Acid-Base Equilibria 1 Will the following salts be acidic, basic or neutral in aqueous solution? 1.NH 4 Cl.NaCl.KC H O 4.NaNO A = acidic B = basic C = neutral Solutions of a Weak Acid or Base The simplest acid-base equilibria are those in which a single acid or base solute reacts with water. In this chapter, we will first look at solutions of weak acids and bases. We must also consider solutions of salts, which can have acidic or basic properties as a result of the reactions of their ions with water. 1
Acid-Ionization Equilibria Acid ionization (or acid dissociation) is the reaction of an acid with water to produce hydronium ion (hydrogen ion) and the conjugate base anion. (See Animation: Acid Ionization Equilibrium) When acetic acid is added to water it reacts as follows. HC H O (aq) HO(aq) HO(l) CHO (aq) Because acetic acid is a weak electrolyte, it ionizes to a small extent in water. 4 Acid-Ionization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acid-ionization constant (also called the acid-dissociation constant). Consider the generic monoprotic acid, HA. HA(aq) H O(l) H O (aq) A (aq) The corresponding equilibrium expression is: K c [H O ][A ] = [HA][H O] 5 Since the concentration of water remains relatively constant, we rearrange the equation to get: [HO ][A ] K = a [HO]K = c [HA] Thus, K a, the acid-ionization constant, equals the constant [H O]K c. [H O ][A ] = [HA] Ka Table 17.1 lists acid-ionization constants for various weak acids. 6
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Experimental Determination of Ka The degree of ionization of a weak electrolyte is the fraction of molecules that react with water to give ions. Electrical conductivity or some other colligative property can be measured to determine the degree of ionization. With weak acids, the ph can be used to determine the equilibrium composition of ions in the solution. A Problem To Consider Nicotinic acid is a weak monoprotic acid with the formula HC 6 H 4 NO. A 0.01 M solution of nicotinic acid has a ph of.9 at 5 C. Calculate the acid-ionization constant for this acid at 5 C. It is important to realize that the solution was made 0.01 M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than 0.01 M. We will abbreviate the formula for nicotinic acid as HNic. 11 Let x be the moles per liter of product formed. HNic(aq) HO(l) HO (aq) Nic (aq) Starting 0.01 0 0 Change -x x x Equilibrium 0.01-x x x The equilibrium-constant expression is: [H O ][Nic ] = [HNic] Ka K Substituting the expressions for the equilibrium concentrations, we get a x = (0.01 x) 1 4
We can obtain the value of x from the given ph. x = [H O ] = anti log( ph) x = antilog(.9) x = 4.1 4 = 0.00041 Substitute this value of x in our equilibrium expression. Note first, however, that ( 0.01 x) = (0.01 0.00041) = 0.01159 0.01 the concentration of unionized acid remains virtually unchanged. 1 Substitute this value of x in our equilibrium expression. x (0.00041) K = = a 1.4 (0.01 x) (0.01) 5 To obtain the degree of dissociation: 0.00041 Degree of dissociation = = 0.04 0.01 The percent ionization is obtained by multiplying by 0, which gives.4%. 14 Calculations With K a Once you know the value of K a, you can calculate the equilibrium concentrations of species HA, A-, and H O for solutions of different molarities. The general method for doing this was discussed in Chapter 15. 15 5
Calculations With K a Note that in our previous example, the degree of dissociation was so small that x was negligible compared to the concentration of nicotinic acid. It is the small value of the degree of ionization that allowed us to ignore the subtracted x in the denominator of our equilibrium expression. The degree of ionization of a weak acid depends on both the K a and the concentration of the acid solution (see Figure 17.). 16 Figure 17.: Variation of Percent Ionization of a Weak Acid with Concentration 17 Calculations With K a How do you know when you can use this simplifying assumption? It can be shown that if the acid concentration, C a, divided by the K a exceeds 0, that is, if C a > Ka 0 then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5%. 18 6
If the simplifying assumption is not valid, you can solve the equilibrium equation exactly by using the quadratic equation. The next example illustrates this with a solution of aspirin (acetylsalicylic acid), HC 9 H 7 O 4, a common headache remedy. The molar mass of HC 9 H 7 O 4 is 180. g. From this we find that the sample contained 0.00180 mol of the acid. Hence, the concentration of the acetylsalicylic acid is 0.00180 mol/0.500 L = 0.006 M (Retain two significant figures, the same number of significant figures in K a ). 19 Note that C 0.006. a = = K 4 a 11 which is less than 0, so we must solve the equilibrium equation exactly. Abbreviate the formula for acetylsalicylic acid as HAcs and let x be the amount of H O formed per liter. The amount of acetylsalicylate ion is also x mol; the amount of nonionized acetylsalicylic acid is (0.006-x) mol. 0 These data are summarized below. HAcs(aq) HO(l) HO (aq) Acs (aq) Starting 0.006 0 0 Change -x x x Equilibrium 0.006-x x x The equilibrium constant expression is O ][Acs ] = K a [H [HAcs] If we substitute the equilibrium concentrations and the K a into the equilibrium constant expression, we get 1 7
O ][Acs ] = K a [H [HAcs] x =. (0.006 x) 4 You can solve this equation exactly by using the quadratic formula. Rearranging the preceding equation to put it in the form ax bx c = 0, we get x 4 6 = (. )x (1. ) 0 Now substitute into the quadratic formula. b ± x = b a 4ac Now substitute into the quadratic formula. (. x = 4 4 6 ) ± (. ) 4(1. The lower sign in ± gives a negative root which we can ignore Taking the upper sign, we get ) x = [HO ] = 9.4 4 Now we can calculate the ph. ph = log(9.4 4 = ).0 Check your answer by substituting the hydrogen Ion concentration into the equilibrium expression and Calculate Ka 4 8
Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. Sulfuric acid, for example, can lose two protons in aqueous solution. H SO (aq) H O(l) 4 HO (aq) HSO4 (aq) HSO 4 (aq) H O(l) H O (aq) SO 4 (aq) The first proton is lost completely followed by a weak ionization of the hydrogen sulfate ion, HSO 4-. 5 For a weak diprotic acid like carbonic acid, H CO, two simultaneous equilibria must be considered. HCO(aq) HO(l) HO (aq) HCO (aq) HCO (aq) H O(l) H O (aq) CO (aq) Each equilibrium has an associated acid-ionization constant. K For the loss of the first proton [HO ][HCO ] 7 = = a1 4. [HCO] 6 Each equilibrium has an associated acid-ionization constant. K For the loss of the second proton [HO ][CO ] 11 = = a 4.8 [HCO ] In general, the second ionization constant, K a, for a polyprotic acid is smaller than the first ionization constant, K a1. In the case of a triprotic acid, such as H PO 4, the third ionization constant, K a, is smaller than the second one, K a. 7 9
When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions. However, reasonable assumptions can be made that simplify these calculations as we show in the next example. 8 A Problem To Consider Ascorbic acid (vitamin C) is a diprotic acid, H C 6 H 6 O 6. What is the ph of a 0. M solution? What is the concentration of the ascorbate ion, C 6 H 6 O 6 -? The acid ionization constants are K a1 = 7.9 x -5 and K a = 1.6 x -1. For diprotic acids, K a is so much smaller than K a1 that the smaller amount of hydronium ion produced in the second reaction can be neglected. The ph can be determined by simply solving the equilibrium problem posed by the first ionization. 9 If we abbreviate the formula for ascorbic acid as H Asc, then the first ionization is: H Asc(aq) HO(l) HO (aq) HAsc (aq) H Asc(aq) HO(l) HO (aq) HAsc (aq) Starting 0. 0 0 Change -x x x Equilibrium 0.-x x x The equilibrium constant expression is ][HAsc ] = Ka1 [HAsc] [H O 0
Substituting into the equilibrium expression x = 7.9 (0. x) 5 Assuming that x is much smaller than 0., you get x (7.9 x.8 5 ) (0.) = 0.008 The hydronium ion concentration is 0.008 M, so ph = log(0.008) =.55 1 The ascorbate ion, Asc -, which we will call y, is produced only in the second ionization of H Asc. HAsc (aq) H O(l) H O (aq) Asc (aq) Assume the starting concentrations for HAsc - and H O to be those from the first equilibrium. HAsc (aq) H O(l) HAsc (aq) H O(l) H O (aq) Asc (aq) H O (aq) Asc Starting 0.008 0.008 0 Change -y y y Equilibrium 0.008-x 0.008y y (aq) The equilibrium constant expression is ][Asc ] = Ka [H O [HAsc ] Substituting into the equilibrium expression (0.008 y) ( 0.008 y)(y) 1 = 1.6 Assuming y is much smaller than 0.008, the equation simplifies to ( 0.008)(y) 1 (0.008) = 1.6 11
Hence, y [Asc ] = 1.6 1 The concentration of the ascorbate ion equals K a. What about bases? 4 Base-Ionization Equilibria Equilibria involving weak bases are treated similarly to those for weak acids. Ammonia, for example, ionizes in water as follows. (aq) HO(l) NH (aq) NH 4 OH (aq) The corresponding equilibrium constant is: [NH ][OH ] K = 4 c [NH][HO] 5 The concentration of water is nearly constant. [NH4 ][OH ] K = [H O]K = b c [NH ] In general, a weak base B with the base ionization (aq) H O(l) B HB (aq) OH (aq) K b has a base ionization constant equal to [HB ][OH = [B] ] Table 17. lists ionization constants for some weak bases. 6 1
A Problem To Consider What is the ph of a 0.0 M solution of pyridine, C 5 H 5 N, in aqueous solution? The K b for pyridine is 1.4 x -9. As before, we will follow the three steps in solving an equilibrium. 1. Write the equation and make a table of concentrations.. Set up the equilibrium constant expression.. Solve for x = [OH - ]. 7 Pyridine ionizes by picking up a proton from water (as ammonia does). C H N(aq) H O(l) C H NH (aq) 5 5 5 5 OH (aq) Starting 0.0 0 0 Change -x x x Equilibrium 0.0-x x x Note that C b = 0.0 = 9 1.4 Kb 1.4 which is much greater than 0, so we may use the simplifying assumption that (0.0-x) (0.0). 8 8 The equilibrium expression is [C 5H5NH ][OH ] = Kb [C5H5N] If we substitute the equilibrium concentrations and the K b into the equilibrium constant expression, we get x = 1.4 (0.0 x) 9 9 1
Using our simplifying assumption that the x in the denominator is negligible, we get x 1.4 (0.0) Solving for x we get x 9 (0.0) (1.4 9 ) x = [OH ] (0.0) (1.4 9 ) = 1.7 5 40 Solving for poh poh = log[oh] = log(1.7 Since ph poh = 14.00 5 = ph = 14.00 poh = 14.00 4.8 = 9. ) 4.8 41 Problem: A solution is 0. M in an unknown acid. The ph of the solution is determined to be.5. What is the Ka of the acid? a..0 x -6 b.1.0 x -5 c..5 x -8 d.5.0 x -6 4 14
Will the following salts be acidic, basic or neutral in aqueous solution? 1.NH 4 Cl.NaCl.KC H O 4.NaNO A = acidic B = basic C = neutral 4 Acid-Base Properties of a Salt Solution One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases. Hydrolysis: the reaction of an ion with water to produce the conjugate acid and hydroxide or the conjugate base and hydronium ion. Consider a solution of sodium cyanide, NaCN. H NaCN(s) O Na (aq) CN (aq) A 0.1 M solution has a ph of 11.1 and is therefore fairly basic. 44 Sodium ion, Na, is unreactive with water, but the cyanide ion, CN -, reacts to produce HCN and OH -. CN (aq) HO(l) HCN(aq) OH (aq) From the Brønsted-Lowry point of view, the CN - ion acts as a base, because it accepts a proton from H O. CN (aq) HO(l) HCN(aq) OH (aq) 45 15
You can also see that OH - ion is a product, so you would expect the solution to have a basic ph. This explains why NaCN solutions are basic. CN (aq) HO(l) HCN(aq) OH (aq) The reaction of the CN - ion with water is referred to as the hydrolysis of CN -. The CN- ion hydrolyzes to give the conjugate acid and hydroxide. CN (aq) HO(l) HCN(aq) OH (aq) 46 The hydrolysis reaction for CN - has the form of a base ionization so you writ the K b expression for it. CN (aq) HO(l) HCN(aq) OH (aq) The NH 4 ion hydrolyzes to the conjugate base (NH ) and hydronium ion. NH (aq) H O(l) NH (aq) 4 HO (aq) This equation has the form of an acid ionization so you write the K a expression for it. NH (aq) H O(l) NH (aq) 4 HO (aq) 47 Predicting Whether a Salt is Acidic, Basic, or Neutral How can you predict whether a particular salt will be acidic, basic, or neutral? The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs. Consequently, the anions of weak acids (poor proton donors) are good proton acceptors. Anions of weak acids therefore, are basic. 48 16
One the other hand, the anions of strong acids (good proton donors) have virtually no basic character, that is, they do not hydrolyze. For example, the Cl - ion, which is conjugate to the strong acid HCl, shows no appreciable reaction with water. Cl (aq) H O(l) no reaction Conversely, the cations of weak bases are acidic. One the other hand, the cations of strong bases have virtually no acidic character, that is, they do not hydrolyze. For example, Na (aq) HO(l) no reaction 49 Predicting Whether a Salt is Acidic, Basic, or Neutral To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt. Consider potassium acetate, KC H O. The potassium ion is the cation of a strong base (KOH) and does not hydrolyze. K (aq) HO(l) no reaction 50 Consider potassium acetate, KC H O. The acetate ion, however, is the anion of a weak acid (HC H O ) and is basic. CHO (aq) HO(l) HCHO OH A solution of potassium acetate is predicted to be basic. 51 17
Predicting Whether a Salt is Acidic, Basic, or Neutral These rules apply to normal salts (those in which the anion has no acidic hydrogen) 1. A salt of a strong base and a strong acid. The salt has no hydrolyzable ions and so gives a neutral aqueous solution. An example is NaCl. 5. A salt of a strong base and a weak acid. The anion of the salt is the conjugate of the weak acid. It hydrolyzes to give a basic solution. An example is NaCN.. A salt of a weak base and a strong acid. The cation of the salt is the conjugate of the weak base. It hydrolyzes to give an acidic solution. An example is NH 4 Cl. 5 How about? 4. A salt of a weak base and a weak acid. Both ions hydrolyze. You must compare the K a of the cation with the K b of the anion. If the K a of the cation is larger the solution is acidic. If the K b of the anion is larger, the solution is basic. 54 18
The ph of a Salt Solution To calculate the ph of a salt solution would require the K a of the acidic cation or the K b of the basic anion. (see Figure 17.8) The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species. Thus the K b for CN - is related to the K a for HCN. 55 56 The ph of a Salt Solution To see the relationship between K a and K b for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN -. HCN(aq) H O(l) CN (aq) H O(l) H O (aq) HCN(aq) CN (aq) OH (aq) K a K b H O(l) H O (aq) OH (aq) K w When these two reactions are added you get the ionization of water. 57 19
HCN(aq) H O(l) CN H O (aq) (aq) H O(l) HCN(aq) CN (aq) OH (aq) K a K b H O(l) H O (aq) OH (aq) K w When two reactions are added, their equilibrium constants are multiplied. HCN(aq) H O(l) H O (aq) CN (aq) H O(l) HCN(aq) CN (aq) OH (aq) K a K b H O(l) H O (aq) OH (aq) K w Therefore, K K = K a 58 b w The ph of a Salt Solution For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases. The only difference is first obtaining the K a or K b for the ion that hydrolyzes. The next example illustrates the reasoning and calculations involved. 59 A Problem To Consider What is the ph of a 0. M NaCN solution at 5 C? The K a for HCN is 4.9 x -. K Only the CN - ion hydrolyzes. CN (aq) H O(l) HCN(aq) OH (aq) The CN - ion is acting as a base, so first, we must calculate the K b for CN -. 14 K w 1.0 5 b = = =.0 Ka 4.9 60 0
Now we can proceed with the equilibrium calculation. Let x = [OH - ] = [HCN], then substitute into the equilibrium expression. [HCN][OH [CN ] ] = K b x =.0 (0. x) 5 61 Solving the equation, you find that Hence, x = [OH ] = 1.4 ph = 14.00 poh = 14.00 log(1.4 = ) 11. As expected, the solution has a ph greater than 7.0. 6 The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. HC H O (aq) H O(l) H O (aq) C H O (aq) Consider a solution of acetic acid (HC H O ), in which you have the following equilibrium. If we were to add NaC H O to this solution, it would provide C H O - ions which are present on the right side of the equilibrium. 6 1
HC H O (aq) H O(l) H O (aq) C H O (aq) The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased. This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect. 64 A Problem To Consider An aqueous solution is 0.05 M in formic acid, HCH O and 0.018 M in sodium formate, NaCH O. What is the ph of the solution. The K a for formic acid is 1.7 x -4. Consider the equilibrium below. HCH O(aq) HO(l) H O (aq) CH O (aq) Starting 0.05 0 0.018 Change -x x x Equilibrium 0.05-x x 0.018x 65 The equilibrium constant expression is: [HO ][CHO ] = Ka [HCHO] Substituting into this equation gives: x(0.018 x) = 1.7 (0.05 x) 4 Assume that x is small compared with 0.018 and 0.05. Then (0.018 x) 0.018 (0.05 x) 0.05 66
The equilibrium equation becomes Hence, x(0.018) 1.7 (0.05) x (1.7 4 4 0.05 ) =.4 0.018 Note that x was much smaller than 0.018 or 0.05. ph = log(.4 4 = ).6 4 For comparison, the ph of 0.05 M formic acid is.69. 67 Which of the following mixtures would form the basis for a good buffered solution? a) HOAc and HCl b) HOAc and KOAc c) HOAc and KCl d) HOAc and H O 68 What happens to the ph as acid concentration incereases? What happens to the degree of ionization? 69
Buffers A buffer is a solution characterized by the ability to resist changes in ph when limited amounts of acid or base are added to it. Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Thus, a buffer contains both an acid species and a base species in equilibrium. 70 Buffers A buffer is a solution characterized by the ability to resist changes in ph when limited amounts of acid or base are added to it. Consider a buffer with equal molar amounts of HA and its conjugate base A -. When H O is added to the buffer it reacts with the base A -. H O (aq) A (aq) HA(aq) HO(l) 71 Buffers A buffer is a solution characterized by the ability to resist changes in ph when limited amounts of acid or base are added to it. Consider a buffer with equal molar amounts of HA and its conjugate base A -. When OH - is added to the buffer it reacts with the acid HA. OH (aq) HA(aq) HO(l) A (aq) Two important characteristics of a buffer are its buffer capacity and its ph. 7 4
Buffer capacity depends on the amount of acid and conjugate base present in the solution. The next example illustrates how to calculate the ph of a buffer. 7 The Henderson-Hasselbalch Equation How do you prepare a buffer of given ph? A buffer must be prepared from a conjugate acid-base pair in which the K a of the acid is approximately equal to the desired H O concentration. To illustrate, consider a buffer of a weak acid HA and its conjugate base A -. The acid ionization equilibrium is: HA(aq) HO(l) HO (aq) A (aq) 74 The Henderson-Hasselbalch Equation How do you prepare a buffer of given ph? The acid ionization constant is: = [H O ][A ] [HA] Ka By rearranging, you get an equation for the H O concentration. [HA] [H O ] = K a [A ] 75 5
How do you prepare a buffer of given ph? Taking the negative logarithm of both sides of the equation we obtain: [HA] log[h O ] = log(k ) log [A ] - a The previous equation can be rewritten ph = pk a [A ] log [HA] Henderson-Hasselbach Equation Calculate Ka at the midpoint 76 http://kinardf.people.cofc.edu/1labchem/1lchemhomepage.htm 77 http://chemlinks.beloit.edu/rain/pages/titr.html 78 6
79 The Henderson-Hasselbalch Equation How do you prepare a buffer of given ph? More generally, you can write ph = pk a [base] log [acid] This equation relates the ph of a buffer to the concentrations of the conjugate acid and base. It is known as the Henderson-Hasselbalch equation. 80 The Henderson-Hasselbalch Equation How do you prepare a buffer of given ph? So to prepare a buffer of a given ph (for example, ph 4.90) we need a conjugate acid-base pair with a pk a close to the desired ph. The K a for acetic acid is 1.7 x -5, and its pk a is 4.77. You could get a buffer of ph 4.90 by increasing the ratio of [base]/[acid]. 81 7
Instructions for making a buffer say to mix 60 ml of 0.0 M NH With 40 ml of 0.0 M NH 4 Cl. What is the ph of the solution? NH (aq) H O (l) NH 4 (aq) OH - (aq) 1. How many moles of NH and NH 4 are are added?. Fill in the concentration table. NH (aq) H O (l) NH 4 (aq) OH - (aq) Start Change Equilibrium 8. Substitute in to equilibrium constant equation. NH (aq) H O (l) NH 4 (aq) OH - (aq) 4. Solve the equation (assume x is small compared with the molar concentrations. ph = 9.4 8 Acid-Ionization Titration Curves An acid-base titration curve is a plot of the ph of a solution of acid (or base) against the volume of added base (or acid). Such curves are used to gain insight into the titration process. You can use titration curves to choose an appropriate indicator that will show when the titration is complete. 84 8
Titration of a Strong Acid by a Strong Base Figure1 shows a curve for the titration of HCl with NaOH. Note that the ph changes slowly until the titration approaches the equivalence point. The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added. 85 Curve for the titration of a strong acid by a strong base. 86 Titration of a Strong Acid by a Strong Base Figure 17.1 shows a curve for the titration of HCl with NaOH. At the equivalence point, the ph of the solution is 7.0 because it contains a salt, NaCl, that does not hydrolyze. However, the ph changes rapidly from a ph of about to a ph of about 11. To detect the equivalence point, you need an acid-base indicator that changes color within the ph range -11. Phenolphthalein can be used because it changes color in the ph range 8.-. (see Figure 16.) 87 9
A Problem To Consider Calculate the ph of a solution in which.0 ml of 0.0 M NaOH is added to 5.0 ml of 0.0 M HCl. Because the reactants are a strong acid and a strong base, the reaction is essentially complete. H O (aq ) OH (aq ) H O(l) H O(l) We get the amounts of reactants by multiplying the volume of each (in liters) by their respective molarities. 88 Mol H O = 0.050L 0.0 mol = L 0.0050 mol Mol OH = 0.00L 0.0 mol L = 0.000 mol All of the OH - reacts, leaving an excess of H O H O = (0.0050 0.000)mol Excess = 0.00150 mol HO You obtain the H O concentration by dividing the mol H O by the total volume of solution (=0.050 L 0.00 L=0.050 L) 0.00150 mol [ H O ] = = 0.049 M 0.050 L 89 Calculate the ph of a solution in which.0 ml of 0.0 M NaOH is added to 5.0 ml of 0.0 M HCl. Hence, ph = log(0.049) = 1.68 90 0
Titration of a Weak Acid by a Strong Base The titration of a weak acid by a strong base gives a somewhat different curve. The ph range of these titrations is shorter. The equivalence point will be on the basic side since the salt produced contains the anion of a weak acid. Figure 1 shows the curve for the titration of nicotinic acid with NaOH. 91 Curve for the titration of a weak acid by a strong base. 9 A Problem To Consider Calculate the ph of the solution at the equivalence point when 5.0 ml of 0. M acetic acid is titrated with 0. M sodium hydroxide. The K a for acetic acid is 1.7 x -5. At the equivalence point, equal molar amounts of acetic acid and sodium hydroxide react to give sodium acetate. First, calculate the concentration of the acetate ion. In this case, 5.0 ml of 0. M NaOH is needed to react with 5.0 ml of 0. M acetic acid. 9 1
The molar amount of acetate ion formed equals the initial molar amount of acetic acid. 5 0. mol acetate ion L soln 1L soln =.5 - mol acetate ion The total volume of the solution is 50.0 ml. Hence, -.5 mol concentration = = molar 50 L 0.050 M The hydrolysis of the acetate ion follows the method given in an earlier section of this chapter. You find the K b for the acetate ion to be 5.9 x - and that the concentration of the hydroxide ion is 5.4 x -6. The ph is 8.7 94 Titration of a Strong Acid by a Weak Base The titration of a weak base with a strong acid is a reflection of our previous example. Figure 14 shows the titration of NH with HCl. In this case, the ph declines slowly at first, then falls abruptly from about ph 7 to ph. Methyl red, which changes color from yellow at ph 6 to red at ph 4.8, is a possible indicator. 95 What would the titration of a weak acid with a weak base appear? What would the titration of a weak base with a weak acid appear? 96
Curve for the titration of a weak base by a strong acid. 97 http://hamers.chem.wisc.edu/chapman/acidbase/titration.html#monodiv 98 Operational Skills Determining K a (or K b ) from the solution ph Calculating the concentration of a species in a weak acid solution using K a Calculating the concentration of a species in a weak base solution using K b Predicting whether a salt solution is acidic, basic, or neutral Obtaining K a from K b or K b from K a Calculating concentrations of species in a salt solution 99
Operational Skills Calculating the common-ion effect on acid ionization Calculating the ph of a buffer from given volumes of solution Calculating the ph of a solution of a strong acid and a strong base Calculating the ph at the equivalence point in the titration of a weak cid with a strong base 0 4