a3s-1 - Chemistry 2000 Spring 2001 Section B Assignment 3...

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Unformatted text preview: Chemistry 2000 Spring 2001 Section B Assignment 3 Solutions 1. A buffer is most effective when the target pH is close to the pKa of the acid. The pKa of hydrofluoric acid is 3.14 so an HF/F buffer would do nicely in at least some applications. (HF and fluoride ions are quite toxic at moderate concentrations so there are some contexts in which this buffer would not be appropriate.) 2. There are at least two possible reactions here: (a) H2 PO4 acts as an acid and HCO3 acts as a base, or (b) HCO3 is the acid and H2 PO4 is the base. Let us take these two possibilities in turn: acid H2 PO4 /base HCO3 : The reaction is found by combining the following processes: ¤ Ka Kb Kw 14 8 8 0 15 ¥ ¤ ¤ K ¦¥ H2 CO3 10 10 ¤ ¢ ¢ ¢ £ ¡ HPO2 4 ¤ ¢ ¡ ¢ ¡ £ HCO3 OH 62 24 10 ¦¥ ¡ ¢ H2 PO4 Ka Kb Kw HPO2 H H2 PO4 4 HCO3 H2 O H2 CO3 H2 O H OH This is not a particularly small equilibrium constant, so we might reason, on this basis alone, that this reaction would occur. We should however think about what happens if we reverse the roles of the two species before we give our final answer (no reference to Who Wants to be a Millionaire intended). acid HCO3 /base H2 PO4 : Bicarbonate (Ka 4 8 10 11 ) is a much weaker acid than H2 PO4 (Ka 6 2 10 8 ). H2 PO4 (Kb 1 3 10 12 ) is also a much weaker base than HCO3 (Kb 2 4 10 8 ). We could calculate the equilibrium constant for this reaction, but it’s clear what the result would be: The reaction of HCO3 acting as an acid with the base H2 PO4 will have a much smaller equilibrium constant than the previously considered case. Accordingly, we would predict that this reaction will not occur. ¦¥ ¦¥ ¤ ¤ ¦¥ ¤ ¦¥ ¤ Our conclusion must therefore be that the reaction ¡ HCO3 HPO2 4 ¢ ¢ H2 PO4 H2 CO3 will occur when the two reactants are mixed. Because of the moderate size of the equilibrium constant, all four species appearing in the reaction will be present in significant quantities at equilibrium. 1 3. HSO4 is, at least in principle, an amphoteric species. However, the conjugate acid of the hydrogen sulfate anion is a strong acid (sulfuric acid) so HSO4 will definitely not act as a base. The acid dissociation is H SO2 HSO4 4 ¡ ¢£ aSO2 4  © ¨ § ¤ aHSO ¥ § aH © Ka with 4 2 The Ka of this acid is 1 2 10 (Kotz & Treichel, 4th ed., Table 17.4, p 799). This Ka is aSO2 . In certainly large enough that we can ignore the dissociation of water. Thus, aH 4 fact, it is so large that a significant amount of the acid can be expected to dissociate. We are therefore going to have to work out the stoichiometry from a little table: ¤¨ ¨ aH 0 x  ¦¥  aHSO aSO2 4 4 initial 0.0053 0 final 0 0053 x x   ¥ Thus we have 2 ¥ ¦ ¥©  ¥  ¥  ¥ ¥ ¥ 10 ¤ § ¦¥ ¢ ¦¥ ¥ ¤¨ ¦¥ ¤¨ ¤  ¤ ¤ ¤ ¤ x2 0 x pH x2 0 0053 x 2 1 2 10 0 0053 x x2 1 2 10 2 x 6 36 10 5 aH 0 0040 (using the quadratic equation). log10 aH 2 40 12 ¥ Ka     4. Sulfuric acid is a strong acid so the first proton will come off quantitatively. If it weren’t for the second proton, the pH would be about 2.3. This is reasonably close to the pKa of the second proton (1.9) so we have to solve the equilibrium problem for the second proton. The technique is very much as in the previous problem except that we have 0.0053 mol/L of protons before the second proton starts to come off. Here, I’ll take the initial state to be what we get just after dissociation of the first proton: ¢ ¥  ¥  10  12 aH 0.0053 0 0053 x ¨ aHSO aSO2 4 4 initial 0.0053 0 final 0 0053 x x The equilibrium relation now reads ¢ § x 0 0053 x 0 0053 x  ¥ ¥ ¤ 2 2 ¥© ¦¥ ¤ Ka § ¥ ¥ ¦ ¥  ¥ ¥ ¥ ¥ ¤ ¤ ¥ ¤¨ ¥ ¤ © ¥ ¢ ¤ ¢ ¤ ¢ ¦¥ aH pH 1 2 10 2 0 0053 x x2 0 0173x 6 36 10 5 0 0031 (using the quadratic equation). 0 0053 x 0 0084 log10 aH 2 07  x 0 x ¥©  ¥     ¨ ¤ § x 0 0053 pH 10  5. If the pH of the buffer is to be 5, we will have ¦ §¥ 5 ©¨ ¤ © ¨ ¤ § ¤ 3 Ka aH 25 ¥¥ ¤ ¤ ¨  ¨ 3 ¤¨ 10 pKa 2 5 10 aC6H5 NH2 aH aC6 H5 NH Ka aC6 H5 NH2 aC6 H5 NH 5 ¥ ¤ The pKa is 4.6 so the Ka of the cation is 10 ¥ aH At this point it is convenient to switch to thinking in terms of the number of moles since we will ultimately want to calculate how many moles of HCl we need. Note that ¤ ¨ 3 ¤ ¨ 25 ¤ ¨ 3 3 (1) ¥¥  nC6 H5 NH2 nC6 H5 NH nC6 H5 NH2 nC6 H5 NH ¨  so we want    £ ¤    3 nC6 H5 NH2 V nC6 H5 NH V C6 H5 NH2 c C6 H5 NH3 c  aC6 H5 NH2 aC6 H5 NH The solution initially contains a total of (0.2 mol/L)(1 L) = 0.2 mol aniline. Reaction with HCl only converts this to the cation, so there is a conservation law: 0 2 mol ¥ ¤¨ ¢ nC6 H5 NH 3 (2) ¥ nC6 H5 NH2 We now have two equations (1 and 2) in the two mole numbers so we can solve for these quantities: 3 ¤ ¥ 0 057 mol 0 143 mol ¥ ¥ ¤ ¥  ¥ ¤ 02 nC6 H5 NH ¥ ¥ 0 057 mol ¨ 3  ¤ and nC6 H5 NH2 3 ¨ ¤ nC6 H5 NH 3 2 5nC6H5 NH ¥¨ ¥ 0 2 mol ¥¨ 2 5nC6H5 NH ¢ nC6 H5 NH2 For each mole of C6 H5 NH3 to be produced, we need one mole of hydrogen ions. The HCl solution must therefore provide 0.057 mol of H . The volume of HCl required is therefore  ¥ ¤ ¥  3 0 011 L 11 mL ¥ 0 057 mol 5 mol L £ £ ¤ VHCl ...
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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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