Chapter 16 Acid–Base Equilibria Lecture Presentation James F. Kirby Quinnipiac University Hamden, CT © 2015 Pearson Education, Inc.

Acids and Bases © 2015 Pearson Education, Inc & 16.2 Some Definitions Arrhenius –An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. –A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions. Brønsted–Lowry –An acid is a proton donor. –A base is a proton acceptor.

Acids and Bases © 2015 Pearson Education, Inc. Brønsted–Lowry Acid and Base A Brønsted–Lowry acid must have at least one removable (acidic) proton (H + ) to donate. A Brønsted–Lowry base must have at least one nonbonding pair of electrons to accept a proton (H + ).

Acids and Bases © 2015 Pearson Education, Inc. What Is Different about Water? Water can act as a Brønsted–Lowry base and accept a proton (H + ) from an acid, as on the previous slide. It can also donate a proton and act as an acid, as is seen below. This makes water amphiprotic.

Acids and Bases © 2015 Pearson Education, Inc. Conjugate Acids and Bases The term conjugate means “joined together as a pair.” Reactions between acids and bases always yield their conjugate bases and acids.

Acids and Bases © 2015 Pearson Education, Inc. Solution 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. (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 − ? Sample Exercise 16.1 Identifying Conjugate Acids and Bases The other conjugate bases are HS − PH 3 CO 3 2−

Acids and Bases © 2015 Pearson Education, Inc. Solution (b)If we add a proton to CN −, we get HCN, its conjugate acid. (b)What is the conjugate acid of CN −, SO 4 2−, H 2 O, HCO 3 − ? Sample Exercise 16.1 Identifying Conjugate Acids and Bases The other conjugate acids are HSO 4 − H3O+H3O+ H 2 CO 3, Notice that the hydrogen carbonate ion (HCO 3 − ) is amphiprotic. It can act as either an acid or a base.

Acids and Bases © 2015 Pearson Education, Inc. 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. Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions

Acids and Bases © 2015 Pearson Education, Inc. 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 a base. We are also asked to identify the conjugate pairs in each equation. 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. Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions 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).

Acids and Bases © 2015 Pearson Education, Inc. Solution (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). (b) in which the ion acts as a base. In both cases identify the conjugate acid–base pairs. Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions

Acids and Bases © 2015 Pearson Education, Inc. Relative Strengths of Acids and Bases Acids above the line with H 2 O as a base are strong acids; their conjugate bases do not act as acids in water. Bases below the line with H 2 O as an acid are strong bases; their conjugate acids do not act as acids in water. The substances between the lines with H 2 O are conjugate acid–base pairs in water.

Acids and Bases © 2015 Pearson Education, Inc. Acid and Base Strength In every acid–base reaction, equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base.  HCl(aq) + H 2 O(l) → H 3 O + (aq) + Cl  (aq)  H 2 O is a much stronger base than Cl , so the equilibrium lies far to the right (K >> 1).  CH 3 COOH(aq) + H 2 O(l) ⇌ H 3 O + (aq) + CH 3 COO – (aq)  Acetate is a stronger base than H 2 O, so the equilibrium favors the left side (K < 1).

Acids and Bases © 2015 Pearson Education, Inc. For the following proton-transfer reaction use Figure 16.4 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) Sample Exercise 16.3 Predicting the Position of a Proton-Transfer Equilibrium

Acids and Bases © 2015 Pearson Education, Inc. 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 the forward 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. HSO 4 − (aq) + CO 3 2 − (aq) SO 4 2 − (aq) + HCO 3 − (aq)

Acids and Bases © 2015 Pearson Education, Inc. 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.

Acids and Bases © 2015 Pearson Education, Inc Autoionization of Water Water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. This is referred to as autoionization.

Acids and Bases © 2015 Pearson Education, Inc. Ion Product Constant The equilibrium expression for this process is K c = [H 3 O + ][OH  ] (The term H 2 O is excluded from the equilibrium constant expression because we exclude the concentrations of pure solids and liquids) This special equilibrium constant is referred to as the ion product constant for water, K w.

Acids and Bases © 2015 Pearson Education, Inc. Ion Product Constant This special equilibrium constant is referred to as the ion product constant for water, K w. At 25 °C, K w = 1.0  10  14 K w = [1.0  10  7 ][1.0  10  7 ] K w = [H 3 O + ][OH  ]

Acids and Bases © 2015 Pearson Education, Inc. Aqueous Solutions Can Be Acidic, Basic, or Neutral If a solution is neutral, [H + ] = [OH – ]. If a solution is acidic, [H + ] > [OH – ]. If a solution is basic, [H + ] < [OH – ].

Acids and Bases © 2015 Pearson Education, Inc. Suppose that equal volumes of the middle and right samples in the figure were mixed. Would the resultant solution be acidic, neutral or basic? basic

Acids and Bases © 2015 Pearson Education, Inc. 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 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. Sample Exercise 16.4 Calculating [H + ] for Pure Water

Acids and Bases © 2015 Pearson Education, Inc. Calculate the concentration of H + (aq) in (a) a solution in which [OH − ] is 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. Sample Exercise 16.5 Calculating [H + ] from [OH − ]

Acids and Bases © 2015 Pearson Education, Inc. 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. Calculate the concentration of H + (aq) in (a) a solution in which [OH − ] is M Sample Exercise 16.5 Calculating [H + ] from [OH − ]

Acids and Bases © 2015 Pearson Education, Inc. Solve (a) Using Equation 16.16, we have This solution is basic because Calculate the concentration of H + (aq) in (a) a solution in which [OH − ] is M Sample Exercise 16.5 Calculating [H + ] from [OH − ]

Acids and Bases © 2015 Pearson Education, Inc. (b) In this instance This solution is acidic because Calculate the concentration of H + (aq) in (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. Sample Exercise 16.5 Calculating [H + ] from [OH − ]

Acids and Bases © 2015 Pearson Education, Inc The pH Scale pH is a method of reporting hydrogen ion concentration. pH = –log[H + ] Neutral pH is Acidic pH is below Basic pH is above 7.00.

Acids and Bases © 2015 Pearson Education, Inc. 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, Equation pH = –log[H + ] (a) In the first instance we found [H+] to be 1.0 × 10 −12 M Calculate the pH values for the two solutions of Sample Exercise Sample Exercise 16.6 Calculating pH from [H + ]

Acids and Bases © 2015 Pearson Education, Inc. Sample Exercise 16.6 Calculating pH from [H + ] Solve pH = –log[H + ] (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) = Because 1.0 × 10 −12 has two significant figures, the pH has two decimal places,

Acids and Bases © 2015 Pearson Education, Inc. (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 to calculate the pH: pH = –log[H + ] Calculate the pH values for the two solutions of Sample Exercise Sample Exercise 16.6 Calculating pH from [H + ]

Acids and Bases © 2015 Pearson Education, Inc. (b) For the second solution, [H + ] = 5.6 × 10 −6 M. 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 Sample Exercise Sample Exercise 16.6 Calculating pH from [H + ]

Acids and Bases © 2015 Pearson Education, Inc (cont) Other “p” Scales The “p” in pH tells us to take the –log of a quantity (in this case, hydrogen ions). Some other “p” systems are  pOH: –log[OH  ]  pK w : –log K w

Acids and Bases © 2015 Pearson Education, Inc. Relating pH and pOH Because [H 3 O + ][OH  ] = K w = 1.0  10  14 we can take the –log of the equation –log[H 3 O + ] + –log[OH  ] = –log K w = which results in pH + pOH = pK w = 14.00

Acids and Bases © 2015 Pearson Education, Inc. Solution 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 pH = –log[H + ] to determine the concentration of H +. A sample of freshly pressed apple juice has a pOH of Calculate [H + ]. Sample Exercise 16.7 Calculating [H + ] from pOH

Acids and Bases © 2015 Pearson Education, Inc. pH = − pOH pH = − = 3.76 A sample of freshly pressed apple juice has a pOH of Calculate [H + ]. Sample Exercise 16.7 Calculating [H + ] from pOH

Acids and Bases © 2015 Pearson Education, Inc. 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 Calculate [H + ]. Sample Exercise 16.7 Calculating [H + ] from pOH

Acids and Bases © 2015 Pearson Education, Inc. How Do We Measure pH? Indicators, including litmus paper, are used for less accurate measurements; an indicator is one color in its acid form and another color in its basic form. pH meters are used for accurate measurement of pH; electrodes indicate small changes in voltage to detect pH.

Acids and Bases © 2015 Pearson Education, Inc. Which of these is best suited to distinguish between a solution that is slightly acidic and one that is slightly basic?

Acids and Bases © 2015 Pearson Education, Inc Strong Acids You will recall that the seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, and HClO 4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution; e.g., HA + H 2 O → H 3 O + + A – So, for the monoprotic strong acids, [H 3 O + ] = [acid]

Acids and Bases © 2015 Pearson Education, Inc. Solution Analyze and Plan Because HClO 4 is a strong acid, it is completely ionized, giving [H + ] = [ClO 4 − ] = 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 M solution of HClO 4 ? Sample Exercise 16.8 Calculating the pH of a Strong Acid

Acids and Bases © 2015 Pearson Education, Inc. Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+ ). Again, these substances dissociate completely in aqueous solution; e.g., MOH(aq) → M + (aq) + OH – (aq) or M(OH) 2 (aq) → M 2+ (aq) + 2 OH – (aq)

Acids and Bases © 2015 Pearson Education, Inc. Strong Bases Which solution has a higher pH, a M solution of NaOH or a M solution of Ba(OH) 2 ? Both NaOH and Ba(OH) 2 are soluble hydroxides. The hydroxide concentration of NaOH are 0.001M and Ba(OH) 2 is M Ba(OH) 2 has a higher hydroxide concentration, it is more basic and has a higher pH.

Acids and Bases © 2015 Pearson Education, Inc. 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 to calculate [H + ] and then use Equation to calculate the pH. Alternatively, we could use [OH − ] to calculate pOH and then use Equation to calculate the pH. What is the pH of (a) a M solution of NaOH, (b) a M solution of Ca(OH) 2 ? Sample Exercise 16.9 Calculating the pH of a Strong Base

Acids and Bases © 2015 Pearson Education, Inc. Solution (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 M. Method 1: What is the pH of (a) a M solution of NaOH, (b) a M solution of Ca(OH) 2 ? Sample Exercise 16.9 Calculating the pH of a Strong Base

Acids and Bases © 2015 Pearson Education, Inc. Solution Method 2: pOH = −log(0.028) = 1.55 pH = − pOH = What is the pH of (a) a M solution of NaOH, (b) a M solution of Ca(OH) 2 ? Sample Exercise 16.9 Calculating the pH of a Strong Base

Acids and Bases © 2015 Pearson Education, Inc. (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 × ( M) = M. Sample Exercise 16.9 Calculating the pH of a Strong Base

Acids and Bases © 2015 Pearson Education, Inc. (b) Method 1: Method 2: pOH = −log(0.0022) = 2.66 pH = − pOH = Sample Exercise 16.9 Calculating the pH of a Strong Base

Acids and Bases © 2015 Pearson Education, Inc Weak Acids For a weak acid, the equation for its dissociation is HA(aq) + H 2 O(l) ⇌ H 3 O + (aq) + A – (aq) Since it is an equilibrium, there is an equilibrium constant related to it, called the acid-dissociation constant, K a : K a = [H 3 O + ][A – ] [HA]

Acids and Bases © 2015 Pearson Education, Inc Weak Acids The greater the value of K a, the stronger is the acid. Based on the table which element is most commonly bonded to the acidic hydrogen?

Acids and Bases © 2015 Pearson Education, Inc. Comparing Strong and Weak Acids What is present in solution for a strong acid versus a weak acid? Strong acids completely dissociate to ions. Weak acids only partially dissociate to ions.

Acids and Bases © 2015 Pearson Education, Inc. Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25  C is Calculate K a for formic acid at this temperature.  We know that [H 3 O + ][HCOO – ] [HCOOH] K a =  To calculate K a, we need the equilibrium concentrations of all three things.  We can find [H 3 O + ], which is the same as [HCOO – ], from the pH.  [H 3 O + ] = [HCOO – ] = 10 –2.38 = 4.2 × 10 –3

Acids and Bases © 2015 Pearson Education, Inc. Calculating K a from pH Now we can set up a table for equilibrium concentrations. We know initial HCOOH (0.10 M) and ion concentrations (0 M); we found equilibrium ion concentrations (4.2 × 10 –3 M); so we calculate the change, then the equilibrium HCOOH concentration. [HCOOH], M[H 3 O + ], M[HCOO  ], M Initially Change  4.2  10   10  3 At equilibrium 0.10  4.2  10  3 = =  10  3

Acids and Bases © 2015 Pearson Education, Inc. Calculating K a from pH [4.2  10  3 ][4.2  10  3 ] [0.10] K a = = 1.8  10  4 This allows us to calculate K a by putting in the equilibrium concentrations.

Acids and Bases © 2015 Pearson Education, Inc. Solution 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. A student prepared a 0.10 M solution of formic acid (HCOOH) and found its pH at 25 °C to be Calculate K a for formic acid at this temperature. Sample Exercise Calculating K a from Measured pH

Acids and Bases © 2015 Pearson Education, Inc. Solution 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 A student prepared a 0.10 M solution of formic acid (HCOOH) and found its pH at 25 °C to be Calculate K a for formic acid at this temperature. Sample Exercise Calculating K a from Measured pH 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

Acids and Bases © 2015 Pearson Education, Inc. 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: Sample Exercise Calculating K a from Measured pH

Acids and Bases © 2015 Pearson Education, Inc. 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. Sample Exercise Calculating K a from Measured pH

Acids and Bases © 2015 Pearson Education, Inc. Calculating Percent Ionization Percent ionization =  100 In this example, [H 3 O + ] eq = 4.2  10  3 M [HCOOH] initial = 0.10 M [H 3 O + ] eq [HA] initial Percent ionization =   10  = 4.2%

Acids and Bases © 2015 Pearson Education, Inc. 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 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. Sample Exercise Calculating Percent Ionization

Acids and Bases © 2015 Pearson Education, Inc. 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. Sample Exercise Calculating Percent Ionization

Acids and Bases © 2015 Pearson Education, Inc. Method to Follow to Calculate pH Using K a 1)Write the chemical equation for the ionization equilibrium. 2)Write the equilibrium constant expression. 3)Set up a table for Initial/Change in/Equilibrium Concentration to determine equilibrium concentrations as a function of change (x). 4)Substitute equilibrium concentrations into the equilibrium constant expression and solve for x. (Make assumptions if you can!)

Acids and Bases © 2015 Pearson Education, Inc. Example Calculate the pH of a 0.30 M solution of acetic acid, HC 2 H 3 O 2, at 25  C. 1) HC 2 H 3 O 2 + H 2 O ⇌ H 3 O + + C 2 H 3 O 2 – 2) K a = [H 3 O + ][C 2 H 3 O 2 – ] / [HC 2 H 3 O 2 ] 3) CH 3 COOH (M)H 3 O + (M)CH 3 COO – (M) Initial Concentration (M) Change in Concentration (M) –x–x+x+x+x+x Equilibrium Concentration (M) 0.30 – xxx

Acids and Bases © 2015 Pearson Education, Inc. Example (concluded) 4) K a = [H 3 O + ][C 2 H 3 O 2 – ] / [HC 2 H 3 O 2 ] = (x)(x) / (0.30 – x) If we assume that x << 0.30, then 0.30 – x becomes The problem becomes easier, since we don’t have to use the quadratic formula to solve it. K a = 1.8 × 10 –5 = x 2 / 0.30, so x = 2.3 × 10 –3 x = [H 3 O + ], so pH = –log(2.3 × 10 –3 ) = 2.64

Acids and Bases © 2015 Pearson Education, Inc. 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.) Sample Exercise Using K a to Calculate pH

Acids and Bases © 2015 Pearson Education, Inc. 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 We proceed as in the example just worked in the text, writing the chemical equation and constructing a table of initial and equilibrium concentrations in which the equilibrium concentration of H + is our unknown. Sample Exercise Using K a to Calculate pH

Acids and Bases © 2015 Pearson Education, Inc. 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: Sample Exercise Using K a to Calculate pH

Acids and Bases © 2015 Pearson Education, Inc. Sample Exercise Using K a to Calculate pH Substituting the equilibrium concentrations into the equilibrium-constant expression yields

Acids and Bases © 2015 Pearson Education, Inc. Sample Exercise Using K a to Calculate pH 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 ≃ Thus,

Acids and Bases © 2015 Pearson Education, Inc. Solving for x, we have Sample Exercise Using K a to Calculate pH 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:

Acids and Bases © 2015 Pearson Education, Inc. Strong vs. Weak Acids— Another Comparison Strong Acid: [H + ] eq = [HA] init Weak Acid: [H + ] eq < [HA] init This creates a difference in conductivity and in rates of chemical reactions.

Acids and Bases © 2015 Pearson Education, Inc. Polyprotic Acids Polyprotic acids have more than one acidic proton. It is always easier to remove the first proton than any successive proton. If the factor in the K a values for the first and second dissociation has a difference of 3 or greater, the pH generally depends only on the first dissociation.

Acids and Bases © 2015 Pearson Education, Inc Weak Bases Ammonia, NH 3, is a weak base. Like weak acids, weak bases have an equilibrium constant called the base dissociation constant. Equilibrium calculations work the same as for acids, using the base dissociation constant instead.

Acids and Bases © 2015 Pearson Education, Inc. Base Dissociation Constants

Acids and Bases © 2015 Pearson Education, Inc. Example What is the pH of 0.15 M NH 3 ? 1) NH 3 + H 2 O ⇌ NH OH – 2) K b = [NH 4 + ][OH – ] / [NH 3 ] = 1.8 × 10 –5 3) NH 3 (M)NH 4 + (M)OH – (M) Initial Concentration (M) Change in Concentration (M) –x–x+x+x+x+x Equilibrium Concentration (M) 0.15 – xxx

Acids and Bases © 2015 Pearson Education, Inc. Example (completed) 4)1.8 × 10 – 5 = x 2 / (0.15 – x) If we assume that x << 0.15, 0.15 – x = Then: 1.8 × 10 –5 = x 2 / 0.15 and:x = 1.6 × 10 –3 Note: x is the molarity of OH –, so –log(x) will be the pOH (pOH = 2.80) and [14.00 – pOH] is pH (pH = 11.20).

Acids and Bases © 2015 Pearson Education, Inc. The solubility of CO 2 in water at 25 °C and 0.1 atm is 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 M solution of H 2 CO 3 ? Sample Exercise Calculating the pH of a Solution of a Polyprotic Acid

Acids and Bases © 2015 Pearson Education, Inc. Solution Analyze We are asked to determine the pH of a M solution of a polyprotic acid. CO 2 (aq) + H 2 O(l) H 2 CO 3 (aq) What is the pH of a M solution of H 2 CO 3 ? Sample Exercise 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 Consequently, the pH can be determined by considering only K a1, thereby treating the acid as if it were a monoprotic acid.

Acids and Bases © 2015 Pearson Education, Inc. Solve Proceeding as in Sample Exercises 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 Solving for x, we have Sample Exercise x = 4.0 × 10 –5 M – x ≃

Acids and Bases © 2015 Pearson Education, Inc. Because we get the same value (to 2 significant figures) our simplifying assumption was justified. The pH is therefore Sample Exercise Calculating the pH of a Solution of a Polyprotic Acid

Acids and Bases © 2015 Pearson Education, Inc. 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 Sample Exercise 16.14

Acids and Bases © 2015 Pearson Education, Inc. Types of Weak Bases Two main categories 1) Neutral substances with an Atom that has a nonbonding pair of electrons that can accept H + (like ammonia and the amines) 2) Anions of weak acids

Acids and Bases © 2015 Pearson Education, Inc Relationship between K a and K b For a conjugate acid–base pair, K a and K b are related in this way: K a × K b = K w Therefore, if you know one of them, you can calculate the other. © 2015 Pearson Education, Inc.

Acids and Bases © 2015 Pearson Education, Inc. Acid–Base Properties of Salts Many ions react with water to create H + or OH –. The reaction with water is often called hydrolysis. To determine whether a salt is an acid or a base, you need to look at the cation and anion separately. The cation can be acidic or neutral. The anion can be acidic, basic, or neutral.

Acids and Bases © 2015 Pearson Education, Inc. Anions Anions of strong acids are neutral. For example, Cl – will not react with water, so OH – can’t be formed. Anions of weak acids are conjugate bases, so they create OH – in water; e.g., C 2 H 3 O 2 – + H 2 O ⇌ HC 2 H 3 O 2 + OH – Protonated anions from polyprotic acids can be acids or bases: If K a > K b, the anion will be acidic; if K b > K a, the anion will be basic.

Acids and Bases © 2015 Pearson Education, Inc. 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!)

Acids and Bases © 2015 Pearson Education, Inc. Hydrated Cations 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.

Acids and Bases © 2015 Pearson Education, Inc. 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.

Acids and Bases © 2015 Pearson Education, Inc. Factors that Affect Acid Strength 1)H—A bond must be polarized with δ + on the H atom and δ – on the A atom 2)Bond strength: Weaker bonds can be broken more easily, making the acid stronger. 3)Stability of A – : More stable anion means stronger acid.

Acids and Bases © 2015 Pearson Education, Inc. Binary Acids Binary acids consist of H and one other element. Within a group, H—A bond strength is generally the most important factor. Within a period, bond polarity is the most important factor to determine acid strength.

Acids and Bases © 2015 Pearson Education, Inc. Oxyacids Oxyacids consist of H, O, and one other element, which is a nonmetal. Generally, as the electronegativity of the nonmetal increases, the acidity increases for acids with the same structure.

Acids and Bases © 2015 Pearson Education, Inc. Oxyacids with Same “Other” Element If an element can form more than one oxyacid, the oxyacid with more O atoms is more acidic; e.g., sulfuric acid versus sulfurous acid. Another way of saying it: If the oxidation number increases, the acidity increases.

Acids and Bases © 2015 Pearson Education, Inc. Carboxylic Acids Carboxylic acids are organic acids containing the —COOH group. Factors contributing to their acidic behavior:  Other O attached to C draws electron density from O—H bond, increasing polarity.  Its conjugate base (carboxylate anion) has resonance forms to stabilize the anion.

Acids and Bases © 2015 Pearson Education, Inc. Lewis Acid/Base Chemistry Lewis acids are electron pair acceptors. Lewis bases are electron pair donors. All Brønsted–Lowry acids and bases are also called Lewis acids and bases. There are compounds which do not meet the Brønsted–Lowry definition which meet the Lewis definition.

Acids and Bases © 2015 Pearson Education, Inc. Comparing Ammonia’s Reaction with H + and BF 3