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Chemistry 14CL Acid-Base Chemistry Review Key 1. Barium hydroxide furnishes two moles of hydroxide in solution for each mole of barium hydroxide that you start with, and HCl donates one mole of protons/mole of acid. Thus the concentration of [OH-] = 0.0800 M. Since equivalents of acid = equivalents of base, (X mL) (0.0800 M) = (15.0 mL)(0.100 M) and X = 18.75 mL. 2. Sulfuric acid furnishes two moles of protons for each mole of acid. Thus the concentration of [H+] = 0.500 M. Since moles of [H+] = moles of [OH-], (X mL)(0.500 M) = (38.40 mL)(0.200 M) and X = 15.4 mL. 3. moles of HCl = moles of NaOH = 0.0125 moles Molecular weight of NaOH = 40.0g, Weight of NaOH required = 0.50 g 4. This is a strong base which dissociates completely in solution. [OH-] = 0.037M; pOH = 1.43; pH = 12.57 5. [H+] = 0.030, pOH = (14-1.5) = 12.5 6. 7. pH = 1.5 [OH-] =3.3 x 10-13 (a) [H+] = 0.05 M pH = 1.3 (b) [H+] = 1.2 x 10-4 pH = 3.9 (a) pOH = 1.3, pH = 12.7, [H+] = 2 x 10-13 (b) pOH = 7, pH = 7, [H+] = 1 x 10- 7 8. (a) [H+] = 3.0 10-3M, pH =2.5 (b) [H+] = 8.9 x 10-5, pH = 4.0 9. This is a limiting reagent problem only. The total moles of nitric acid involved in the titration is 2.00 x 10-3. Therefore, half way through the titration, there are 1.00 x 1-3 moles left untitrated. The volume of the solution at this point in the titration can be calculated from the original 10 mL of nitric acid and the added 10 mL of base solution giving a solution volume of 20 mL. Therefore, the [H+] of the solution at this point in the titration is 1.00 x 10 3 mol 2 5.0x10 M and the pH is 1.3 0.020L 10. The calculations differs from #9 only in the fact that you have a different amount of unreacted nitric acid and solution volume. 1.60 x 10 3 mol 0.114M and the theoretical pH is 0.94 0.014L 11. This is also a limiting reagent problem. However, since it also involves a weak acid, the equilibrium expression must be used to calculate [H+] after the stoichiometry is determined from the limiting reagent calculations as was done in the problem above. Like problem 9, half way through the titration the reaction has produced 1.00 x 10 -3 moles of chloroacetate salt and 1.00 x 10 -3 moles of acid remain untitrated. Since all the reagents and products are in the same flask and have the same volume, the concentration of chloroacetate salt must equal the concentration of acid at this point in the titration. These terms cancel in the equilibrium expression, which then reduces to Ka [H ][chloroacetate] 1.5 x 10 [chloroacetic acid] 3 Ka = [H+] or pH = pKa. [H+] = 1.5 x 10-3 and the pH = 2.8 12. This differs from problem 11 only in the difference in the ratio of the chloroacetate and chloroacetic acid. [H ] 1.5 x 10 3 4 [H+] = 6.0 x 10-3 and pH = 2.2 Ka 13. Sketch the titration curves for each of the following systems. Calculate and indicate on each sketch, the equivalence point volume, the initial pH of the solution, and at least one other pH on the titration curve. Also be sure to indicate the approximate pH of the equivalence point if this is not the second point that you choose. (a) 10.00 mL of 0.0.0500 M HNO3 (beaker) titrated with 0.100 M NaOH (buret) The titration of a strong acid (HNO3) with a strong base (NaOH) yields a titration curve with a characteristic slow-rising, flat portion followed by a steep rise that levels off once again. The equivalence point volume for the base is 5.00 mL. The initial pH is due to the complete dissociation of the nitric acid. [H +] = 0.0500 M, pH = 1.30 The equivalence point pH is 7 Calculation of other points requires more arithmetic. Any other pH can be calculated prior to the equivalence point by considering the untitrated acid in the flask. For example, at the 2.5mL point in the titration, 5 mL of acid are unreacted in a solution volume of 12.5 mL. [H+] = 0.0200; pH = 1.70 After the equivalence point, you consider the concentration of the unreacted excess hydroxide. Chemistry 14CL 2 Acid-Base Review Key (b) 10.00 mL of 0.0500 M benzoic acid (Ka = 6.46 x 10-5) (beaker) titrated with 0.100 M NaOH (buret). The titration of a weak acid (benzoic acid) with a strong base (NaOH) yields a titration curve with a shape that resembles your titrations from the lab when you titrated vinegar with NaOH. The equivalence point volume for the base is 5.00 mL. The initial pH is due to partial dissociation of the benzoic acid. [H +] = 0.00180 M, pH = 2.75 Half way through the titration, after 2.5 mL of base have been added, the pH = pK a = 4.19. Because of the hydrolysis of the benzoate salt at the equivalence point, the pH will be greater than 7. Calculation of [OH-] at the equivalence point gives 2.26 x 10-6M or pOH = 5.65 and pH = 8.35 (c) 10.00 mL of 0.0500 M TRIS (Kb = 1.20 x 10-6) beaker titrated with 0.1000 M HCl. The titration of a weak base (TRIS) with a strong acid (HCl) yields approximately a mirror image of the curve in (b) where the mirror is placed horizontally at pH 7. The initial pH is due to partial reaction of the TRIS, which is an ammonia base, with water to give [OH-] = 2.475 x 10-4, pOH = 3.61 and pH = 10.39 Halfway through the titration, after 2.5 mL of acid has been added, the pOH = pKb = 5.92 and pH = 8.08 At the equivalence point the conjugate acid hydrolyzes to give a pH below 7. Calculation gives [H+] = 1.67 x 10-5, pH = 4.78 14. Acid base indicators are usually weak acids, which can exist in solution as either the undissociated acid or the conjugate base. The two species have different colors. As the pH of the titration changes, the predominant form of the indicator changes from one form to the other. Complete the following table. SA = strong acid, SB = strong base; WA = weak acid, WB= weak base pH color of color of Appropriate indicator for titration Indicator range of undissociated conjugate of 0.1000 M reagents of acid and color indicator base base (Explain below) change SA-SB WA-SB WB-SA Bromocresol 3.8 5.4 yellow blue X green Phenol8.2 colorless pinkX X phthalein 10.0 magenta Methyl red 4.8 6.0 red yellow X <a href="/keyword/methyl-orange/" >methyl orange</a> 3.2 4.2 red yellow X Chemistry 14CL 3 Acid-Base Review Key Give your reasoning for choice of titration. The pH of the equivalence point occurs below 7 in the titration of a weak base with a strong acid due to the hydrolysis of the conjugate acid of the base. Thus, to make the end point of the indicator correspond to the equivalence point of the titration, the indicator choice must change color below pH 7. Bromocresol green, methyl red and <a href="/keyword/methyl-orange/" >methyl orange</a> are appropriate indicators. The equivalence point pH in the titration of a strong acid with a strong base occurs at 7. Phenolphthalein can be used for this titration if the strong base has a concentration greater than 0.1M. Because the change in pH at the equivalence point with these concentrations M goes from 5 9 within 1 drop the end point will correspond to the equivalence point. If the solutions are more dilute, this is not an appropriate indicator. (See lecture notes from 6/1/05). The pH of the equivalence point occurs above 7 in the titration of a weak aced with a strong base due to the hydrolysis of the conjugate base of the acid. Thus, to make the end point of the indicator correspond to the equivalence point of the titration, the indicator choice must change color above pH 7. Phenolphthalein meets this criterion. 15. The transformation of the mass action equilibrium expression to the HendersonHasselbalch equation which is useful if you are making buffere solutions is accomplished by taking the log of the mass action expression and assuming that dissociation is negligible. The subscripts &quot;initial&quot; are misleading if you are titrating to a given pH as they refer to the molar concentration of acid and base after the acid-base stoichiometric reaction has occurred. 16. The Henderson-Hasselbalch equation can only be used to calculate the pH of a buffer solution where there are large known amounts of both the acid and base present. Chemistry 14CL 4 Acid-Base Review Key ... View Full Document

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