CHAPTER 17: ACIDS AND BASES

INTRODUCTION CHAPTER 17: ACIDS AND BASES In this chapter you will learn about the properties of acids and bases. You know about some of the properties of acids already. Substances such as lemon juice and vinegar contain acids, and their sour taste is from the acid each contains. You will learn how to determine the properties of acids and how to measure the acidity of a solution. GOALS FOR THIS CHAPTER 1. Know the Arrhenius and Bronsted-Lowry models of acids and bases, and be able to identify a conjugate acid-base pair. (Section 17.1) 2. Be able to state the properties of weak and strong acids. (Section 17.2) 3. Understand how water can be both an acid and a base, and how the ion-product constant is derived. (Section 17.3) 4. Understand the ph scale and be able to use a calculator to calculate the ph or poh of a solution, given [H+] or [OH-]. (Section 17.4) 5. Be able to use a calculator to calculate [H+] or [OH-], given the ph or por. (Section 17.4) 6. Be able to calculate the ph ofa strong acid solution ofknown molarity. (Section 17.5) 7. Know the characteristics of a buffer, and how a buffer works to resist changes in ph. (Section 17.6) QUICK DEFINITIONS Alkalis Alkalis have a bitter taste and a slippery feel. called bases. (Section 17.1) They are also Arrhenius concept of acid and base Bronsted-Lowry model Conjugate acid Conjugate base An acid produces hydrogen ions and a base produces hydroxide ions when in solution. (Section 17.1) An acid is a proton (H+) donor, and a base is a proton acceptor. (Section 17.1) When a strong acid is dissolved in water, the species which bonds to the proton is called the conjugate acid. (Section 17.1) When a strong acid is dissolved in water, the part of the acid molecule which is left after the proton is gone is called the conjugate base. (Section 17.1) 355

Conjugate acid-base pair Hydronium ion Strong acid Weak acid Diprotic acid Oxyacids Organic acids Carboxyl group Two substances related to each other by donating and accepting a proton, H+. (Section 17.1) The ion H 30+, that is a water molecule that has accepted a proton, H+. (Section 17.1) An acid that completely dissociates in water to become ions. The equilibrium HA(aq) + HzO(l)!:::; H 30+(aq) + A'(aq) lies far to the right for a strong acid. (Section 17.2) An acid that does not completely dissociate in water to form ions. The equilibrium HA(aq) + HzO(l)!::; H 30+(aq) + A'(aq) lies far to the left for a weak acid. (Section 17.2) An acid which can donate two protons for each acid molecule. (Section 17.2) Acids which have the acidic proton bonded to an oxygen atom. (Section 17.2) Acids which contain a carbon skeleton, and often a carboxyl group. (Section 17.2) A carboxyl group is the acidic part of molecules with carbon chains. It consists of a carbon atom doubly bonded to an oxygen and also bonded to an OH. Amphoteric substance Ionization of water Ion-product constant -C-OH " A substance which can behave as an acid or a base. (Section 17.3) Dissociation of a water molecule, with the transfer of the proton to a second water molecule to produce H 30+ + OB-. (Section 17.3) A constant K, whose value is 1.0 x 10-14, and is equal to the molar concentration of hydrogen ion times the molar concentration of hydroxide ion, [H+][OB-]. (Section 17.3) 356 17 Acids and Bases

Neutral solution Acidic solution Basic solution ph scale Buffered solution A solution where the hydrogen ion concentration equals the hydroxide ion concentration, [H+] = [OH-]. The ph of a neutral solution is 7.0. (Section 17.3) A solution where the hydrogen ion concentration is greater than the hydroxide ion concentration, [H+] > [OH]. The ph is less than 7.0. (Section 17.3) A solution where the hydrogen ion concentration is less than the hydroxide ion concentration, [H+] < [OH-]. The ph is greater than 7.0. (Section 17.3) A convenient way to express small hydrogen ion concentrations. ph =-log[h+]. (Section 17.4) A solution which resists ph change even when a strong base or strong acid is added. A buffered solution contains a weak acid, plus the conjugate base of the weak acid. (Section 17.6) PRETEST 1. When nitrous acid, HN0 2, reacts with water, which two substances are the acid-conjugate base pair? 2. When the weak acid HI reacts with water, toward which side of the reaction does the equilibrium lie? 3. At 25 C what is the [OH-] of a solution for which [H+] =4.5 X 10-3 M? 4. What is the [OH-] of a 0.25 M solution of HCI in water at 25 C? 5. What is the ph of a solution in which [H+] = 1.1 X 10-4 M? 6. What is the ph of a solution in which [OH-] =8.23 X 10-6 M? 7. What is the poh of a solution in which [OH-] =5.2 x 10-2 M? Pretest 357

8. A solution of NaOH in water has a ph of 10.3. What is the [H+]? 9. What is the ph of a 2.1 x 10-3 M solution of HCl? 10. A buffer was made by adding together solutions of N~P04 and H 3P04. Show what reaction occurs when H+ is added to the buffer solution. PRETEST ANSWERS 1. When HNO z reacts with water, HNO z is the acid and NO z-is the conjugate base. (17.1) 2. When the weak acid HI reacts with water, the equilibrium lies toward the left side of the reaction. Most of the HI molecules remain undissociated. (17.2) 3. If we are given [H+] we can find [OH-] if we remember that [H+] [OH-] = 1.0 x 10-14, which is K w Rearrange to isolate [OH-] on one side. Now substitute into the equation [OH-] = 1.0 x 10-14 4.5 X 10-3 (17.4) 4. When HCl dissolyes in water, each mole of HCl forms a mole ofw ions and a mole ofcr ions. A solution which is 0.25 M HCl is also 0.25 M H+. So [H+] =0.25. We can use the K, expression to find [OH-]. 358 17 Acids and Bases

Rearrange the equation to isolate [OH-] on one side. Now substitute values into the equation. (17.4) 5. To find the ph of a solution in which [H+] equals 1.1 x 10-4 M, we can use the formula ph ::;: -log [H+]. First, find the log of 1.1 x 10-4 Enter 1.1 X 10-4 on your calculator and press the log button. log (1.1 x 10-4 ) ::;: -3.96 Now press the ±key to multiply -3.96 times -1. The result is 3.96. So the ph of a solution which contains 1.1 x 10-4 M H+ is 3.96. (17.4) 6. We need to know [H+] to find the ph. We can use the K; expression, ~::;: [H+] [OH-], to find [H+] if we know [OH-]. [H+]::;: 1.0 x 10-14 8.23 X 10-6 Now that we know [H+], we can use the ph formula, ph::;:-log [H+] to find the ph. ph::;:-log [H+] log 1.2 x 10-9 ::;: -8.92 Press the ± key to multiply -8.92 times -1. The ph is 8.92. (17.4) Pretest Answers 359

7. The formula for finding poh is poh =-log [OH-]. First enter the [OH-] on your calculator and press the log key. log 5.2 X 10-2 =-1.28 Now press the ± key to multiply -1.28 times -1. The poh is 1.28. (17.4) 8. We want to try to find the [H+] of a NaOH solution from its ph by working backwards. Find the minus log by undoing the log {W]. We can do this by finding the inverse log of -ph. [H+] =inverse log (-ph) First, enter the ph in your calculator. Then press ± key to multiply the ph by -1. Then use the keys on your calculator to generate the inverse log. [H+] = inverse log (-ph) [H+] = inverse log (-10.3) [H+] =5.0 X 10-11 M (17.4) 9. Because HCI is a strong acid, a solution that is 2.1 x 10-3 Min HCI is also 2.1 x 10-3 M in H+. ph =-log[h+] ph =-log (2.1 x 10-3 ) ph =2.68 (17.5) 10. A buffer is often made from a weak acid and the salt of a weak acid. In this buffer, the weak acid is ~P04 and the salt of the weak acid is Na 3P04. When H+ is added to a solution containing this buffer, the ph doesn't change much because the buffer reacts to remove H+ from solution. The pot ion has a high affinity for H+ and reacts with H+ to form H 3PO 4' thereby removing it from the solution. (17.6) 360 17 Acids and Bases

CHAPTER REVIEW 17.1 ACIDS AND BASES What Are Acids and Bases? Arrhenius proposed that an acid was anything that produced hydrogen ions in aqueous solution, and a base was anything that produced hydroxide ions. Scientists discovered that this definition of acids and bases was too restrictive, that there were other bases besides hydroxide ions. Bronsted and Lowry proposed that an acid was a proton (hydrogen ion) donor, and a base was a proton acceptor. This was a much broader definition of acids and bases. WhatHappens When an AcidIs Dissolvedin Water? According to the Bronsted-Lowry model of acids and bases, when an acid dissolves in water, a reaction occurs between the acid (the proton donor) and water (the proton acceptor). Two products are formed as a result of the reaction, a water molecule with an extra proton called a hydonium ion, and the remains ofthe acid after the proton has left. HA(aq) + HzO(l)... H30+ (aq) + A- (aq) t hydronium ion The protonated water is called a conjugate add, and the remaining part of the acid (A-) is called the conjugate base. HA and A- form a pair called the acid-conjugate base pair. HzO and H 30+ form a pair called the base-conjugate acid pair. When an acid or a base react, it is useful to identify the conjugate base or the conjugate acid. What is the conjugate base of HzS? HzS can donate a proton to water by the following reaction. Hz S + HzO...H30+ + HSt conjugate acid "conjugate base The conjugate base of HzS is HS-. 17.2 ACID STRENGTH Why Are Some Acids Weak and Some Strong? The reaction of an acid with water is a reversible reaction. water molecule, or to the conjugate base. The proton can be attached to the Chapter Review 361

The relative attractions of ~O and A- for the proton determines whether HA or H 30+ predominates. If A- attracts the proton more strongly than does H 20, then the equlibrium lies to the left, and there is relatively little H 30+. Acids in which the equlibrium lies to the left are called weak acids because they exist mostly in the HA form. If H 2 0 attracts the proton more strongly than does A-, then the equilibrium lies to the right. These are called strong acids because virtually all ofthe protons have dissociated from the conjugate base and are attached to a water molecule. 17.3 WATER AS AN ACID AND A BASE How Can Water Be Both an Acid and a Base? Water is a substance which can act both as an acid and a base. Such substances are said to be amphoteric. The amphoteric nature ofwater can be seen when two water molecules react. One water molecule donates a proton; it is acting as an acid. The other water molecule accepts the proton; it is acting as a base. H 20 t base t acid t OH-"- + conjugate base conjugate acid We can write an equilibrium expression for the ionization of water by showing the reaction of just one water molecule. The equilibrium constant for this reaction is K; and is called the ion-product constant, or dissociation constant. Liquid water does not appear in the equilibrium expression because the concentration of water changes very little. It has been shown experimentally that K; has the value 1.0 x 10-14 mol 2 jl2. For any solution, the product of [H+] and [OH-] is always 1.0 x 10-14 For an aqueous solution of a strong acid where the [H+] is very high, the [OH-] must be very low, so that when the two concentrations are multiplied, the result is 1.0 x 10-14 If we know the concentration of either hydrogen ion or hydroxide ion, we can calculate the other, using the equilibrium expression for the ionization of water. 362 17 Acids and Bases

Example: If the hydroxide ion concentration of a solution is 2.0 x 10-2 M, what is the hydrogen ion concentration? First, write the equilibrium expression for the ionization ofwater. Then isolate the unknown value, [H+], on one side of the equation by dividing both sides by [OH-]. Then, substitute the values into the equation. The concentration of hydrogen ion is 5.0 x 10-13 M. multiplying (2.0 x 10-2 ) (5.0 x 10-13 ). You can check your calculations by (2.0 x 10-2)(5.0 x 10-13 ) = 10 X 10-15 = 1.0 X 10-14 The result is 1.0 x 10-14 17.4 THE ph SCALE How Can We Use Logarithms to Calculate the ph ofa Solution? The hydrogen ion concentration of a solution is represented by very small numbers. Using scientific notation is a good way to represent small numbers, but calculating and using the ph of a solution is another easy way to represent small numbers. Taking the p of any number means we take the logarithm (log) of that number, and multiply the result by -1. pn = (-1) x log N When we find the ph of a solution, we take the log of the hydrogen ion concentration in mol/l and multiply by -1. Chapter Review 363

Example: What is the ph of a solution with a hydrogen ion concentration of 2.3 x 10-2 mol/l? First, find the log of 2.3 x 10-2 by entering the number on your calculator, and pressing the key for log. log 2.3 X 10-2 = -1.638 Then, press the ± key to multiply -1.638 by -1. Multiplying any number by -1 changes the sign, but not the value of the number. -1.638 x -1 =1.638 Next, express the number to the correct number of significant figures. When using logarithms, the part of the number to the right of the decimal point should have the same number of digits as the hydrogen ion concentration has significant figures. The number 2.3 X 10-2 has two significant figures, so the ph should have two numbers to the right of the decimal point. 1.638 would become 1.64. The ph of a solution with 2.3 x 10-2 mol/l hydrogen ion is 1.64. Can a Logarithmic Scale Be Used to Express Hydroxide Ion Concentration? A log scale can also be used to express the hydroxide ion concentration. poh is calculated the same way as ph. Start with the hydroxide ion concentration in mol/l, Find the log of that number, then multiply by -1. For example, if the [OH-] is 6.5 x 10-5 mol/l, the poh is poh =-log (6.5 x 10-5 ) =-4.19-4.19 x -1 =4.19 The poh is 4.19. Because 6.5 x 10-5 has two significant figures, there should be two digits to the right ofthe decimal pointin the poh value. What Is the Relationship Between ph andpoh? [H+] and [OH-] are related to each other through the equilibrium expression for the dissociation of water, and the ion product constant, K w ' If we take the p of each of the parts of this equation we have 364 17 Acids and Bases

or, ph + poh = 14.00 This means that for any solution, the sum of ph and poh will always equal 14. If we know the pli, we can easily find the poh, and vice versa. Example: If the poh of a solution is 8.23, what is the ph? Use the equation ph + poh = 14 to solve this problem. First, isolate the unknown quantity, pfl, on one side of the equation. Then, substitute with the values we know. The ph of the solution is 5.77. ph= 14-pOH ph = 14.00-8.23 = 5.77 17.5 CALCULATING THE ph OF STRONG ACID SOLUTIONS How Can We Calculate the ph ofa Strong Acid? We can calculate the ph of any solution if we know the hydrogen ion concentration. When strong acids dissolve in water, they dissociate completely into H+ and A-. If we know the molarity of the acid solution, we know the [H+]. 0.5 M HCI --.. 0.5 M H+ + 0.5 M cr We know that the [H+] is 0.5 mol/l because all of the HCI dissociates into H+ and cr. From the [H+] we can calculate the ph. ph = -log (0.5) = 0.30 The ph is 0.30. This is a strongly acidic solution. 17.6 BUFFERED SOLUTIONS What Is a Buffered Solution and How Does It Work? A buffered solution resists large changes in ph even when strong acid or strong base is added. The addition of even small amounts of strong acid or base can greatly lower or raise the ph of a nonbuffered solution. Living systems contain many differents kinds of buffers to help keep fluids and tissues at the correct pl.i, even under stressful conditions. Most buffers are made from a weak acid, plus the soluble salt of the weak acid, the conjugate base. Acetic acid and sodium acetate are often used for this purpose. When strong acid is added to a buffered solution containing acetic acid and sodium acetate, the following reaction occurs. Chapter Review 365

The excess hydrogen ion, which would ordinarily lower the ph of the solution, attaches to the acetate anion to form acetic acid. Remember that weak acids form strong conjugate bases, so the acetate anion has a high affinity for hydrogen ions. The hydrogen ions are removed from the solution, and the ph does not decrease as it would if there were no buffer present. When strong base is added to the solution, the H+reacts with the acetate ion in the following way. The excess hydroxide ion from the strong base reacts with acetic acid to form a water molecule and the conjugate base, the acetate anion. The ph does not rise very much, because the OH- ions are no longer free in solution. Almost any weak acid and its conjugate base will act as a buffer. LEARNING REVIEW 1. Explain the differences between the Arrhenius concept of an acid and a base, and the concept of Bronsted and Lowry. 2. For the reaction of perbromic acid with water: a. Which two substances are an acid-conjugate base pair? b. Which two substances are a base-conjugate acid pair? 3. Write equations to show what happens when each of the acids below reacts with water, 4. Show how acetic acid, HC 2H302, reacts with water by drawing Lewis structures for water and its conjugate acid. 5. When formic acid, HCOOH, is mixed with water, the resulting solution weakly conducts an electric current. a. Is formic acid a strong or a weak acid? b. Toward which side of the reaction does the equilibrium lie? c. Which species is the stronger base, H 20 or HCOO-? 366 17 Acids and Bases

6. Which of the acids below are strong acids and which are weak acids? a. HF b. H 2SO4 c. HC 2H3 0 2 d. HClO 4 7. The expression for the dissociation of water is K, =[H+][OH-]. Why does liquid water not appear in this expression? 8. All the aqueous solutions below are at a temperature of 25 "C, a. What is the [H+] of a solution for which [OH-] = 1.5 X 10-6 M? b. What is the [H+] of a solution for which [OH-] =6.3 x 10-3 M? c. What is the [OH-] of a solution for which [H+] = 3.25 X 10-1 M? 9. What is the [H+] of a 0.1 M solution of NaOH in water at 25 CC? 10. What is the ph of each of the solutions below? a. A solution in which [H+] =3.0 x 10-3 M b. A solution in which [H+] =5.2 x 10-6 M c. A solution in which [OH-] = 1.4 x 10-1 M 11. What is the poh ofeach solution below? a. [OH-] =4.89 x 10-10 M b. [OH-] =3.2 x 10-5 M c. [H+] = 1.6 x 10-8 M 12. For any solution, what is the relationship between ph and poh? 13. The ph oflemon juice is 2.1. What is the poh? 14. The poh of black coffee is 9.0. What is the ph? 15. Calculate the [H+] or [OH-] for the solutions below. a. Milk has a ph of6.9. What is the [H+]? b. Oven cleaner has a ph of 13.4. What is the [OH-]? c. A phosphate-containing detergent has a poh of 4.7. What is the [OH-]? Learning Review 367

16. What is the ph of each of the following solutions? a. 0.02 MHCI b. 3.5 x 10-3 M HN0 3 17. HCI is added to a solution containing ~C03 and NaHC0 3. a. Use an equation to show what would happen to the hydrogen ions from the HCl. b. Why would the ph of this solution not change drastically when the HCI is added? ANSWERS TO LEARNING REVIEW 1. Arrhenius's model of acids and bases proposes that acids produce hydrogen ions in aqueous solution, while bases produce hydroxide ions. The Bronsted-Lowry model of acids and bases proposes that acids are proton donors, while bases are proton acceptors. 2. a. For the reaction of perbromic acid with water, perbromic acid/perbromate (HBrOiBr04-) is the acid-conjugate base pair. b. Water/hydronium ion (H20/H30+) is the base-conjugate acid pair. 3. One model of an acid postulates that acids donate protons in aqueous solutions. The proton acceptor is the base, water. In aqueous solutions, acids donate protons to the base water to form the hydroniumion and the conjugate base. a. H 2S(aq) + H 20(1) HS-(aq) + H 30+(aq) b. HNOiaq) + H 20(1) N0 2-(aq) + H 30+(aq) 4..r>; H-y: + HC,H,O, ---fl...c,h,o,- ~ [H-O-H]+ H 5. a. b. When formic acid is mixed with water, a weak electric current is generated. An electrical current requires ions. We know that only a few ions are in this solution because the current is weak. So formic acid is a weak acid. Because there are only a few formate and hydronium ions formed, the equilibrium lies toward the left, that is, toward undissociated formic acid. 368 17 Acids and Bases