a3s - Chemistry 2000B Spring 2002 Assignment 3 Solutions 1....

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Unformatted text preview: Chemistry 2000B Spring 2002 Assignment 3 Solutions 1. We first have to find the equivalence point: 12 10 equivalence point pH 8 pKa 6 4 0 5 10 20 15 V(OH) (mL) 25 30 35 When half the acid has been neutralized, the pH equals the pKa . Since the equivalence point is at 18 mL, the pKa is the pH at 9 mL of hydroxide solution added. From the graph, we find 48 1 58 10 5 ¡¤ 48 10 ¦¡£ ¥¤ ¡¡ £ pKa Ka ¢ 2. The weak base nitrite will react with the strong acid stoichiometrically: HNO2 We must determine which reagent is in excess. ¡ ¡ £  ¡ £ ¡ £ nH 2 mNaNO2 3 75 g 0 0544 mol MNaNO2 68 9953 g mol 0 85 mol L 0 065 L 0 0553 mol ¡£ nNO ¡ ¤ © ¨ § H ¡ NO2 ¡ ¦¡£ ¤ ¦ ¡ £ ¨   ¨ ¤¦¡ ¡   ¡ £  The hydrogen ions are in excess by 8 98 10 4 mol. Normally in these problems, the excess protons determine the pH. In other words, we ignore the dissociation of the acid. If we can do this, the concentration of H in solution is just 1 1 38 10 ¤ 8 98 10 4 mol 0 065 L 2 mol L ¡ H Is it legitimate to ignore the dissociation of the acid? The Ka expression for nitrous acid is  aHNO2 Ka aH aHNO2 Assuming that aH is approximately 1 38 10 2 (as calculated above), we have ¡ £ ¤¤ ¦¦ ¡¡ £  ¦¡ ¤ 4 5 10 1 38 10 2  aHNO2 4 0 033 2 ¡ 2  £ aNO aNO aH 2 ¡ £ or aNO   Ka In other words, about 3% of the nitrous acid formed will dissociate. The concentration of nitrous acid, again neglecting dissociation, is 0 0544 mol 0 065 L 0 84 mol L so dissociation of the acid would product about 0 027 mol L of additional protons. This is bigger than the number of excess protons, so we can’t really neglect this. There are two ways to deal with this problem, both of which revolve around the Ka expression:  ¡£ ¡ ¨  ¡ ¡ (a) Go back to the beginning. The initial concentrations of H and of nitrite are, respectively, 0.85 mol/L and 0 0544 mol 0 065 L 0 836 mol L.  ¡  aH ¡¡ ¡ ¡¡§ ¡ ¡§ ¡ ¡ ¡  £    £ ¤ ¦ ¡ 10 4 2 0 836 aHNO2 x ¨ 0 711 1 686x x2 x2 1 687x 0 711 0 864 or 0 823 x 0 85 x ¡  ¡   ¡ £ £ £ £ 10 4 x 0 x  ¢ ¢ ¤ ¦¡ ¢ 45 aNO  ¡ 45 Ka aNO aHNO2 2 0.836 0 x x 0 836 x x  ¡ Therefore, ¡£ Initial Change Final  aH 0.85 x 0 85 x The second solution has to be the right one, otherwise the concentrations of H and of NO2 turn out negative. We therefore have 0 823 0 0273 ¡£ 0 85 1 56 ¡ ¡¡ ¡¡ aH pH £ £¢ ¤ (b) The other approach is to assume that the reaction goes to completion, and then to calculate how much dissociation occurs “after” that. Of course, it doesn’t actually 2 happen that way, but it works just as well from the point of view of calculating the concentrations at equilibrium. In this case, we would take H 1 38 10 2 mol L and HNO2 0 836 mol L (the concentration generated by the reaction if it goes to completion). Then our table is ¦ ¡ £¨   ¡ ¡ ¡  § ¤ ¦ ¡  £   aH x 1 38 10 2 0 836 x  ¡¤ ¦ ¡ ¤ ¦¡ ¡¤ ¦ ¡  ¤ ¦ ¡ § ¡ § ¤ ¦ ¡  ¦¡ aHNO2 £¤ £ x 1 38 10 2 x x2 1 43 10 2x 3 76 10 2 78 10 2 or 1 35 10 2 x 4  ¢ £ £¢ £  ¢ ¡¤ ¦¡ ¢ Again, the second solution is the correct one. It follows that aH 1 35 10 2 2 74 10 2 , so that the pH is 1.56. 1 38 ¦ ¡ £ 2 10 §¤ aNO  x 0 x §¤ ¦ ¡ ¦¡ 0 836 ¤ 4 4 10    10 45  ¡£ 45 Ka x ¤ Initial Change Final aHNO2 0.836 x 0 836 x aNO 2 0 x x 2  aH 1 38 10 x 1 38 10 2 2 ¦¡£¤ ¤ ¦¡ 3. The sodium nitrate is irrelevant1 since nitrate is the conjugate base of a strong acid. Accordingly, log10 aH log10 0 85 0 07 pH 0 135 mol L ¦ ¡   nacid . The molar mass of the acid is therefore 74 4 g mol ¡ ¡ £ 10 05 g 0 135 mol ¡ ¡£ £  ¡ !£  (a) The Ka of the ammonium ion is 5 6 10 10 . Since we want the pH to be 9, the activity 10 9 . We can calculate the desired ratio of ammonia of the hydrogen ion will be aH and ammonium: aNH3 aH aNH  ¡ ¡ £ ¤¤¦¡ £  ¡   ¦¡ £ £ £ 4 ¤ aNH3 aNH ¤ Ka ¢ 1 3 ¡ £ ¤ M 10 ¡ £ 1 43 mol L 94 40 At the equivalence, point, nOH 5.  £ nOH ¡ ¡ £ ¡  4. The titration used 4 Ka aH 56 10 10 9 10 0 56 There can be a small effect since solutes in general change the volume of the solution, but this is an extremely small effect. 3 Suppose that we take 1 L of the ammonia solution. This would contain 0.0435 mol of ammonia. Since the ratio of the activities is the same as the ratio of the number of moles, we want nNH3 0 0435 mol nNH 0 0777 mol 4 0 56 0 56 From the number of moles of ammonium, we can calculate the mass of ammonium chloride required: ¡ 4 16 g ¡ ¡ ¡ ¡£ ¡£ ¡ £ £ 0 0777 mol 53 4910 g mol  ¡   ¡ "£ mNH4 Cl The procedure for making our buffer is therefore as follows: i. Measure 1 L of the ammonia solution. ii. Dissolve 4.16 g of ammonium chloride into the ammonia solution. iii. Adjust the pH with HCl or ammonia solution, as appropriate. (b) The buffer solution contains 0.0435 mol/L ammonia and 0.0777 mol/L ammonium ion. 50 mL of this solution therefore contains ¦ ¡ £  mol mol mol ¡ H2 O ¤ ¡  § © NH3 3 ¡ ¡ £  ¡ £ ¤  ¡ "£  §¤ OH NH4 ¦ ¡   ¡   The hydroxide will react with the ammonium: 3 ¤ 10 0 005 L 0 20 mol L 3 ¡ ¡ £ ¡ £ ¦ 10 5 mL of 0.20 mol/L sodium hydroxide solution contains nOH 10 10 ¤ 4 2 175 3 884 ¡ 0 0435 mol L 0 050 L 0 0777 mol L 0 050 L nNH3 nNH The final numbers of ammonium and of hydroxide ions after reaction are therefore   #  ¦¡  ¤¦ ¦¡ §¤ ¦ 5 09 ¦¡£ 3 mol 10 10 ¡¤ ¤ 3 mol ¡£ ¡ ¡ ¤ ¤ ¦ ¡ ¡ £ ¦¡ ¡ ¤ £ £  £ £ £ ¦¡ £ log10 aH 10 10 ¦ 2 884 3 175 9 29 ¤ 10 ¡ ¤ 10 ¡£ 56 pH 1 0 10 3 mol 3 175 10 3 mol 3 mol 1 0 10 3 mol 2 884 10 3 mol aNH3 aH aNH 4 aNH 4 Ka aNH3 But since Ka aH 3 ¦ 4 10 10 ¤ 2 175 3 884 ¡ nNH3 nNH ¢ 6. H2 PO4 and HCO3 are both amphoteric species. The question we must first ask ourselves is which one is the strongest base and which the strongest acid? We look up these two species in the table of ionization constants (table 17.4) in the textbook. We find the following: 4 ¤ ¤ 11 ¤ ¦¡ ¦¡ ¤ ¦¡ ¦¡ ¤ 12 ¤ ¤ Ka Kb HCO3 4 8 10 2 4 10 8 ¤ H2 PO4 6 2 10 1 3 10 8 H2 PO4 is the strongest acid (largest Ka ) and HCO3 is the strongest base (largest Kb ). Thus, in the first, instance, we might see these two species react together. To analyze this reaction, we consider the relevant ionization reactions: H2 CO3 K1 6 2 10 8 K2 2 4 10 8 K3 1 Kw 1014 K K1 K2 K3 0 149 ¡ © ©¤ §¤ ¤§ ¨ §¤ H OH £ ¡£ ££ ¤ ¦¡ £ ¦¡ £ ¤ HPO2 4 H2 CO3 H2 O HPO2 4 ¤ ¤ © H2 PO4 HCO3 H2 O OH H H2 PO4 HCO3 ©¤ §¤ §¨ §¤ ¤ We now have the overall reaction and its equilibrium constant. The rest is a standard equilibrium problem, give or take that we need to take the dilution on mixing the two solutions into account. Initially, we have  c1V1 0 043 mol L 15 mL 0 0161 mol L c2 15 25 mL 0 54 mol L 25 mL HCO3 0 338 mol L 40 mL We can now construct the following table: ¡ ¡ £   ¡ ¡ £   § ¡  £  aH2CO3 x2 x 0 338 ¡£ ¡£ #  ¡ ¡  ¤¦¡ ¡§ §¡   £     ¡ ¡  ¡ ¤ ¡     £ £ 0 149 0 0161 x 0 338 x 0 149 5 44 10 3 0 354x x2 8 10 10 4 0 0526x 0 149x2 0 851x2 0 0526x 8 10 10 4 0 0526 0 0526 2 4 0 851 8 10 2 0 851 H2 CO3 0 0128 mol L 0 0161 0 0128 mol L 0 0034 mol L 0 338 0 0128 mol L 0 325 mol L 0 0161 x ¡ ¦ ¡  ¡ # ¡   5 3  aHCO ¡ ¡ ¡ ¡£   ¡$%§ § ¤ ¦ ¦ ¡  ¡ £ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¢ £ £ £ £ x 4 ¡   0 aH2 PO 10 4 ¡ £ ¤ ¡ 4  aHPO2 0 149 x2 ¤ £ ¤ £ ¤   £ ¢ ¢  ¢ HPO2 4 H2 PO4 and HCO3   K 0 0128 ¡ ¤ ¤ £ ¤ The equilibrium condition is therefore H2 CO3 0 x x ¡  ¤ £ Initial Change Final ¤ H2 PO4 HCO3 HPO2 4 0.0161 0.338 0 x x x 0 0161 x 0 338 x x ¡ H2 PO4  ...
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This note was uploaded on 03/03/2012 for the course CHEM 2000 taught by Professor Roussel during the Fall '06 term at Lethbridge College.

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