Solution Analyze We are asked to give the conjugate base for several acids and the conjugate acid for several bases. Plan The conjugate base of a substance.

Solution Analyze We are asked to give the conjugate base for several acids and the conjugate acid for several bases. Plan The conjugate base of a substance is simply the parent substance minus one proton, and the conjugate acid of a substance is the parent substance plus one proton. Solve (a)If we remove a proton from HClO 4, we obtain ClO 4 −, which is its conjugate base. The other conjugate bases are HS −, PH 3, and CO 3 2−. (b)If we add a proton to CN −, we get HCN, its conjugate acid. The other conjugate acids are HSO 4 −, H 3 O +, and H 2 CO 3. Notice that the hydrogen carbonate ion (HCO 3 − ) is amphiprotic. It can act as either an acid or a base. (a)What is the conjugate base of HClO 4, H 2 S, PH 4 +, HCO 3 − ? (b)What is the conjugate acid of CN −, SO 4 2−, H 2 O, HCO 3 − ? Identifying Conjugate Acids and Bases

Solution Analyze and Plan We are asked to write two equations representing reactions between HSO 3 − and water, one in which HSO 3 − should donate a proton to water, thereby acting as a Brønsted–Lowry acid, and one in which HSO 3 − should accept a proton from water, thereby acting as abase. We are also asked to identify the conjugate pairs in each equation. Solve (a) HSO 3 − (aq) + H 2 O(l) SO 3 2 − (aq) + H 3 O + (aq) The conjugate pairs in this equation are HSO 3− (acid) and SO 3 2 − (conjugate base), and H 2 O (base) and H 3 O + (conjugate acid). (b) HSO 3 − (aq) + H 2 O(l) H 2 SO 3 (aq) + OH − (aq) The conjugate pairs in this equation are H 2 O (acid) and OH − (conjugate base), and HSO 3 − (base) and H 2 SO 3 (conjugate acid). The hydrogen sulfite ion (HSO 3 − ) is amphiprotic. Write an equation for the reaction of HSO 3 – with water (a) in which the ion acts as an acid and (b) in which the ion acts as a base. In both cases identify the conjugate acid–base pairs. Writing Equations for Proton-Transfer Reactions

Solution Analyze We are asked to predict whether an equilibrium lies to the right, favoring products, or to the left, favoring reactants. Plan This is a proton-transfer reaction, and the position of the equilibrium will favor the proton going to the stronger of two bases. The two bases in the equation are CO 3 2 −, the base in theforward reaction, and SO 4 2 −, the conjugate base of HSO 4 −. We can find the relative positions of these two bases in Figure 16.4 to determine which is the stronger base. For the following proton-transfer reaction use Figure below to predict whether the equilibrium lies to the left (K c 1): HSO 4 − (aq) + CO 3 2 − (aq) SO 4 2 − (aq) + HCO 3 − (aq) Predicting the Position of a Proton-Transfer Equilibrium

Solve The CO 3 2 − ion appears lower in the right-hand column in Figure 16.4 and is therefore a stronger base than SO 4 2 −. Therefore, CO 3 2− will get the proton preferentially to become HCO 3 −, while SO 4 2 − will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is, K c > 1): Comment Of the two acids HSO 4 – and HCO 3 –, the stronger one (HSO 4 – ) gives up a proton more readily, and the weaker one (HCO 3 – ) tends to retain its proton. Thus, the equilibrium favors the direction in which the proton moves from the stronger acid and becomes bonded to the stronger base. Continued Predicting the Position of a Proton-Transfer Equilibrium

Solution Analyze We are asked to determine the concentrations of H + and OH − ions in a neutral solution at 25 °C. Plan We will use Equation 16.16 and the fact that, by definition, [H + ] = [OH − ] in a neutral solution. Solve We will represent the concentration of H + and OH − in neutral solution with x. This gives [H + ][OH − ] = (x)(x) = 1.0 × 10 −14 x 2 = 1.0 × 10 −14 x = 1.0 × 10 −7 M = [H + ] = [OH − ] In an acid solution [H + ] is greater than 1.0 × 10 −7 M; in a basic solution [H + ] is less than 1.0 × 10 −7 M. Calculate the values of [H + ] and [OH − ] in a neutral aqueous solution at 25 °C. Calculating [H + ] for Pure Water

Solution Analyze We are asked to calculate the [H + ] concentration in an aqueous solution where the hydroxide concentration is known. Plan We can use the equilibrium-constant expression for the autoionization of water and the value of K w to solve for each unknown concentration. Solve (a) Using Equation 16.16, we have This solution is basic because (b) In this instance This solution is acidic because Calculate the concentration of H + (aq) in (a) a solution in which [OH − ] is 0.010 M, (b) a solution in which [OH − ] is 1.8 × 10 −9 M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 °C. Calculating [H + ] from [OH − ]

Solution Analyze We are asked to determine the pH of aqueous solutions for which we have already calculated [H + ]. Plan We can calculate pH using its defining equation, (a) In the first instance we found [H+] to be 1.0 × 10 −12 M, so that pH = −log(1.0 × 10 −12 ) = −(−12.00) = 12.00 Because 1.0 × 10 −12 has two significant figures, the pH has two decimal places, 12.00. (b) For the second solution, [H + ] = 5.6 × 10 −6 M. Before performing the calculation, it is helpful to estimate the pH. To do so, we note that [H + ] lies between 1 × 10 −6 and 1 × 10 −5. Thus, we expect the pH to lie between 6.0 and 5.0. We use Equation 16.17 to calculate the pH: pH = −log(5.6 × 10 −6 ) = 5.25 Check After calculating a pH, it is useful to compare it to your estimate. In this case the pH, as we predicted, falls between 6 and 5. Had the calculated pH and the estimate not agreed, we should have reconsidered our calculation or estimate or both. Calculate the pH values for the two solutions of the last example. Calculating pH from [H + ]

Analyze We need to calculate [H + ] from pOH. Plan We will first use Equation 16.20, pH + pOH = 14.00, to calculate pH from pOH. Then we will use Equation 16.17 to determine the concentration of H +. Solve From Equation 16.20, we have pH = 14.00 − pOH pH = 14.00 − 10.24 = 3.76 Next we use Equation 16.17: pH = −log[H + ] = 3.76 Thus, log[H + ] = –3.76 To find [H + ], we need to determine the antilogarithm of −3.76. Your calculator will show this command as 10 x or INV log (these functions are usually above the log key). We use this function to perform the calculation: [H + ] = antilog (−3.76) = 10 −3.76 = 1.7 × 10 −4 M Comment The number of significant figures in [H+] is two because the number of decimal places in the pH is two. Check Because the pH is between 3.0 and 4.0, we know that [H+] will be between 1.0 × 10 −3 M and 1.0 × 10 −4 M. Our calculated [H + ] falls within this estimated range. A sample of freshly pressed apple juice has a pOH of 10.24. Calculate [H + ]. Calculating [H + ] from pOH

Solution Analyze and Plan Because HClO 4 is a strong acid, it is completely ionized, giving [H + ] = [ClO 4 − ] = 0.040 M. Solve pH = −log(0.040) = 1.40 Check Because [H + ] lies between 1 × 10 −2 and 1 × 10 −1, the pH will be between 2.0 and 1.0. Our calculated pH falls within the estimated range. Furthermore, because the concentration has two significant figures, the pH has two decimal places. What is the pH of a 0.040 M solution of HClO 4 ? Calculating the pH of a Strong Acid

Solution Analyze We are asked to calculate the pH of two solutions of strong bases. Plan We can calculate each pH by either of two equivalent methods. First, we could use Equation 16.16 to calculate [H + ] and then use Equation 16.17 to calculate the pH. Alternatively, we could use [OH − ] to calculate pOH and then use Equation 16.20 to calculate the pH. Solve (a)NaOH dissociates in water to give one OH − ion per formula unit. Therefore, the OH − concentration for the solution in (a) equals the stated concentration of NaOH, namely 0.028 M. Method 1: Method 2: pOH = −log(0.028) = 1.55 pH = 14.00 − pOH = 12.45 What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH) 2 ? Calculating the pH of a Strong Base

(b) Ca(OH) 2 is a strong base that dissociates in water to give two OH – ions per formula unit. Thus, the concentration of OH – (aq) for the solution in part (b) is 2 × (0.0011 M) = 0.0022 M. Method 1: Method 2: pOH = −log(0.0022) = 2.66 pH = 14.00 − pOH = 11.34 Continued Calculating the pH of a Strong Base

Analyze We are given the molar concentration of an aqueous solution of weak acid and the pH of the solution, and we are asked to determine the value of K a for the acid. Plan Although we are dealing specifically with the ionization of a weak acid, this problem is very similar to the equilibrium problems we encountered in Chapter 15. We can solve this problem using the method first outlined in Sample Exercise 15.8, starting with the chemical reaction and a tabulation of initial and equilibrium concentrations. Solve The first step in solving any equilibrium problem is to write the equation for the equilibrium reaction. The ionization of formic acid can be written as HCOOH(aq) H + (aq) + HCOO – (aq) The equilibrium-constant expression is From the measured pH, we can calculate [H + ]: pH = –log [H + ] = 2.38 log[H + ] = –2.38 [H + ] = 10 –2.38 = 4.2 × 10 –3 M A student prepared a 0.10 M solution of formic acid (HCOOH) and found its pH at 25 °C to be 2.38. Calculate K a for formic acid at this temperature. Calculating K a from Measured pH

To determine the concentrations of the species involved in the equilibrium, we imagine that the solution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acid into H + and HCOO –. For each HCOOH molecule that ionizes, one H + ion and one HCOO – ion are produced in solution. Because the pH measurement indicates that [H + ] = 4.2 × 10 –3 M at equilibrium, we can construct the following table: Notice that we have neglected the very small concentration of H + (aq) due to H 2 O autoionization. Notice also that the amount of HCOOH that ionizes is very small compared with the initial concentration of the acid. To the number of significant figures we are using, the subtraction yields 0.10 M: (0.10 – 4.2 × 10 –3 ) M ≃ 0.10 M We can now insert the equilibrium concentrations into the expression for K a : Check The magnitude of our answer is reasonable because K a for a weak acid is usually between 10 –2 and 10 –10. Calculating K a from Measured pH

Solution Analyze We are given the molar concentration of an aqueous solution of weak acid and the equilibrium concentration of H + (aq) and asked to determine the percent ionization of the acid. Plan The percent ionization is given by Equation 16.27. Solve As calculated in Sample Exercise 16.10, a 0.10 M solution of formic acid (HCOOH) contains 4.2 × 10 –3 M H + (aq). Calculate the percentage of the acid that is ionized. Calculating Percent Ionization

Solution Analyze We are given the molarity of a weak acid and are asked for the pH. From Table 16.2, K a for HCN is 4.9 × 10 –10. Plan Write the chemical equation and constructing a table of initial and equilibrium concentrations in which the equilibrium concentration of H + is our unknown. Calculate the pH of a 0.20 M solution of HCN. (Refer to Table 16.2 or Appendix D for the value of K a.) Using K a to Calculate pH

Solve Writing both the chemical equation for the ionization reaction that forms H + (aq) and the equilibrium-constant (K a ) expression for the reaction: Next, we tabulate the concentrations of the species involved in the equilibrium reaction, letting x = [H + ] at equilibrium: Substituting the equilibrium concentrations into the equilibrium-constant expression yields We next make the simplifying approximation that x, the amount of acid that dissociates, is small compared with the initial concentration of acid, 0.20 – x ≃ 0.20. Thus, Continued Using K a to Calculate pH

Solving for x, we have A concentration of 9.9 × 10 –6 M is much smaller than 5% of 0.20, the initial HCN concentration. Our simplifying approximation is therefore appropriate. We now calculate the pH of the solution: Continued Using K a to Calculate pH

Solution Analyze We are asked to calculate the percent ionization of a solution of HF. From Appendix D, we find K a = 6.8 × 10 –4. Plan We approach this problem as for previous equilibrium problems: We write the chemical equation for the equilibrium and tabulate the known and unknown concentrations of all species. We then substitute the equilibrium concentrations into the equilibrium-constant expression and solve for the unknown concentration of H +. Solve The equilibrium reaction and equilibrium concentrations are as follows: The equilibrium-constant expression is Calculate the pH and percentage of HF molecules ionized in a 0.10 M HF solution. Using the Quadratic Equation to Calculate pH and Percent Ionization

When we try solving this equation using the approximation 0.10 – x ≃ 0.10 (that is, by neglecting the concentration of acid that ionizes), we obtain Because this approximation is greater than 5% of 0.10 M, however, we should work the problem in standard quadratic form. Rearranging, we have Substituting these values in the standard quadratic formula gives Of the two solutions, only the positive value for x is chemically reasonable. From that value, we can determine [H + ] and hence the pH Continued Using the Quadratic Equation to Calculate pH and Percent Ionization x = 8.2 × 10 –3 M x 2 = (0.10 − x)(6.8 × 10 −4 ) = 6.8 × 10 −5 – (6.8 × 10 −4 )x x 2 + (6.8 × 10 −4 )x – 6.8 × 10 −5 = 0 x = [H + ] = [F – ] = 7.9 × 10 –3 M, so pH = –log[H + ] = 2.10

From our result, we can calculate the percent of molecules ionized: Continued Using the Quadratic Equation to Calculate pH and Percent Ionization

Solution Analyze We are asked to determine the pH of a 0.0037 M solution of a polyprotic acid. Plan H 2 CO 3 is a diprotic acid; the two acid-dissociation constants, K a1 and K a2 (Table 16.3), differ by more than a factor of 10 3. Consequently, the pH can be determined by considering only K a1, thereby treating the acid as if it were a monoprotic acid. The solubility of CO 2 in water at 25 °C and 0.1 atm is 0.0037 M. The common practice is to assume that all the dissolved CO 2 is in the form of carbonic acid (H 2 CO 3 ), which is produced in the reaction CO 2 (aq) + H 2 O(l) H 2 CO 3 (aq) What is the pH of a 0.0037 M solution of H 2 CO 3 ? Calculating the pH of a Solution of a Polyprotic Acid

Solve Proceeding as in Sample Exercises 16.12 and 16.13, we can write the equilibrium reaction and equilibrium concentrations as The equilibrium-constant expression is Solving this quadratic equation, we get Alternatively, because K a1 is small, we can make the simplifying approximation that x is small, so that Thus, Continued Calculating the pH of a Solution of a Polyprotic Acid x = 4.0 × 10 –5 M 0.037 – x ≃ 0.0037

Solving for x, we have Because we get the same value (to 2 significant figures) our simplifying assumption was justified. The pH is therefore Comment If we were asked for [CO 3 2− ] we would need to use K a2. Let’s illustrate that calculation. Using our calculated values of [HCO 3 − ] and [H + ] and setting [CO 3 2− ] = y, we have Assuming that y is small relative to 4.0 × 10 −5, we have Continued Calculating the pH of a Solution of a Polyprotic Acid

We see that the value for y is indeed very small compared with 4.0 × 10 −5, showing that our assumption was justified. It also shows that the ionization of HCO 3 − is negligible relative to that of H 2 CO 3, as far as production of H + is concerned. However, it is the only source of CO 3 2−, which has a very low concentration in the solution. Our calculations thus tell us that in a solution of carbon dioxide in water, most of the CO 2 is in the form of CO 2 or H 2 CO 3, only a small fraction ionizes to form H + and HCO 3 −, and an even smaller fraction ionizes to give CO 3 2−. Notice also that [CO 3 2− ] is numerically equal to K a2. Continued Calculating the pH of a Solution of a Polyprotic Acid

Solution Analyze We are given the concentration of a weak base and asked to determine the concentration of OH −. Plan We will use essentially the same procedure here as used in solving problems involving the ionization of weak acids—that is, write the chemical equation and tabulate initial and equilibrium concentrations. Calculate the concentration of OH− in a 0.15 M solution of NH 3. Using K b to Calculate OH − Solve The ionization reaction and equilibrium-constant expression are Ignoring the concentration of H 2 O because it is not involved in the equilibrium-constant expression, the equilibrium concentrations are

Continued Using K b to Calculate OH − Inserting these quantities into the equilibrium- constant expression gives Because K b is small, the amount of NH 3 that reacts with water is much smaller than the NH 3 concentration, and so we can neglect x relative to 0.15 M. Then we have Check The value obtained for x is only about 1% of the NH 3 concentration, 0.15 M. Therefore, neglecting x relative to 0.15 was justified. Comment You may be asked to find the pH of a solution of a weak base. Once you have found [OH − ], you can proceed as in Sample Exercise 16.9, where we calculated the pH of a strong base. In the present sample exercise, we have seen that the 0.15 M solution of NH 3 contains [OH − ] = 1.6 × 10 −3 M. Thus, pOH = −log(1.6 × 10 −3 ) = 2.80, and pH = 14.00 − 2.80 = 11.20. The pH of the solution is above 7 because we are dealing with a solution of a base.

Continued Using K b to Calculate OH −

Solution Analyze We are asked to determine dissociation constants for F −, the conjugate base of HF, and NH 4 +, the conjugate acid of NH 3. Plan We can use the tabulated K values for HF and NH 3 and the relationship between K a and K b to calculate the dissociation constants for their conjugates, F − and NH 4 +. Calculate (a) K b for the fluoride ion, (b) K a for the ammonium ion. Calculating K a or K b for a Conjugate Acid–Base Pair

Solve (a) For the weak acid HF K a = 6.8 × 10 −4. To calculate K b for the conjugate base, F − : (b) For NH 3, K b = 1.8 ×10 −5, and K a for the conjugate acid, NH 4 + : Continued Calculating K a or K b for a Conjugate Acid–Base Pair

Solution Analyze We are asked to predict whether a solution of Na 2 HPO 4 is acidic or basic. This substance is an ionic compound composed of Na + and HPO 4 2− ions. Plan We need to evaluate each ion, predicting whether it is acidic or basic. Because Na + is a cation of group 1A, it has no influence on pH. Thus, our analysis of whether the solution is acidic or basic must focus on the behavior of the HPO 4 2− ion. We need to consider that HPO 4 2− can act as either an acid or a base: As acid HPO 4 2– (aq) H + (aq) + PO 4 3– (aq) [16.46] As base HPO 4 2– (aq) + H 2 O H 2 PO 4 – (aq) + OH – (aq) [16.47] Of these two reactions, the one with the larger equilibrium constant determines whether the solution is acidic or basic. Predict whether the salt Na 2 HPO 4 forms an acidic solution or a basic solution when dissolved in water. Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic

Solve The value of K a for Equation 16.46 is K a3 for H 3 PO 4 : 4.2 × 10 −13 (Table 16.3). For Equation 16.47, we must calculate K b for the base HPO 4 2− from the value of K a for its conjugate acid, H 2 PO 4 −, and the relationship K a × K b = K w (Equation 16.40). The relevant value of K a for H 2 PO 4 − is Ka 2 for H 3 PO 4 : 6.2 × 10 –8 (from Table 16.3). We therefore have Continued Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic This K b value is more than 10 5 times larger than K a for HPO 4 2− ; thus, the solution is basic.

Solution Analyze We are asked to arrange two sets of compounds in order from weakest acid to strongest acid. In (a), the substances are binary compounds containing H, and in (b) the substances are oxyacids. Arrange the compounds in each series in order of increasing acid strength: (a) AsH 3, HBr, KH, H 2 Se; (b) H 2 SO 4, H 2 SeO 3, H 2 SeO 4. Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic Plan For the binary compounds, we will consider the electronegativities of As, Br, K, and Se relative to the electronegativity of H. The higher the electronegativity of these atoms, the higher the partial positive charge on H and so the more acidic the compound. For the oxyacids, we will consider both the electronegativities of the central atom and the number of oxygen atoms bonded to the central atom. Solve (a)Because K is on the left side of the periodic table, it has a very low electronegativity (0.8, from Figure 8.7, p. 310). As a result, the hydrogen in KH carries a negative charge. Thus, KH should be the least acidic (most basic) compound in the series.

Arsenic and hydrogen have similar electronegativities, 2.0 and 2.1, respectively. This means that the As—H bond is nonpolar, and so AsH 3 has little tendency to donate a proton in aqueous solution. The electronegativity of Se is 2.4, and that of Br is 2.8. Consequently, the H—Br bond is more polar than the H—Se bond, giving HBr the greater tendency to donate a proton. (This expectation is confirmed by Figure 16.18, where we see that H 2 Se is a weak acid and HBr a strong acid.) Thus, the order of increasing acidity is KH < AsH 3 < H 2 Se < HBr. (b)The acids H 2 SO 4 and H 2 SeO 4 have the same number of O atoms and the same number of OH groups. In such cases, the acid strength increases with increasing electronegativity of the central atom. Because S is slightly more electronegative than Se (2.5 vs 2.4), we predict that H 2 SO 4 is more acidic than H 2 SeO 4. Continued Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic

For acids with the same central atom, the acidity increases as the number of oxygen atoms bonded to the central atom increases. Thus, H 2 SeO 4 should be a stronger acid than H 2 SeO 3. We predict the order of increasing acidity to be H 2 SeO 3 < H 2 SeO 4 < H 2 SO 4. Continued Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic

Chapter 15 Acid-Base Equilibria

15.1 Common Ion Cations  Group I or Group II (Ca 2+, Sr 2+, or Ba 2+ ) metal cations are neutral.  Polyatomic cations are typically the conjugate acids of a weak base; e.g., NH 4 +.  Transition and post-transition metal cations are acidic. Why? (There are no H atoms in these cations!) Hydrated Cations

15.1 Common Ion  Transition and post-transition metals form hydrated cations.  The water attached to the metal is more acidic than free water molecules, making the hydrated ions acidic. Hydrated Cations

15.1 Common Ion Salt Solutions— Acidic, Basic, or Neutral? 1)Group I/II metal cation with anion of a strong acid: neutral 2)Group I/II metal cation with anion of a weak acid: basic (like the anion) 3)Transition/Post-transition metal cation or polyatomic cation with anion of a strong acid: acidic (like the cation) 4)Transition/Post-transition metal cation or polyatomic cation with anion of a weak acid: compare K a and K b ; whichever is greater dictates what the salt is.

Solution Analyze We are given the chemical formulas of five ionic compounds (salts) and asked whether their aqueous solutions will be acidic, basic, or neutral. Plan We can determine whether a solution of a salt is acidic, basic, or neutral by identifying the ions in solution and by assessing how each ion will affect the pH. Solve (a)This solution contains barium ions and acetate ions. The cation is an ion of a heavy alkaline earth metal and will therefore not affect the pH. The anion, CH 3 COO −, is the conjugate base of the weak acid CH 3 COOH and will hydrolyze to produce OH − ions, thereby making the solution basic (combination 2). (b)In this solution, NH 4 + is the conjugate acid of a weak base (NH 3 ) and is therefore acidic. Cl − is the conjugate base of a strong acid (HCl) and therefore has no influence on the pH of the solution. Because the solution contains an ion that is acidic (NH 4 + ) and one that has no influence on pH (Cl − ), the solution of NH 4 Cl will be acidic (combination 3). Determine whether aqueous solutions of each of these salts are acidic, basic, or neutral: (a) Ba(CH 3 COO) 2, (b) NH 4 Cl, (c) CH 3 NH 3 Br, (d) KNO 3, (e) Al(ClO 4 ) 3. Determining Whether Salt Solutions Are Acidic, Basic, or Neutral

(c)Here CH 3 NH 3 + is the conjugate acid of a weak base (CH 3 NH 2, an amine) and is therefore acidic, and Br − is the conjugate base of a strong acid (HBr) and therefore pH neutral. Because the solution contains one ion that is acidic and one that has no influence on pH, the solution of CH 3 NH 3 Br will be acidic (combination 3). (d)This solution contains the K + ion, which is a cation of group 1A, and the NO 3 − ion, which is the conjugate base of the strong acid HNO 3. Neither of the ions will react with water to any appreciable extent, making the solution neutral (combination 1). (e)This solution contains Al 3+ and ClO 4 − ions. Cations, such as Al 3+, that have a charge of 3+ or higher are acidic. The ClO 4 − ion is the conjugate base of a strong acid 1(HClO 4 ) and therefore does not affect pH. Thus, the solution of Al(ClO 4 ) 3 will be acidic (combination 3). Continued Determining Whether Salt Solutions Are Acidic, Basic, or Neutral

Solution Analyze NaClO is an ionic compound consisting of Na+ and ClO – ion. As such, it is a strong electrolyte that completely dissociates in solution into Na+, a spectator ion, and ClO − ion, a weak base with K b = 3.3 × 10 −7 Given this information we must calculate the number of moles of NaClO needed to increase the pH of 2.00-L of water to 10.50. Plan From the pH, we can determine the equilibrium concentration of OH −. We can then construct a table of initial and equilibrium concentrations in which the initial concentration of ClO − is our unknown. We can calculate [ClO − ] using the expression for K b. Solve A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of 10.5 Calculate the number of moles of NaClO added to the water. Using pH to Determine the Concentration of a Salt pOH = 14.00 – pH = 14.00 – 10.50 = 3.50 [OH – ] = 10 –3.50 = 3.2 × 10 –4 M

This concentration is high enough that we can assume there is only one source of OH − ; that is, we can neglect any OH − produced by the autoionization of H 2 O. We now assume a value of x for the initial concentration of ClO − and solve the equilibrium problem in the usual way. We now use the expression for the base- dissociation constant to solve for x: We say that the solution is 0.31 M in NaClO even though some of the ClO − ions have reacted with water. Because the solution is 0.31 M in NaClO and the total volume of solution is 2.00 L, 0.62 mol of NaClO is the amount of the salt that was added to the water. Continued Using pH to Determine the Concentration of a Salt

15.1 Common Ion Copyright © Cengage Learning. All rights reserved 43 Common Ion Effect  Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction.  An application of Le Châtelier’s principle.

15.1 Common Ion Example HCN(aq) + H 2 O(l) H 3 O + (aq) + CN - (aq)  Addition of NaCN will shift the equilibrium to the left because of the addition of CN -, which is already involved in the equilibrium reaction.  A solution of HCN and NaCN is less acidic than a solution of HCN alone.

Solution Analyze We are asked to determine the pH of a solution of a weak electrolyte (CH 3 COOH) and a strong electrolyte (CH 3 COONa) that share a common ion, CH 3 COO –. Plan In any problem in which we must determine the pH of a solution containing a mixture of solutes, it is helpful to proceed by a series of logical steps: (1)Consider which solutes are strong electrolytes and which are weak electrolytes, and identify the major species in solution. (2)Identify the important equilibrium reaction that is the source of H + and therefore determines pH. (3)Tabulate the concentrations of ions involved in the equilibrium. (4)Use the equilibrium-constant expression to calculate [H + ] and then pH. Solve First, because CH 3 COOH is a weak electrolyte and CH 3 COONa is a strong electrolyte, the major species in the solution are CH 3 COOH (a weak acid), Na + (which is neither acidic nor basic and is therefore a spectator in the acid–base chemistry), and CH 3 COO – (which is the conjugate base of CH 3 COOH). What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? Calculating the pH When a Common Ion Is Involved

Continued Second, [H+] and, therefore, the pH are controlled by the dissociation equilibrium of CH 3 COOH: CH 3 COOH(aq) H + (aq) + CH 3 COO – (aq) (We have written the equilibrium using H + (aq) rather than H 3 O + (aq), but both representations of the hydrated hydrogen ion are equally valid.) Third, we tabulate the initial and equilibrium concentrations as we did in solving other equilibrium problems in Chapters 15 and 16: The equilibrium concentration of CH 3 COO – (the common ion) is the initial concentration that is due to CH 3 COONa (0.30 M) plus the change in concentration (x) that is due to the ionization of CH 3 COOH. Now we can use the equilibrium-constant expression: Calculating the pH When a Common Ion Is Involved

Continued The dissociation constant for CH 3 COOH at 25 °C is from Table 16.2, or Appendix D; addition of CH 3 COONa does not change the value of this constant. Substituting the equilibrium-constant concentrations from our table into the equilibrium expression gives: Because K a is small, we assume that x is small compared to the original concentrations of CH 3 COOH and CH 3 COO – (0.30 M each). Thus, we can ignore the very small x relative to 0.30 M, giving Calculating the pH When a Common Ion Is Involved

Continued The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem.x = 1.8 × 10 –5 M = [H + ] Finally, we calculate the pH from the equilibrium concentration of H + (aq): pH = –log(1.8 × 10 –5 ) = 4.74 Comment In Section 16.6, we calculated that a 0.30 M solution of CH 3 COOH has a pH of 2.64, corresponding to [H + ] = 2.3 × 10 –3 M. Thus, the addition of CH 3 COONa has substantially decreased [H + ], as we expect from Le Châtelier’s principle. Calculating the pH When a Common Ion Is Involved

Solution Analyze We are asked to determine the concentration of F – and the pH in a solution containing the weak acid HF and the strong acid HCl. In this case the common ion is H +. Plan We can again use the four steps outlined in Sample Exercise 17.1. Solve Because HF is a weak acid and HCl is a strong acid, the major species in solution are HF, H +, and Cl –. The Cl –, which is the conjugate base of a strong acid, is merely a spectator ion in any acid–base chemistry. The problem asks for [F – ], which is formed by ionization of HF. Thus, the important equilibrium isHF(aq) H + (aq) + F – (aq) The common ion in this problem is the hydrogen (or hydronium) ion. Now we can tabulate the initial and equilibrium concentrations of each species involved in this equilibrium: Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Calculating Ion Concentrations When a Common Ion Is Involved

The equilibrium constant for the ionization of HF, from Appendix D, is 6.8 × 10 –4. Substituting the equilibrium-constant concentrations into the equilibrium expression gives If we assume that x is small relative to 0.10 or 0.20 M, this expression simplifies to This F – concentration is substantially smaller than it would be in a 0.20 M solution of HF with no added HCl. The common ion, H +, suppresses the ionization of HF. The concentration of H + (aq) is[H + ] = (0.10 + x) M ≃ 0.10 M Thus,pH = 1.00 Comment Notice that for all practical purposes, the hydrogen ion concentration is due entirely to the HCl; the HF makes a negligible contribution by comparison. Continued Calculating Ion Concentrations When a Common Ion Is Involved

Section 15.2 Buffered Solutions Key Points about Buffered Solutions  Buffered Solution – resists a change in pH.  They are weak acids or bases containing a common ion.  After addition of strong acid or base, deal with stoichiometry first, then the equilibrium. Copyright © Cengage Learning. All rights reserved 51

Section 15.2 Buffered Solutions Adding an Acid to a Buffer Copyright © Cengage Learning. All rights reserved 52 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE

Section 15.2 Buffered Solutions Buffers Copyright © Cengage Learning. All rights reserved 53 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE

Section 15.2 Buffered Solutions Henderson–Hasselbalch Equation  For a particular buffering system (conjugate acid–base pair), all solutions that have the same ratio [A – ] / [HA] will have the same pH. Copyright © Cengage Learning. All rights reserved 57

Section 15.2 Buffered Solutions  Kahn Academy Video Kahn Academy Video

Solution Analyze We are asked to calculate the pH of a buffer containing lactic acid (HC 3 H 5 O 3 ) and its conjugate base, the lactate ion (C 3 H 5 O 3 – ). Plan We will first determine the pH using the method described in Section 17.1. Because HC 3 H 5 O 3 is a weak electrolyte and NaC 3 H 5 O 3 is a strong electrolyte, the major species in solution are HC 3 H 5 O 3, Na +, and C 3 H 5 O 3 –. The Na + ion is a spectator ion. The HC 3 H 5 O 3 /C 3 H 5 O 3 – conjugate acid–base pair determines [H + ] and, thus, pH; [H + ] can be determined using the acid-dissociation equilibrium of lactic acid. Solve The initial and equilibrium concentrations of the species involved in this equilibrium are The equilibrium concentrations are governed by the equilibrium expression: Because K a is small and a common ion is present, we expect x to be small relative to either 0.12 or 0.10 M. Thus, our equation can be simplified to give What is the pH of a buffer that is 0.12 M in lactic acid [CH 3 CH(OH)COOH, or HC 3 H 5 O 3 ] and 0.10 M in sodium lactate [CH 3 CH(OH)COONa or NaC 3 H 5 O 3 ]? For lactic acid, K a = 1.4 × 10 –4. Calculating the pH of a Buffer

Solving for x gives a value that justifies our approximation: Then, we can solve for pH: Alternatively, we can use the Henderson–Hasselbalch equation with the initial concentrations of acid and base to calculate pH directly: Continued Calculating the pH of a Buffer

Solution Analyze We are asked to calculate the pH of a buffer two different ways. We know the initial concentrations of the weak acid and its conjugate base, and the K a of the weak acid. Plan We will first use the Henderson–Hasselbalch equation, which relates pK a and ratio of acid–base concentrations to the pH. This will be straightforward. Then, we will redo the calculation making no assumptions about any quantities, which means we will need to write out the initial/change/equilibrium concentrations, as we have done before. In addition, we will need to solve for quantities using the quadratic equation (since we cannot make assumptions about unknowns being small). Solve (i)The Henderson–Hasselbalch equation is We know the K a of the acid (1.8 × 10 –5 ), so we know pK a (pK a = –log K a = 4.742. We know the initial concentrations of the base, sodium acetate, and the acid, acetic acid, which we will assume are the same as the equilibrium concentrations. Calculate the pH of a buffer that initially contains 1.00 × 10 –3 M CH 3 COOH and 1.00 × 10 –4 M CH 3 COONa in the following two ways: (i) using the Henderson–Hasselbalch equation and (ii) making no assumptions about quantities (which means you will need to use the quadratic equation). The K a of CH 3 COOH is 1.80 × 10 –5. Calculating pH When the Henderson–Hasselbalch Equation May Not Be Accurate

Continued Therefore, we have Therefore, pH = 4.74 – 1.00 = 3.74 (ii)Now we will redo the calculation, without making any assumptions at all. We will solve for x, which represents the H + concentration at equilibrium, in order to calculate pH. Calculating pH When the Henderson–Hasselbalch Equation May Not Be Accurate

Continued Comment In Sample Exercise 17.3, the calculated pH is the same whether we solve exactly using the quadratic equation or make the simplifying assumption that the equilibrium concentrations of acid and base are equal to their initial concentrations. The simplifying assumption works because the concentrations of the acid–base conjugate pair are both a thousand times larger than K a. In this Sample Exercise, the acid–base conjugate pair concentrations are only 10–100 as large as K a. Therefore, we cannot assume that x is small compared to the initial concentrations (that is, that the initial concentrations are essentially equal to the equilibrium concentrations). The best answer to this Sample Exercise is pH = 4.06, obtained without assuming x is small. Thus we see that the assumptions behind the Henderson–Hasselbalch equation become less accurate as the acid/base becomes stronger and/or its concentration becomes smaller. Calculating pH When the Henderson–Hasselbalch Equation May Not Be Accurate

Solution Analyze We are asked to determine the amount of NH 4 + ion required to prepare a buffer of a specific pH. Plan The major species in the solution will be NH 4 +, Cl –, and NH 3. Of these, the Cl – ion is a spectator (it is the conjugate base of a strong acid). Thus, the NH 4 + /NH 3 conjugate acid–base pair will determine the pH of the buffer. The equilibrium relationship between NH 4 + and NH 3 is given by the base-dissociation reaction for NH 3 : The key to this exercise is to use this K b expression to calculate [NH 4 + ]. Solve We obtain [OH – ] from the given pH: pOH = 14.00 – pH = 14.00 – 9.00 = 5.00 and so [OH – ] = 1.0 × 10 –5 M Because K b is small and the common ion [NH 4 + ] is present, the equilibrium concentration of NH 3 essentially equals its initial concentration: [NH 3 ] = 0.10 M How many moles of NH 4 Cl must be added to 2.0 L of 0.10 M NH 3 to form a buffer whose pH is 9.00? (Assume that the addition of NH 4 Cl does not change the volume of the solution.) Calculating a Buffer

We now use the expression for K b to calculate [NH 4 + ]: Thus, for the solution to have pH = 9.00, [NH 4 + ] must equal 0.18 M. The number of moles of NH 4 Cl needed to produce this concentration is given by the product of the volume of the solution and its molarity: (2.0 L)(0.18 mol NH 4 Cl/L) = 0.36 mol NH 4 Cl Comment Because NH 4 + and NH 3 are a conjugate acid–base pair, we could use the Henderson–Hasselbalch equation (Equation 17.9) to solve this problem. To do so requires first using Equation 16.41 to calculate pK a for NH 4 + from the value of pK b for NH 3. We suggest you try this approach to convince yourself that you can use the Henderson–Hasselbalch equation for buffers for which you are given K b for the conjugate base rather than K a for the conjugate acid. Continued Calculating a Buffer

Solution Analyze We are asked to determine the pH of a buffer after addition of a small amount of strong base and to compare the pH change with the pH that would result if we were to add the same amount of strong base to pure water. Plan Solving this problem involves the two steps outlined in Figure 17.3. First we do a stoichiometry calculation to determine how the added OH – affects the buffer composition. Then we use the resultant buffer composition and either the Henderson–Hasselbalch equation or the equilibrium-constant expression for the buffer to determine the pH. A buffer is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.000 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH(aq) solution is added. (b) For comparison, calculate the pH of a solution made by adding 5.0 mL of 4.0 M NaOH(aq) solution to 1.000 L of pure water. Calculating pH Changes in Buffers

Solve (a)Stoichiometry Calculation: The OH – provided by NaOH reacts with CH 3 COOH, the weak acid component of the buffer. Since volumes are changing, it is prudent to figure out how many moles of reactants and products would be produced, then divide by the final volume later to obtain concentrations. Prior to this neutralization reaction, there are 0.300 mol each of CH 3 COOH and CH 3 COO –. The amount of base added is 0.0050 L × 4.0 mol/L = 0.020 mol. Neutralizing the 0.020 mol OH – requires 0.020 mol of CH 3 COOH. Consequently, the amount of CH 3 COOH decreases by 0.020 mol, and the amount of the product of the neutralization, CH 3 COO –, increases by 0.020 mol. We can create a table to see how the composition of the buffer changes as a result of its reaction with OH – : Equilibrium Calculation: We now turn our attention to the equilibrium for the ionization of acetic acid, the relationship that determines the buffer pH: CH 3 COOH(aq) H + (aq) + CH 3 COO – (aq) Continued Calculating pH Changes in Buffers

Using the quantities of CH 3 COOH and CH 3 COO – remaining in the buffer after the reaction with strong base, we determine the pH using the Henderson–Hasselbalch equation. The volume of the solution is now 1.000 L + 0.0050 L = 1.005 L due to addition of the NaOH solution: (b)To determine the pH of a solution made by adding 0.020 mol of NaOH to 1.000 L of pure water, we first determine the concentration of OH – ions in solution, [OH – ] = 0.020 mol/1.005 L = 0.020 M We use this value in Equation 16.18 to calculate pOH and then use our calculated pOH value in Equation 16.20 to obtain pH: pOH = –log[OH–] = –log(0.020) = +1.70 pH = 14 – (+1.70) = 12.30 Comment Note that the small amount of added NaOH changes the pH of water significantly. In contrast, the pH of the buffer changes very little when the NaOH is added. Continued Calculating pH Changes in Buffers

Section 15.2 Buffered Solutions Buffered Solution Characteristics  Buffers contain relatively large concentrations of a weak acid and corresponding conjugate base.  Added H + reacts to completion with the weak base.  Added OH - reacts to completion with the weak acid. Copyright © Cengage Learning. All rights reserved 70

Section 15.2 Buffered Solutions Buffered Solution Characteristics  The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A – or B and BH + ) are large compared with amounts of H + or OH – added. Copyright © Cengage Learning. All rights reserved 71

Section 15.3 Buffering Capacity  The amount of protons or hydroxide ions the buffer can absorb without a significant change in pH.  Determined by the magnitudes of [HA] and [A – ].  A buffer with large capacity contains large concentrations of the buffering components. Copyright © Cengage Learning. All rights reserved 72

Section 15.3 Buffering Capacity  Optimal buffering occurs when [HA] is equal to [A – ].  It is for this condition that the ratio [A – ] / [HA] is most resistant to change when H + or OH – is added to the buffered solution. Copyright © Cengage Learning. All rights reserved 73

Section 15.3 Buffering Capacity Choosing a Buffer  pK a of the weak acid to be used in the buffer should be as close as possible to the desired pH. Copyright © Cengage Learning. All rights reserved 74

Section 15.4 Titrations and pH Curves Titration Curve  Plotting the pH of the solution being analyzed as a function of the amount of titrant added.  Equivalence (Stoichiometric) Point – point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated. Copyright © Cengage Learning. All rights reserved 75

Section 15.4 Titrations and pH Curves Neutralization of a Strong Acid with a Strong Base Copyright © Cengage Learning. All rights reserved 76 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERECLICK HERE

Solution Analyze We are asked to calculate the pH at two points in the titration of a strong acid with a strong base. The first point is just before the equivalence point, so we expect the pH to be determined by the small amount of strong acid that has not yet been neutralized. The second point is just after the equivalence point, so we expect this pH to be determined by the small amount of excess strong base. Plan (a) As the NaOH solution is added to the HCl solution, H + (aq) reacts with OH – (aq) to form H 2 O. Both Na + and Cl – are spectator ions, having negligible effect on the pH. To determine the pH of the solution, we must first determine how many moles of H + were originally present and how many moles of OH – were added. We can then calculate how many moles of each ion remain after the neutralization reaction. To calculate [H + ], and hence pH, we must also remember that the volume of the solution increases as we add titrant, thus diluting the concentration of all solutes present. Therefore, it is best to deal with moles first, and then convert to molarities using total solution volumes (volume of acid plus volume of base). Solve The number of moles of H+ in the original HCl solution is given by the product of the volume of the solution and its molarity: Likewise, the number of moles of OH –, in 49.0 mL of 0.100 M NaOH is Calculate the pH when (a) 49.0 mL and (b) 51.0 mL of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution. Calculations for a Strong Acid–Strong Base Titration

Continued Because we have not reached the equivalence point, there are more moles of H + present than OH –. Therefore, OH – is the limiting reactant. Each mole of OH – reacts with 1 mol of H +. Using the convention introduced in Sample Exercise 17.6, we have The volume of the reaction mixture increases as the NaOH solution is added to the HCl solution. Thus, at this point in the titration, the volume in the titration flask is50.0 mL + 49.0 mL = 99.0 mL = 0.0990 L Thus, the concentration of H + (aq) in the flask is The corresponding pH is –log(1.0 × 10 –3 ) = 3.00 Calculations for a Strong Acid–Strong Base Titration

Continued Plan (b) We proceed in the same way as we did in part (a) except we are now past the equivalence point and have more OH – in the solution than H +. As before, the initial number of moles of each reactant is determined from their volumes and concentrations. The reactant present in smaller stoichiometric amount (the limiting reactant) is consumed completely, leaving an excess of hydroxide ion. Solve In this case, the volume in the titration flask is 50.0 mL + 51.0 mL = 101.0 mL = 0.1010 L Hence, the concentration of OH – (aq) in the flask is and we have pOH = –log(1.0 × 10 –3 ) = 3.00 pH = 14.00 – pOH = 14.00 – 3.00 = 11.00 Calculations for a Strong Acid–Strong Base Titration

Continued Comment Note that the pH increased by only two pH units, from 1.00 (Figure 17.7) to 3.00, after the first 49.0 mL of NaOH solution was added, but jumped by eight pH units, from 3.00 to 11.00, as 2.0 mL of base solution was added near the equivalence point. Such a rapid rise in pH near the equivalence point is a characteristic of titrations involving strong acids and strong bases.. Calculations for a Strong Acid–Strong Base Titration

Solution Analyze We are asked to calculate the pH before the equivalence point of the titration of a weak acid with a strong base. Plan We first must determine the number of moles of CH 3 COOH and CH 3 COO – present after the neutralization reaction (the stoichiometry calculation). We then calculate pH using K a, [CH 3 COOH], and [CH 3 COO – ] (the equilibrium calculation). Solve Stoichiometry Calculation: The product of the volume and concentration of each solution gives the number of moles of each reactant present before the neutralization: The 4.50 × 10 –3 of NaOH consumes 4.50 × 10 –3 of CH 3 COOH: Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH (K a = 1.8 × 10 –5 ). Calculations for a Weak Acid–Strong Base Titration

Continued The total volume of the solution is 45.0 mL + 50.0 mL = 95.0 mL = 0.0950 L The resulting molarities of CH 3 COOH and CH 3 COO – after the reaction are therefore Equilibrium Calculation: The equilibrium between CH 3 COOH and CH 3 COO – must obey the equilibrium- constant expression for CH 3 COOH: Solving for [H + ] gives Comment We could have solved for pH equally well using the Henderson–Hasselbalch equation in the last step. Calculations for a Weak Acid–Strong Base Titration

Solution Analyze We are asked to determine the pH at the equivalence point of the titration of a weak acid with a strong base. Because the neutralization of a weak acid produces its anion, a conjugate base that can react with water, we expect the pH at the equivalence point to be greater than 7. Plan The initial number of moles of acetic acid equals the number of moles of acetate ion at the equivalence point. We use the volume of the solution at the equivalence point to calculate the concentration of acetate ion. Because the acetate ion is a weak base, we can calculate the pH using K b and [CH 3 COO – ]. Solve The number of moles of acetic acid in the initial solution is obtained from the volume and molarity of the solution: Moles = M × L = (0.100 mol/L)(0.0500 L) = 5.00 × 10 –3 mol CH 3 COOH Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. Calculating the pH at the Equivalence Point

Hence, 5.00 × 10 –3 mol of CH 3 COO – is formed. It will take 50.0 mL of NaOH to reach the equivalence point (Figure 17.9). The volume of this salt solution at the equivalence point is the sum of the volumes of the acid and base, 50.0 mL + 50.0 mL = 100.0 mL = 0.1000 L. Thus, the concentration of CH 3 COO – is Continued Calculating the pH at the Equivalence Point

The CH 3 COO – ion is a weak base: CH 3 COO – (aq) + H 2 O(l) CH 3 COOH(aq) + OH – (aq) The K b for CH 3 COO – can be calculated from the K a value of its conjugate acid, K b = K w /K a = (1.0 × 10 –14 )/(1.8 × 10 –5 ) = 5.6 × 10 –10. Using the K b expression, we have Making the approximation that 0.0500 – x ≃ 0.0500, and then solving for x, we have x = [OH – ] = 5.3 × 10 –6 M, which gives pOH = 5.28 and pH = 8.72. Check The pH is above 7, as expected for the salt of a weak acid and strong base. Continued Calculating the pH at the Equivalence Point

Section 15.4 Titrations and pH Curves Weak Acid–Strong Base Titration Step 1:A stoichiometry problem (reaction is assumed to run to completion) then determine concentration of acid remaining and conjugate base formed. Step 2: An equilibrium problem (determine position of weak acid equilibrium and calculate pH). Copyright © Cengage Learning. All rights reserved 88

Section 15.4 Titrations and pH Curves Consider a solution made by mixing 0.10 mol of HCN (K a = 6.2 × 10 –10 ) with 0.040 mol NaOH in 1.0 L of aqueous solution. What are the major species immediately upon mixing (that is, before a reaction)? HCN, Na +, OH –, H 2 O Copyright © Cengage Learning. All rights reserved 89 CONCEPT CHECK!

Section 15.4 Titrations and pH Curves Let’s Think About It…  Why isn’t NaOH a major species?  Why aren’t H + and CN – major species?  List all possibilities for the dominant reaction. Copyright © Cengage Learning. All rights reserved 90

Section 15.4 Titrations and pH Curves Let’s Think About It… The possibilities for the dominant reaction are: 1.H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) 2.HCN(aq) + H 2 O(l) H 3 O + (aq) + CN – (aq) 3.HCN(aq) + OH – (aq) CN – (aq) + H 2 O(l) 4.Na + (aq) + OH – (aq) NaOH 5.Na + (aq) + H 2 O(l) NaOH + H + (aq) Copyright © Cengage Learning. All rights reserved 91

Section 15.4 Titrations and pH Curves Let’s Think About It…  How do we decide which reaction controls the pH? H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) HCN(aq) + H 2 O(l) H 3 O + (aq) + CN – (aq) HCN(aq) + OH – (aq) CN – (aq) + H 2 O(l)

Section 15.4 Titrations and pH Curves Let’s Think About It… HCN(aq) + OH – (aq) CN – (aq) + H 2 O(l)  What are the major species after this reaction occurs? HCN, CN –, H 2 O, Na + Copyright © Cengage Learning. All rights reserved 93

Section 15.4 Titrations and pH Curves Let’s Think About It…  Now you can treat this situation as before.  List the possibilities for the dominant reaction.  Determine which controls the pH. Copyright © Cengage Learning. All rights reserved 94

Section 15.4 Titrations and pH Curves Calculate the pH of a solution made by mixing 0.20 mol HC 2 H 3 O 2 (K a = 1.8 × 10 –5 ) with 0.030 mol NaOH in 1.0 L of aqueous solution. Copyright © Cengage Learning. All rights reserved 95 CONCEPT CHECK!

Section 15.4 Titrations and pH Curves Let’s Think About It…  What are the major species in solution? Na +, OH –, HC 2 H 3 O 2, H 2 O  Why isn’t NaOH a major species?  Why aren’t H + and C 2 H 3 O 2 – major species? Copyright © Cengage Learning. All rights reserved 96

Section 15.4 Titrations and pH Curves Let’s Think About It…  What are the possibilities for the dominant reaction? 1.H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) 2.HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 – (aq) 3.HC 2 H 3 O 2 (aq) + OH – (aq) C 2 H 3 O 2 – (aq) + H 2 O(l) 4.Na + (aq) + OH – (aq) NaOH(aq) 5.Na + (aq) + H 2 O(l) NaOH + H + (aq)  Which of these reactions really occur? Copyright © Cengage Learning. All rights reserved 97

Section 15.4 Titrations and pH Curves Let’s Think About It…  Which reaction controls the pH? H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH – (aq) HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 – (aq) HC 2 H 3 O 2 (aq) + OH – (aq) C 2 H 3 O 2 – (aq) + H 2 O(l)  How do you know? Copyright © Cengage Learning. All rights reserved 98

Section 15.4 Titrations and pH Curves Let’s Think About It… K = 1.8 × 10 9 Copyright © Cengage Learning. All rights reserved 99 HC 2 H 3 O 2 (aq) +OH – C 2 H 3 O 2 – (aq)+ H 2 O Before0.20 mol 0.030 mol0 Change–0.030 mol +0.030 mol After0.17 mol 0 0.030 mol

Section 15.4 Titrations and pH Curves Steps Toward Solving for pH K a = 1.8 × 10 –5 pH = 3.99 Copyright © Cengage Learning. All rights reserved 100 HC 2 H 3 O 2 (aq) +H2OH2OH3O+H3O+ + C 2 H 3 O 2 - (aq) Initial 0.170 M~00.030 M Change –x+x Equilibrium 0.170 – xx0.030 + x

Section 15.4 Titrations and pH Curves Calculate the pH of a 100.0 mL solution of 0.100 M acetic acid (HC 2 H 3 O 2 ), which has a K a value of 1.8 x10 –5. pH = 2.87 Copyright © Cengage Learning. All rights reserved 101 EXERCISE!

Solution Analyze We are asked to calculate the pH before the equivalence point of the titration of a weak acid with a strong base. Plan We first must determine the number of moles of CH 3 COOH and CH 3 COO – present after the neutralization reaction (the stoichiometry calculation). We then calculate pH using K a, [CH 3 COOH], and [CH 3 COO – ] (the equilibrium calculation). Solve Stoichiometry Calculation: The product of the volume and concentration of each solution gives the number of moles of each reactant present before the neutralization: The 4.50 × 10 –3 of NaOH consumes 4.50 × 10 –3 of CH 3 COOH: Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH 3 COOH (K a = 1.8 × 10 –5 ). Calculations for a Weak Acid–Strong Base Titration

Continued The total volume of the solution is 45.0 mL + 50.0 mL = 95.0 mL = 0.0950 L The resulting molarities of CH 3 COOH and CH 3 COO – after the reaction are therefore Equilibrium Calculation: The equilibrium between CH 3 COOH and CH 3 COO – must obey the equilibrium- constant expression for CH 3 COOH: Solving for [H + ] gives Comment We could have solved for pH equally well using the Henderson–Hasselbalch equation in the last step. Calculations for a Weak Acid–Strong Base Titration

Solution Analyze We are asked to determine the pH at the equivalence point of the titration of a weak acid with a strong base. Because the neutralization of a weak acid produces its anion, a conjugate base that can react with water, we expect the pH at the equivalence point to be greater than 7. Plan The initial number of moles of acetic acid equals the number of moles of acetate ion at the equivalence point. We use the volume of the solution at the equivalence point to calculate the concentration of acetate ion. Because the acetate ion is a weak base, we can calculate the pH using K b and [CH 3 COO – ]. Solve The number of moles of acetic acid in the initial solution is obtained from the volume and molarity of the solution: Moles = M × L = (0.100 mol/L)(0.0500 L) = 5.00 × 10 –3 mol CH 3 COOH Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH 3 COOH with 0.100 M NaOH. Calculating the pH at the Equivalence Point

Hence, 5.00 × 10 –3 mol of CH 3 COO – is formed. It will take 50.0 mL of NaOH to reach the equivalence point (Figure 17.9). The volume of this salt solution at the equivalence point is the sum of the volumes of the acid and base, 50.0 mL + 50.0 mL = 100.0 mL = 0.1000 L. Thus, the concentration of CH 3 COO – is Continued Calculating the pH at the Equivalence Point

The CH 3 COO – ion is a weak base: CH 3 COO – (aq) + H 2 O(l) CH 3 COOH(aq) + OH – (aq) The K b for CH 3 COO – can be calculated from the K a value of its conjugate acid, K b = K w /K a = (1.0 × 10 –14 )/(1.8 × 10 –5 ) = 5.6 × 10 –10. Using the K b expression, we have Making the approximation that 0.0500 – x ≃ 0.0500, and then solving for x, we have x = [OH – ] = 5.3 × 10 –6 M, which gives pOH = 5.28 and pH = 8.72. Check The pH is above 7, as expected for the salt of a weak acid and strong base. Continued Calculating the pH at the Equivalence Point

Section 15.5 Acid-Base Indicators  Marks the end point of a titration by changing color.  The equivalence point is not necessarily the same as the end point (but they are ideally as close as possible). Copyright © Cengage Learning. All rights reserved 110

Solution Analyze We are asked to determine the pH of a solution of a weak electrolyte (CH 3 COOH) and a strong electrolyte (CH 3 COONa) that share a common ion, CH 3 COO –. Plan In any problem in which we must determine the pH of a solution containing a mixture of solutes, it is helpful to proceed by a series of logical steps: (1)Consider which solutes are strong electrolytes and which are weak electrolytes, and identify the major species in solution. (2)Identify the important equilibrium reaction that is the source of H + and therefore determines pH. (3)Tabulate the concentrations of ions involved in the equilibrium. (4)Use the equilibrium-constant expression to calculate [H + ] and then pH. Solve First, because CH 3 COOH is a weak electrolyte and CH 3 COONa is a strong electrolyte, the major species in the solution are CH 3 COOH (a weak acid), Na + (which is neither acidic nor basic and is therefore a spectator in the acid–base chemistry), and CH 3 COO – (which is the conjugate base of CH 3 COOH). What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? Calculating the pH When a Common Ion Is Involved

Continued Second, [H+] and, therefore, the pH are controlled by the dissociation equilibrium of CH 3 COOH: CH 3 COOH(aq) H + (aq) + CH 3 COO – (aq) (We have written the equilibrium using H + (aq) rather than H 3 O + (aq), but both representations of the hydrated hydrogen ion are equally valid.) Third, we tabulate the initial and equilibrium concentrations as we did in solving other equilibrium problems in Chapters 15 and 16: The equilibrium concentration of CH 3 COO – (the common ion) is the initial concentration that is due to CH 3 COONa (0.30 M) plus the change in concentration (x) that is due to the ionization of CH 3 COOH. Now we can use the equilibrium-constant expression: Calculating the pH When a Common Ion Is Involved

Continued The dissociation constant for CH 3 COOH at 25 °C is from Table 16.2, or Appendix D; addition of CH 3 COONa does not change the value of this constant. Substituting the equilibrium-constant concentrations from our table into the equilibrium expression gives: Because K a is small, we assume that x is small compared to the original concentrations of CH 3 COOH and CH 3 COO – (0.30 M each). Thus, we can ignore the very small x relative to 0.30 M, giving Calculating the pH When a Common Ion Is Involved

Continued The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem.x = 1.8 × 10 –5 M = [H + ] Finally, we calculate the pH from the equilibrium concentration of H + (aq): pH = –log(1.8 × 10 –5 ) = 4.74 Comment In Section 16.6, we calculated that a 0.30 M solution of CH 3 COOH has a pH of 2.64, corresponding to [H + ] = 2.3 × 10 –3 M. Thus, the addition of CH 3 COONa has substantially decreased [H + ], as we expect from Le Châtelier’s principle. Calculating the pH When a Common Ion Is Involved

Solution Analyze We are asked to determine the concentration of F – and the pH in a solution containing the weak acid HF and the strong acid HCl. In this case the common ion is H +. Plan We can again use the four steps outlined in Sample Exercise 17.1. Solve Because HF is a weak acid and HCl is a strong acid, the major species in solution are HF, H +, and Cl –. The Cl –, which is the conjugate base of a strong acid, is merely a spectator ion in any acid–base chemistry. The problem asks for [F – ], which is formed by ionization of HF. Thus, the important equilibrium isHF(aq) H + (aq) + F – (aq) The common ion in this problem is the hydrogen (or hydronium) ion. Now we can tabulate the initial and equilibrium concentrations of each species involved in this equilibrium: Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Calculating Ion Concentrations When a Common Ion Is Involved

The equilibrium constant for the ionization of HF, from Appendix D, is 6.8 × 10 –4. Substituting the equilibrium-constant concentrations into the equilibrium expression gives If we assume that x is small relative to 0.10 or 0.20 M, this expression simplifies to This F – concentration is substantially smaller than it would be in a 0.20 M solution of HF with no added HCl. The common ion, H +, suppresses the ionization of HF. The concentration of H + (aq) is[H + ] = (0.10 + x) M ≃ 0.10 M Thus,pH = 1.00 Comment Notice that for all practical purposes, the hydrogen ion concentration is due entirely to the HCl; the HF makes a negligible contribution by comparison. Continued Calculating Ion Concentrations When a Common Ion Is Involved

Solution Analyze We are asked to give the conjugate base for several acids and the conjugate acid for several bases. Plan The conjugate base of a substance.

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Solution Analyze We are asked to give the conjugate base for several acids and the conjugate acid for several bases. Plan The conjugate base of a substance.

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Presentation on theme: "Solution Analyze We are asked to give the conjugate base for several acids and the conjugate acid for several bases. Plan The conjugate base of a substance."— Presentation transcript:

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