AP FREE RESPONSE QUESTIONS ACIDS/BASES 199 D A chemical reaction occurs when 100. milliliters of 0.200molar HCl is added dropwise to 100. milliliters of 0.100molar Na 3 P0 solution. (a) Write the two net ionic equations for the formation of the major products. (b) Identify the species that acts as both a Brønsted acid and as a Brønsted base in the equation in (a), Draw the Lewis electrondot diagram for this species. Sketch a graph using the axes provided, showing the shape of the titration curve that results when 100. milliliters of the HCl solution is added slowly from a buret to the Na 3PO solution. Account for the shape of the curve. p H 0 m L H C l Write the equation for the reaction that occurs if a few additional milliliters of the HCl solution are added to the solution resulting from the titration in. 3 (a) PO + H + 2 2 HPO ; HPO + H + H 2PO 2 (b) HPO.. : O :...... O.. :: P : O : H.... : O.. : p H 0 P O 3 + H + H P O 2 H P O 2 + H + H 2 P O m L H C l H + + H 2PO H 3PO
1996 A AP FREE RESPONSE QUESTIONS ACIDS/BASES HOCl OCl + H + Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. The aciddissociation constant, a, for the reaction represented above is 3.2 10 8. (a) Calculate the [H + ] of a 0.1molar solution of HOCl. (b) Write the correctly balanced net ionic equation for the reaction that occurs when NaOCl is dissolved in water and calculate the numerical value of the equilibrium constant for the reaction. Calculate the ph of a solution made by combining 0.0 milliliters of 0.1molar HOCl and 10.0 milliliters of 0.56molar NaOH. How many millimoles of solid NaOH must be added to 50.0 milliliters of 0.20molar HOCl to obtain a buffer solution that has a ph of 7.9? Assume that the addition of the solid NaOH results in a negligible change in volume. (e) Household bleach is made by dissolving chlorine gas in water, as represented below. Cl 2(g) + H 2O H + + Cl + HOCl(aq) Calculate the ph of such a solution if the concentration of HOCl in the solution is 0.065 molar. (a) a = (b) + - [H ][OCl ] [HOCl] = 3.2 10 8 X = amount of acid that ionizes = [OCl ] = [H + ] (0.1 X) = [HOCl] that remains unionized 3.2 10 8 = ; X = 6.7 10 5 M = [H + ] 0. 1 X NaOCl(s) + H 2O Na + (aq) + HOCl(aq) + OH (aq) b = w a 1 1 10 3.2 10 8 = 3.1 10 7 [ ] o after dilution but prior to reaction: 0 ml [HOCl] = 0.1 M = 0.11 M 5 0 ml [OH ] = 0.56 M X 2 1 0 ml 5 0 ml = 0.11 M Equivalence point reached. [OH ] ~ [HOCl] [O H ] 2 b = = 3.1 10 7 0.1 1 [OH ] = 1.8 10 ; poh = 3.7 ph = 1 3.7 = 10.3 at ph 7.9, the [H + ] = 10 7.9 = 3.2 10 8 M [O Cl ] when the solution is halfneutralized, ph = p a and = 1 [H OC l] 0.20 mol HOC l 1 L 50.0 ml = 10.0 mmol HOCl half this amount, or 5.0 mmol of NaOH added. (e) 1 mol H+ for every 1 mole of HOCl produced [H + ] ~ [HOCl] = 0.065 M ph = log (0.065) = 1.2
AP FREE RESPONSE QUESTIONS ACIDS/BASES 1997 A The overall dissociation of oxalic acid, H 2C 2O, is represented below. The overall dissociation constant is also indicated. (a) (b) H 2C 2O 2 H + + C 2O 2 = 3.78 10 6 What volume of 0.00molar NaOH is required to neutralize completely a 5.00 10 3 mole sample of pure oxalic acid? Give the equations representing the first and second dissociations of oxalic acid. Calculate the value of the first dissociation constant, 1, for oxalic acid if the value of the second dissociation constant, 2, is 6.0 10 5. To a 0.015molar solution of oxalic acid, a strong acid is added until the ph is 0.5. Calculate the [C 2O 2 ] in the resulting solution. (Assume the change in volume is negligible.) Calculate the value of the equilibrium constant, b, for the reaction that occurs when solid Na 2C 2O is dissolved in water. 2 mo l H + 1 mo l O H 100 0. ml NaOH (a) 5.00 10 3 mol oxalic acid = 25.0 ml NaOH 1 mo l o x alic aci d 1 mo l H + 0. 00 mol NaOH (b) H 2C 2O H + + HC 2O HC 2O H + + C 2O 2 = 1 2 1 = = 3.78 10 6 5 6.0 10 = 5.91 102 X = amt. ionized [H 2C 2O ] = 0.015 X [H + ] = 10 ph = 10 0.5 = 0.316 M [C 2O 2 ] = X a = 2 + 2 2- [H ] [C2O ] [H C O ] w b = = 2 2 2 1.0 10 = 3.78 10 6 = 1 6 6.0 10 = 1.56 1010 2 (0.316) ( X ) (0.015 X ) ; X = 5.67 107 M
AP FREE RESPONSE QUESTIONS ACIDS/BASES 1998 D (Required) An approximately 0.1molar solution of NaOH is to be standardized by titration. Assume that the following materials are available. Clean, dry 50 ml buret 250 ml Erlenmeyer flask Wash bottle filled with distilled water Analytical balance Phenolphthalein indicator solution Potassium hydrogen phthalate, HP, a pure solid monoprotic acid (to be used as the primary standard) (a) Briefly describe the steps you would take, using the materials listed above, to standardize the NaOH solution. (b) Describe (i.e., set up) the calculations necessary to determine the concentration of the NaOH solution. After the NaOH solution has been standardized, it is used to titrate a weak monoprotic acid, HX. The equivalence point is reached when 25.0 ml of NaOH solution has been added. In the space provided at the right, sketch the titration curve, showing the ph changes that occur as the volume of NaOH solution added increases from 0 to 35.0 ml. Clearly label the equivalence point on the curve. Describe how the value of the aciddissociation constant, a, for the weak acid HX could be determined from the titration curve in part. (e) The graph below shows the results obtained by titrating a different weak acid, H 2Y, with the standardized NaOH solution. Identify the negative ion that is present in the highest concentration at the point in the titration represented by the letter A on the curve. Answer
1998 D AP FREE RESPONSE QUESTIONS ACIDS/BASES Answer each of the following using appropriate chemical principles. (b) When NH 3 gas is bubbled into an aqueous solution of CuCl 2, a precipitate forms initially. On further bubbling, the precipitate disappears. Explain these two observations. In each case, justify your choice. Answer (b) A small amount of NH 3 in solution causes an increase in the [OH ]. NH 3 + H 2O NH + + OH This, in turn, causes the sp of copper(ii) hydroxide to be exceeded and the solution forms a precipitate of Cu(OH) 2. With the addition of more NH 3, you form the soluble tetraamminecopper(ii) complex ion, [Cu(NH 3) ] 2+, which will cause the precipitate to dissolve.