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FINAL_REVIEW - REVIEW FOR ALL MATERIAL AFTER EXAM 3 I...

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REVIEW FOR ALL MATERIAL AFTER EXAM 3 I. Titration Curves: 1.These curves plot the pH change during an acid-base neutralization. 2.Titration is one of the most important ways of accurately finding the quantity of an acid, a base or some other substance in a mixture or of determining the purity of a substance. 3.Acid-base behavior can be seen with titration curves. Types of reactions we will deal with: a. Rxn of strong acid w/ strong base b. Rxn of weak acid w/ strong base 4.Things to remember: a. Six Strong Acids (all other acids are weak acids): HCl HI HBr HNO 3 H 2 SO 4 HClO 4 b. Strong Bases: Soluble ionic compounds that directly release OH - such as NaOH or Ba(OH) 2 Ionic compounds that have either H - or NH 2 - such as CaH 2 or NaNH 2 Organics with a negatively charged carbon c. When dealing with titration problems, treat each situation as “BRAND-NEW” 5.Strong acid/Strong base: a. Titrate 15mL of 1.0M HBr w/ 0.400 M NaOH. Find pH after the addition of: No base 20mL of base 37.5mL of base 42mL of base HBr + NaOH NaBr + H 2 O No Base—Point A: Since no base is added yet, [HX] i = [H 3 O + ] f HX=strong acid Therefore, for this problem: 1.00 M HBr = 1.00M H 3 O + So, pH = -log(1.00) This titration curve is the same for all strong acid/strong base titrations. pH = 0
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20mL of Base—Region B In this region, the pH increases very slowly—pH is a log scale so to go up by one pH unit, 90% of the remaining acid must be neutralized. So if you go from pH of 1 to a pH of 2, you used up 90% of the acid If you go from pH of 1 to pH of 3, you used up 99% of the acid, etc. Also, in this region, moles base added < initial moles of acid. You can recognize you are in region B b/c these conditions apply. Then: Since moles base added < initial moles of acid, all of the strong acid is not consumed. So what is left in the solution?—excess acid and NaBr(this has no effect on pH) So, have to do limiting reactant problem to find moles of acid left over which is equal to the moles of H 3 O + left over. (mol HX left over = moles H 3 O + left over) Then, divide these moles of H 3 O + left over by total volume in liters to get [H 3 O + ] pH
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37.5mL Base Added—Point C: Here, initial moles acid = moles base added—equivalence point. Equivalence point—the point where one reactant has been completely consumed by the addition of another reactant—happens in acid-base titrations ONLY when strong acid titrated w/ strong base. What is in solution here?—only have NaBr and H 2 O; no strong acid or strong base left over. Therefore, pH = 7—this is only true if and only if rxn is between a strong acid and a strong base. 42mL Base Added—Region D: Here, moles base added > initial moles acid. Do a limiting reactant problem to find excess moles of OH - Divide these moles by the total volume in liters [OH - ] pOH pH
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6.Titration of weak acid with a strong base: a. Titrate 15mL of 1.2M HF (Ka = 7.2 × 10 -4 ) with a 2.0M NaOH solution. Find pH: When no base added 4.5mL of base added 6.25mL of base added
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