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Example 1 Nicotinic Acid Is a Monoprotic Acid

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Example 1 Nicotinic Acid Is a Monoprotic Acid Weak acids and bases Example 1 Nicotinic acid is a monoprotic acid. A solution that is 0.012 M in nicotinic acid has a pH of 3.39. What is the acid ionization constant (Ka)? What is the degree of ionization of nicotinic acid in this solution? Solution When we say the solution is 0.012 M, this refers to how the solution is prepared. The solution is made up by adding 0.012 mol of substance to enough water to give a liter of solution. Once the solution is prepared, some molecules ionize, so the actual concentration is somewhat less than 0.012 M. - + We start with 0.012 M of acid (HNic) before ionization, i.e. [Nic ]=0 before ionization. The H3O concentration at the start is that from the self-ionization of water. It is usually much smaller than that obtained from the acid (unless the solution is extremely dilute or Ka is quite small), so you can write + + – [H3O ] = ~ 0 (meaning approximately zero). If x mol HNic ionizes, x mol each of H3O and Nic is formed, leaving (0.012 - x) mol HNic in solution. You can summarize the situation in the table: + – Concentration (M) HNic(aq) + H2O(l) <==> H3O (aq) + Nic (aq) Starting 0.012 ~0 0 Change -x +x +x Equilibrium 0.012 - x x x The equilibrium-constant equation is + – [H3O ][Nic ] Ka = [HNic] When you substitute the expressions for the equilibrium concentrations, you get x2 Ka = (0.012 - x) The value of x equals the numerical value of the molar hydronium-ion concentration, and can be obtained from the pH of the solution. + –pH –3.39 –4 x = [H3O ] = 10 = 10 = 4.1 x 10 = 0.00041 You can substitute this value of x into the above equation. x2 (0.00041)2 –5 Ka = = = 1.45 x 10 (0.012 - x) 0.012-0.00041 To obtain the degree of ionization, note that x mol out of 0.012 mol of nicotinic acid ionizes. Hence, x 0.00041 Degree of ionization = = = 0.034 0.012 0.012 The percent ionization is obtained by multiplying this by 100, which gives 3.4%. Example 2 What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0.10 M nicotinic acid (HC6H4NO2), at 25°C? What is the pH of the solution? What is the degree of ionization of nicotinic acid? The acid-ionization constant is 1.4 x 10-5. Solution As above, at the start (before ionization), the concentration of nicotinic acid, HNic, is 0.10 M and that of - + its conjugate base, Nic , is 0. The concentration of H3O is essentially zero (~ 0), assuming that the contribution from the self-ionization of water can be neglected. In 1 L of solution, the nicotinic acid + - ionizes to give x mol H3O and x mol Nic , leaving (0.10 - x) mol of nicotinic acid. These data are summarized in the following table: + - Concentration (M) HNic(aq) + H2O(l) <==> H3O (aq) + Nic (aq) Starting 0.10 ~0 0 Change -x +x +x Equilibrium 0.10 - x x x + – The equilibrium concentrations of HNic, H3O , and Nic are (0.10 - x), x, and x, respectively. Now substitute these concentrations and the value of Ka into the equilibrium-constant equation for acid ionization: + - 2 [H3O ][Nic ] x -5 = Ka or = 1.4 x 10 [HNic] (0.10 - x) This is actually a quadratic equation, but it can be simplified so that the value of x is easily found. Because the acid-ionization constant is small ( Ka <<< cHNic, the value of x is small. Assume that x is much smaller than 0.10, so that 0.10 - x ~ 0.10. The equilibrium-constant equation becomes x2 = 1.4 x 10-5 0.10 x 1.4 10 -5 ××= 0.10 Hence, x = 0.0012 . Now you can substitute the value of x into the last line of the table to find the concentrations of species. The concentrations of nicotinic acid, hydronium ion, and nicotinate ion are 0.10 M, 0.0012 M, and 0.0012 + M, respectively. The pH of the solution is pH = –log [H3O ] = –log (0.0012) = 2.92 The degree of ionization equals the amount per liter of nicotinic acid that ionizes (x = 0.0012) divided by the total amount per liter of nicotinic acid initially present (0.10). Thus, the degree of ionization is 0.0012/0.10 = 0.012 = 1.2%. Example 3 –6 What is the pH of a 0.300 M solution of morphine? Kb = 1.62 x 10 These are the important equations: + Mor + H2O <==> MorH + OH¯ + –6 Kb = ( [MorH ] [OH¯] ) / [Mor] = 1.62 x 10 Mor refers to the morphine molecule and MorH+ refers to the molecule after accepting a proton. It is completely unimportant what its formula is. We need the [OH¯]. So we have: [OH¯] = x and [MorH+] = x This is because of the one-to-on molar ratio between [OH¯] and [MorH+] that is created as Mor molecules react with the water. Remember, the [Mor] started at 0.300 M and went down as Mor molecules reacted. In fact, due to the one-to-one ratio, it went down by 'x' amount and wound up at an ending value of 0.300 - x. x2 = 1.62 x 10-6 0.30-x Kb <<< cHNic, the value of x is small. So that 0.30 - x ~ 0.30. The equilibrium-constant equation becomes x2 = 1.62 x 10-6 0.30 0.30 10 1.62 x 1.62 10 -6 ××= 0.30 x = 6.97 x 10–4 M We take the pOH to get 3.157. Converting to pH (remember pH + pOH = 14), we get a pH = 10.843 Exercises 1, A 0.025 M solution of lactic acid has a pH of 2.75. What is the acid ionization constant (Ka)? What is the degree of ionization? (1.36×10–4, 7.11%) 2, What are the concentrations of acetic acid, hydrogen ion, and acetate ion in a solution of 0.10 M acetic acid? What is the pH of the solution? What is the degree of ionization? The acid-ionization constant is 1.86 x 10-5. (0.0986 M, 1.36×10–3 M, 1.36×10–3 M, 1.36%) 3, Calculate the acid ionization constant for propionic acid (C2H5–COOH). The pH of a 0.012 M aqueous solution is 3.40. (1.37×10–5) –4 4, A solution of 0.020 M nitrous acid (HNO2) has a pH of 2.53. What is the value of Ka? (5.1×10 ) 5, What is the pH of a 0.20 M solution of formic acid? What is the degree of ionization of formic acid in –4 this solution? Ka = 1.7×10 . (2.234, 2,92%) 6, Calculate the concentration of hydrogen ion and barbiturate ion in a 0.20 M solution of barbituric acid. –5 –3 –3 The value of Ka is 9.8×10 . (4.43×10 M, 4.43×10 M) 7, A 0.15 M aqueous solution of ethanolamine has a pH of 11.34. What is Kb for ethanolamine? − 2 – –2.66 []OH (pOH = 2.66, [OH ] = 10 , K = =3.24 × 10−5 ) b c−[] OH − 8, What is the concentration of hydroxide ion in a 0.50 M aqeous solution of methylamine? What is the –4 –2 pH? Kb = 4.4×10 . (1.48×10 M, 12.17) –5 9, A solution of 0.25 M trimethylamine has a pH of 11.63. What is the value of Kb? (7.41×10 ) 10, What is the concentration of hydroxide ion in a 0.15 M aqeous solution of hydroxylamine? What is –8 –5 the pH? Kb = 1.1×10 . (4.06×10 M, 9.61) .
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