Chapter 15 Acids and Bases - Tamalpais Union High School ...

Chapter 15 Acids and Bases - Tamalpais Union High School ...

Lecture Presentation Chapter 16 AcidBase Equilibria 2015 Pearson Education, Inc. James F. Kirby Quinnipiac University Hamden, CT 16.1 & 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. BrnstedLowry An acid is a proton donor. A base is a proton acceptor. Acids and Bases 2015 Pearson Education, Inc. BrnstedLowry Acid and Base A BrnstedLowry acid must have at least one removable (acidic) proton (H+) to donate. A BrnstedLowry 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 BrnstedLowry 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. Sample Exercise 16.1 Identifying Conjugate Acids and Bases

(a) What is the conjugate base of HClO4, H2S, PH4+, HCO3? (b) What is the conjugate acid of CN, SO42 , H2O, HCO Solution 3 ? 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 HClO4, we obtain ClO4, which is its conjugate base. The other conjugate bases are HS PH3 Acids 2

CO3 and Bases 2015 Pearson Education, Inc. Sample Exercise 16.1 Identifying Conjugate Acids and Bases (b) What is the conjugate acid of CN, SO42 , H2O, HCO3? Solution (b)If we add a proton to CN, we get HCN, its conjugate acid. The other conjugate acids are HSO4 H3O+ H2CO3, Notice that the hydrogen carbonate ion

(HCO3) is amphiprotic. It can act as either an acid or a base. Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO3) is amphiprotic. Write an equation for the reaction of HSO3 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 acidbase pairs.

Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO3) 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 acidbase pairs. Solution Analyze and Plan We are asked to write two equations representing reactions between HSO3 and water, one in which HSO3 should donate a proton to water, thereby acting as a BrnstedLowry acid, and one in which HSO3 should

accept a proton from water, thereby acting as a base. We are also asked to identify the conjugate pairs in each equation. Solve (a) HSO3 (aq) + H2O(l) SO32 (aq) + H3O+(aq) The conjugate pairs in this equation are HSO3 (acid) and SO32 (conjugate base), and H2O (base) and H3O+ (conjugate Acids acid). and Bases 2015 Pearson Education, Inc. Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions (b) in which the ion acts as a base. In both cases identify the conjugate acidbase pairs.

Solution (b) HSO3 (aq) + H2O(l) H2SO3(aq) + OH (aq) The conjugate pairs in this equation are H2O (acid) and OH (conjugate base), and HSO3 (base) and H2SO3 (conjugate acid). Acids and Bases 2015 Pearson Education, Inc. Relative Strengths of Acids

Acids above the and Bases line with H2O as a base are strong acids; their conjugate bases do not act as acids in water. Bases below the line with H2O as an acid are strong bases; The substances between the their conjugate lines with H2O are conjugate acids do not act as acidbase pairs in water. Acids acids in water. and

Bases 2015 Pearson Education, Inc. Acid and Base Strength In every acidbase 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) + H2O(l) H3O+(aq) + Cl(aq) H2O is a much stronger base than Cl, so the equilibrium lies far to the right (K >> 1). CH3COOH(aq) + H2O(l) H H3O+(aq) + CH3COO(aq) Acetate is a stronger base than H2O, so the equilibrium favors the left side (K < 1). Acids and Bases

2015 Pearson Education, Inc. Sample Exercise 16.3 Predicting the Position of a Proton-Transfer Equilibrium For the following proton-transfer reaction use Figure 16.4 to predict whether the equilibrium lies to the left (Kc < 1) or to the right (Kc > 1): HSO4 (aq) + CO32 (aq) SO42 (aq) + HCO3 (aq) Acids and Bases 2015 Pearson Education, Inc.

HSO4 (aq) + CO32 (aq) Solution SO42 (aq) + HCO3 (aq) 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 CO32 , the base in the forward reaction, and SO42 , the conjugate base of HSO4. We can find the relative positions of these two bases in Figure 16.4 to determine which is the stronger base. Acids and Bases

2015 Pearson Education, Inc. Solve The CO32 ion appears lower in the right-hand column in Figure 16.4 and is therefore a stronger base than SO42 . Therefore, CO32 will get the proton preferentially to become HCO3, while SO42 will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is, Kc > 1): Comment Of the two acids HSO4 and HCO3, the stronger one (HSO4) 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 Acids proton moves from the stronger acid and becomes bonded to the and stronger base. Bases 2015 Pearson Education, Inc.

16.3 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 Kc = [H3O+][OH] (The term H2O 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 Acids for water, Kw. and Bases 2015 Pearson Education, Inc. Ion Product Constant This special equilibrium constant is referred to as the ion product constant for water, Kw. At 25 C, Kw = 1.0 1014 Kw = [1.0 107][1.0 107]

Kw = [H3O+][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 2015 Pearson Education, Inc. Acids and Bases Sample Exercise 16.4 Calculating [H+] for Pure Water Calculate the values of [H+] and [OH] in a neutral aqueous solution at 25 C. 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 1014 x2 = 1.0 1014 x = 1.0 107M = [H+] = [OH] In an acid solution [H+] is greater than 1.0 107 M; in a basic solution [H+] Acids is less than 1.0 107 M. and Bases 2015 Pearson Education, Inc. Sample Exercise 16.5 Calculating [H+] from [OH]

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 109M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 C. Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.5 Calculating [H+] from [OH] Calculate the concentration of H+(aq) in (a) a solution in which [OH] is 0.010 M

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 Kw to solve for each unknown concentration. Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.5 Calculating [H+] from [OH] Calculate the concentration of H+(aq) in (a) a solution in which [OH] is 0.010 M

Solve (a) Using Equation 16.16, we have This solution is basic because Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.5 Calculating [H+] from [OH] Calculate the concentration of H+(aq) in (b) a solution in which [OH] is 1.8 109M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 C.

(b) In this instance This solution is acidic because Acids and Bases 2015 Pearson Education, Inc. 16.4 The pH Scale pH is a method of reporting hydrogen ion concentration. pH = log[H+]

Neutral pH is 7.00. Acidic pH is below 7.00. Basic pH is above 7.00. Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.6 Calculating pH from [H+] Calculate the pH values for the two solutions of Sample Exercise 16.5. 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 16.17.

pH = log[H+] (a) In the first instance we found [H+] to be 1.0 1012 M 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 1012 M, so that pH = log(1.0 1012) = (12.00) = 12.00 Because 1.0 1012 has two significant

figures, the pH has two decimal places, 12.00. Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.6 Calculating pH from [H+] Calculate the pH values for the two solutions of Sample Exercise 16.5. (b) For the second solution, [H+] = 5.6 106 M. Before performing the calculation, it is helpful to estimate the pH. To do so, we note that [H+] lies between 1 106 and 1 105. Thus, we expect the pH to lie between 6.0 and 5.0. We use Equation 16.17 to calculate the pH: pH = log[H+]

Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.6 Calculating pH from [H+] Calculate the pH values for the two solutions of Sample Exercise 16.5. (b) For the second solution, [H+] = 5.6 106 M. pH = log(5.6 106) = 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. Acids

and Bases 2015 Pearson Education, Inc. 16. 4 (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] pKw: log Kw Acids and Bases 2015 Pearson Education, Inc. Relating pH and pOH

Because [H3O+][OH] = Kw = 1.0 1014 we can take the log of the equation log[H3O+] + log[OH] = log Kw = 14.00 which results in pH + pOH = pKw = 14.00 Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.7 Calculating [H+] from pOH A sample of freshly pressed apple juice has a pOH of 10.24. Calculate [H+].

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+. Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.7 Calculating [H+] from pOH A sample of freshly pressed apple juice has a pOH of 10.24. Calculate [H+].

pH = 14.00 pOH pH = 14.00 10.24 = 3.76 Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.7 Calculating [H+] from pOH A sample of freshly pressed apple juice has a pOH of 10.24. Calculate [H +]. 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 10x or INV log (these functions are usually above the log key). We use this function to perform the calculation: [H+] = antilog (3.76) = 103.76 = 1.7 104 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 1.0 103 M and 1.0 104 M. Our calculated [H+] falls within this estimated range. 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. 16.5 Strong Acids You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution; e.g., HA + H2O H3O+ + A

So, for the monoprotic strong acids, [H3O+] = [acid] Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.8 Calculating the pH of a Strong Acid What is the pH of a 0.040 M solution of HClO4? Solution Analyze and Plan Because HClO4 is a strong acid, it is completely ionized, giving [H+] = [ClO4] = 0.040 M. Solve pH = log(0.040) = 1.40 Check Because [H+] lies between 1 102 and 1 101, 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. 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 (Ca2+, Sr2+, and Ba2+). Again, these substances dissociate completely in aqueous solution; e.g., MOH(aq) M+(aq) + OH(aq) or

M(OH)2(aq) M2+(aq) + 2 OH(aq) Acids and Bases 2015 Pearson Education, Inc. Strong Bases Which solution has a higher pH, a 0.001 M solution of NaOH or a 0.001 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 0.002 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. Sample Exercise 16.9 Calculating the pH of a Strong Base What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? 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. Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.9 Calculating the pH of a Strong Base What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? 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 0.028 M. Method 1: Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.9 Calculating the pH of a Strong Base What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? Solution Method 2: pOH = log(0.028) = 1.55 pH = 14.00 pOH = 12.45

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

and Bases 2015 Pearson Education, Inc. Sample Exercise 16.9 Calculating the pH of a Strong Base (b) Method 1: Method 2: pOH = log(0.0022) = 2.66 pH = 14.00 pOH = 11.34 Acids and Bases 2015 Pearson Education, Inc.

16. 6 Weak Acids For a weak acid, the equation for its dissociation is HA(aq) + H2O(l) H H3O+(aq) + A(aq) Since it is an equilibrium, there is an equilibrium constant related to it, called the acid-dissociation constant, Ka: Ka = [H3O+][A] [HA] Acids and Bases 2015 Pearson Education, Inc.

16. 6 Weak Acids The greater the value of Ka, the stronger is the acid. Based on the table which element is most commonly bonded to the acidic hydrogen? 2015 Pearson Education, Inc. Acids and Bases 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 Ka from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25 C is 2.38. Calculate Ka for formic acid at this temperature. We know that [H3O+][HCOO] Ka = [HCOOH] To calculate Ka, we need the equilibrium

concentrations of all three things. We can find [H3O+], which is the same as [HCOO], from the pH. [H3O+] = [HCOO] = 102.38 = 4.2 103 Acids and Bases 2015 Pearson Education, Inc. Calculating Ka 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 (4.2 103 M); so we calculate the change, then the equilibrium HCOOH concentration. [HCOOH], M

[H3O+], M [HCOO], M Initially 0.10 0 0 Change 4.2 103 +4.2 103

+4.2 103 0.10 4.2 103 = 0.0958 = 0.10 4.2 103 4.2 103 At equilibrium Acids and Bases 2015 Pearson Education, Inc.

Calculating Ka from pH This allows us to calculate Ka by putting in the equilibrium concentrations. [4.2 103][4.2 103] Ka = [0.10] = 1.8 104 Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.10 Calculating Ka from Measured pH A student prepared a 0.10 M solution of formic acid (HCOOH) and found its pH at 25 C to be 2.38. Calculate Ka for formic acid at this temperature. 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 Ka 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. Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.10 Calculating Ka from Measured pH A student prepared a 0.10 M solution of formic acid (HCOOH) and found its pH at 25 C to be 2.38.

Calculate Ka for formic acid at this temperature. 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 From the measured pH, we can calculate [H+] pH = log [H +] = 2.38 log[H+] = 2.38 [H+] = 102.38 = 4.2 103 M 2015 Pearson Education, Inc. Acids

and Bases Sample Exercise 16.10 Calculating Ka 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+] = 4.2 103 M at equilibrium, we can construct the following table: Acids and Bases 2015 Pearson Education, Inc.

Sample Exercise 16.10 Calculating Ka from Measured pH Notice that we have neglected the very small concentration of H +(aq) due to H2O 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 4.2 103) M 0.10 0.10 M We can now insert the equilibrium concentrations into the expression for Ka: Check The magnitude of our answer is reasonable because Ka for a weak Acids 2 10 acid is usually between 10 and 10 . and Bases

2015 Pearson Education, Inc. Calculating Percent Ionization [H3O+]eq Percent ionization = 100 [HA]initial In this example, [H3O+]eq = 4.2 103 M [HCOOH]initial = 0.10 M 4.2 103 Percent ionization = 100 0.10 = 4.2% 2015 Pearson Education, Inc.

Acids and Bases Sample Exercise 16.11 Calculating Percent Ionization As calculated in Sample Exercise 16.10, a 0.10 M solution of formic acid (HCOOH) contains 4.2 103 M H+(aq). Calculate the percentage of the acid that is ionized. 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. Acids

and Bases 2015 Pearson Education, Inc. Sample Exercise 16.11 Calculating Percent Ionization As calculated in Sample Exercise 16.10, a 0.10 M solution of formic acid (HCOOH) contains 4.2 103 M H+(aq). Calculate the percentage of the acid that is ionized. Solve Acids and Bases 2015 Pearson Education, Inc. Method to Follow to Calculate pH Using Ka 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, HC2H3O2, at 25 C. 1) HC2H3O2 + H2O H H3O+ + C2H3O2

2) Ka = [H3O+][C2H3O2] / [HC2H3O2] 3) CH3COOH (M) H3O+ (M) CH3COO (M) Initial Concentration (M) 0.30 0 0

Change in Concentration (M) x +x +x Equilibrium Concentration (M) 0.30 x x

x 2015 Pearson Education, Inc. Acids and Bases Example (concluded) 4) Ka = [H3O+][C2H3O2] / [HC2H3O2] = (x)(x) / (0.30 x) If we assume that x << 0.30, then 0.30 x becomes 0.30. The problem becomes easier, since we dont have to use the quadratic formula to solve it. Ka = 1.8 105 = x2 / 0.30, so x = 2.3 103 x = [H3O+], so pH = log(2.3 103) = 2.64

Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.12 Using Ka to Calculate pH Calculate the pH of a 0.20 M solution of HCN. (Refer to Table 16.2 or Appendix D for the value of Ka.) Acids and

Bases 2015 Pearson Education, Inc. Sample Exercise 16.12 Using Ka to Calculate pH Solution Analyze We are given the molarity of a weak acid and are asked for the pH. From Table 16.2, Ka for HCN is 4.9 1010. 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. Acids and

Bases 2015 Pearson Education, Inc. Sample Exercise 16.12 Using Ka to Calculate pH Solve Writing both the chemical equation for the ionization reaction that forms H+(aq) and the equilibrium-constant (Ka) expression for the reaction: Next, we tabulate the concentrations of the species involved in the equilibrium reaction, letting x = [H+] at equilibrium: Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.12 Using Ka to Calculate pH

Substituting the equilibrium concentrations into the equilibrium-constant expression yields Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.12 Using Ka 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 0.10 0.20. Thus, Acids and

Bases 2015 Pearson Education, Inc. Sample Exercise 16.12 Using Ka to Calculate pH Solving for x, we have A concentration of 9.9 106 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 Ka 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. 16.7 Weak Bases Ammonia, NH3, 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 NH3? 1) NH3 + H2O H NH4+ + OH 2) Kb = [NH4+][OH] / [NH3] = 1.8 105 3) NH3 (M)

NH4+ (M) OH (M) Initial Concentration (M) 0.15 0 0 Change in Concentration (M)

x +x +x Equilibrium Concentration (M) 0.15 x x x Acids and

Bases 2015 Pearson Education, Inc. Example (completed) 4) 1.8 10 5 = x2 / (0.15 x) If we assume that x << 0.15, 0.15 x = 0.15. Then: 1.8 105 = x2 / 0.15 and: x = 1.6 103 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. Sample Exercise 16.14 Calculating the pH of a Solution

of a Polyprotic Acid The solubility of CO2 in water at 25 C and 0.1 atm is 0.0037 M. The common practice is to assume that all the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced in the reaction CO2(aq) + H2O(l) H2CO3(aq) What is the pH of a 0.0037 M solution of H2CO3? Acids and Bases 2015 Pearson Education, Inc.

Sample Exercise 16.14 CO2(aq) + H2O(l) H2CO3(aq) What is the pH of a 0.0037 M solution of H2CO3? Solution Analyze We are asked to determine the pH of a 0.0037 M solution of a polyprotic acid. Plan H2CO3 is a diprotic acid; the two acid-dissociation constants, Ka1 and Ka2 (Table 16.3), differ by more than a factor of 103. Consequently, the pH can be determined by

considering only Ka1, thereby treating the acid as if it were a Acids and monoprotic acid. Bases 2015 Pearson Education, Inc. Sample Exercise 16.14 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

x = 4.0 105 M Alternatively, because Ka1 is small, we can make the simplifying approximation that x is small, so that 0.037 x 0.0037 Solving for x, we have Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.14 Calculating the pH of a Solution of a Polyprotic Acid

Because we get the same value (to 2 significant figures) our simplifying assumption was justified. The pH is therefore Acids and Bases 2015 Pearson Education, Inc. Sample Exercise 16.14 Comment If we were asked for [CO32] we would need to use Ka2. Lets illustrate that calculation. Using our calculated values of [HCO 3] and [H+] and setting [CO32] = y, we have Assuming that y is small relative to 4.0 105, we have

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. 16.8 Relationship between Ka and Kb For a conjugate acidbase pair, Ka and Kb are related in this way: Ka Kb = Kw Therefore, if you know one of them, you can calculate the other. 2015 Pearson Education, Inc. 2015 Pearson Education, Inc. Acids and Bases

AcidBase 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 Acids neutral. and Bases 2015 Pearson Education, Inc. Anions Anions of strong acids are neutral. For

example, Cl will not react with water, so OH cant be formed. Anions of weak acids are conjugate bases, so they create OH in water; e.g., C2H3O2 + H2O H HC2H3O2 + OH Protonated anions from polyprotic acids can be acids or bases: If Ka > Kb, the anion will be acidic; if Kb > Ka, the anion Acids will be basic. and Bases 2015 Pearson Education, Inc. Cations

Group I or Group II (Ca2+, Sr2+, or Ba2+) metal cations are neutral. Polyatomic cations are typically the conjugate acids of a weak base; e.g., NH4+. 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 Ka and Kb; whichever is greater Acids and dictates what the salt is. Bases 2015 Pearson Education, Inc. Factors that Affect Acid Strength 1) HA bond must be polarized with + on the H atom and 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, HA bond strength is generally the most important factor. Within a period, bond polarity is the most important factor to

determine acid strength. 2015 Pearson Education, Inc. Acids and Bases 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. 2015 Pearson Education, Inc. Acids and Bases 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 OH 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 BrnstedLowry acids and bases are also called Lewis acids and bases. There are compounds which do not meet the BrnstedLowry definition which meet the Lewis definition. Acids and Bases 2015 Pearson Education, Inc. Comparing Ammonias Reaction with H+ and BF3

Acids and Bases 2015 Pearson Education, Inc.

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