 must be added to neutralize 47.9 mL of 1.233 M KOH? 29.6 mL

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