CHAPTER 16: ACIDS AND BASES

CHAPTER 16: ACIDS AND BASES Active Learning: 4, 6, 14; End-of-Chapter Problems: 2-25, 27-58, 66-68, 70, 75-77, 83, 90-91, 93-104 Chapter 15 End-of-Chapter Problems: 69-74, 125, 129, 133 16.1 ACIDS AND BASES ACID: A substance that produces H + ions in water Some acids are monoprotic (release only H + per molecule) e.g. HCl, HBr, HI, HNO 3, HClO 4 Some acids are polyprotic (release more than one H + per molecule) e.g. H 2 SO 4 and H 2 CO 3 are both diprotic; H 3 PO 4 is triprotic. 2 properties of acids: taste sour, neutralize bases, react with CO 3 or HCO 3 to produce CO 2 (g), react with metals to produce H 2, turn blue litmus paper red We can show the ionization or dissociation of acids as follows: HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl (aq) HC 2 H 3 O 2 (aq) + H 2 O(l) H 3 O + (aq) + C 2 H 3 O 2 (aq) Note: H 3 O + is the hydronium ion, a hydrated proton: H + + H 2 O = H 3 O + Since H 3 O + H + + H 2 O, we will abbreviate it as H +, but recognize that it actually exists as H 3 O +. HCl(aq) HC 2 H 3 O 2 (aq) H + (aq) + Cl (aq) H + (aq) + C 2 H 3 O 2 (aq) BASE: A substance that produces OH ions in water e.g. NaOH, Ba(OH) 2 properties of bases: taste bitter, feel soapy or slippery, neutralize acids, turn red litmus paper blue We can show the ionization of a base as follows: NaOH(aq) Ba(OH) 2 (aq) Na + (aq) + OH (aq) Ba +2 (aq) + 2 OH (aq) ARRHENIUS DEFINITIONS OF ACIDS AND BASES Arrhenius acid: A substance that produces H + ions in water e.g. HCl, HNO 3, H 2 SO 4 Arrhenius base: A substance that produces OH ions in water e.g. NaOH, Ba(OH) 2 The general equation for an Arrhenius acid-base neutralization reaction is shown below: HX(aq) + MOH(aq) H 2 O(l) + MX(aq) acid base water salt CHEM139: Zumdahl Chapter 16 page 1 of 10

BRØNSTED-LOWRY DEFINITIONS OF ACIDS AND BASES Why is H + called a proton? Brønsted-Lowry acid: A substance that donates a proton (H + ) i.e., a proton donor Brønsted-Lowry base: A substance that accepts a proton (H + ) i.e., a proton acceptor Unlike an Arrhenius base, a B-L base need not contain OH. A Brønsted-Lowry acid-base reaction simply involves a proton (H + ) transfer, as shown in the example below: NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH (aq) Note in this reaction simply involves H 2 O donating a H + ion to NH 3 to produce NH + 4 and OH. In this reaction, H 2 O is the Brønsted-Lowry acid, and NH 3 is the Brønsted-Lowry base. The conjugate acid-base pairs differ only by a H +. + In this reaction, the conjugate acid-base pairs are NH 3 and NH 4 and H 2 O and OH. conjugate acid-base pairs: a Brønsted-Lowry acid/base and its conjugate differ by a H + For the reaction above, when HA donates H + to H 2 O, it leaves behind A, which can act as a base for the reverse reaction. An acid and base that differ only by the presence of H + are conjugate acid-base pairs. Thus, HA is the conjugate acid of A, and A is the conjugate base of HA. CHEM139: Zumdahl Chapter 16 page 2 of 10

The general reaction for the dissociation (or ionization) of an acid can be represented as, where the double-arrow indicates both the forward and reverse reactions occur: HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) Brønsted-Lowry acid = Brønsted-Lowry base= Note: The double arrow ( ) indicates the reaction is reversible (goes in both directions). Example: Determine the Brønsted-Lowry acid and base in each of the following reactions, and fill in the blanks to below indicate the conjugate acid/base pairs in the reaction: a. H 2 O(l) + H 2 SO 4 (aq) HSO 4 (aq) + H 3 O + (aq) Brønsted-Lowry acid = Brønsted-Lowry base= H 2 O(l) is the conjugate acid base of. circle one H 2 SO 4 (aq) is the conjugate acid base of. circle one b. H 2 O(l) + H 2 PO 4 (aq) H 3 PO 4 (aq) + OH (aq) Brønsted-Lowry acid = Brønsted-Lowry base= H 3 PO 4 (aq) is the conjugate of. OH (aq) is the conjugate of. Note that H 2 O is amphoteric since it can behave as an acid or a base. CHEM139: Zumdahl Chapter 16 page 3 of 10

16.2 ACID STRENGTH A strong acid ionizes completely in water to form hydronium ion. Almost all of the acid is broken up into ions in solution. A forward arrow (not a double arrow) is used for the dissociation of a strong acid like HNO 3 (aq) since virtually all the HNO 3 ionizes to H + and NO 3, leaving almost not HNO 3. HNO 3 (aq) + H 2 O(l) H 3 O + (aq) + NO 3 (aq) Similarly, a strong base dissociates (breaks up) completely. Almost all of the base is broken up into ions in solution. A forward arrow is used for the dissociation of a strong base like Ca(OH) 2 (aq) since virtually all the Ca(OH) 2 ionizes to Ca 2+ and OH, leaving almost not Ca(OH) 2. Ca(OH) 2 (aq) Ca 2+ (aq) + 2 OH (aq) In contrast, a weak acid ionizes only to a very small extent. Most of the acid remains intact as a molecule in solution. Thus, a double arrow is used for the dissociation of a weak acid like HF(aq) since mostly HF exists at equilibrium, with very few H + or F ions in solution. HF(aq) + H 2 O(l) H 3 O + (aq) + F (aq) Know the following acids and bases. All other acids and bases are weak! Strong Acids Strong Bases HCl, HBr, HI, HNO 3, HClO 4, H 2 SO 4 LiOH, NaOH, KOH, Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 Note: H 2 SO 4 (aq) is a strong acid and diprotic (able to release 2 H + ions), but it generally ionizes to release only one H +, as follows: H 2 SO 4 (aq) + H 2 O(l) H 3 O + (aq) + HSO 4 (aq) The resulting HSO 4 (aq) is a weak acid: HSO 4 (aq) + H 2 O(l) H 3 O + (aq) + SO 4 2 (aq) CHEM139: Zumdahl Chapter 16 page 4 of 10

The strength of an acid is inversely related to the strength of its conjugate base. The weaker an acid, the stronger its conjugate base. The conjugate bases of weak acids (F,NO 2, C 2 H 3 O 2, etc.) are much stronger bases compared to H 2 O. These bases can react with a H 2 O molecule by removing a H + ion from the water molecule to form its conjugate acid and OH. F (aq) + H 2 O(l) HF(aq) + OH (aq) The conjugate bases of strong acids (Cl, Br, I, NO 3, ClO 3, ClO 4 ) are much weaker bases compared to H 2 O. These bases are not strong enough to react with a H 2 O molecule. Cl (aq) + H 2 O(l) No Reaction Ex. 1: Circle all of the ions below that would react with water to produce hydroxide ion: Br C 2 H 3 O 2 NO 2 NO 3 SO 3 2 PO 4 3 16.3 WATER AS AN ACID AND A BASE Consider the autoionization of water, H 2 O + H 2 O H 3 O + + OH which has the equilibrium expression: K w = [H 3 O + ] [ OH ] = [H + ] [ OH ] where K w is called the ion-product constant or dissocation constant for water. Experiment shows that at 25 C, [H + ]=[ OH ]=1.0 10-7 M so K w = [H + ] [ OH ] = (1.0 10-7 ) (1.0 10-7 ) = 1.0 10-14. (2 sig figs) Acidic, Basic, and Neutral A solution is acidic when A solution is basic when it contains H + ions. it contains OH ions. [H + ] > [OH ] [OH ] > [H + ] A substance is neutral when [H + ]=[OH ], so water is neutral. If a substance does not produce H + or OH ions in water, it will have [H + ]=[OH ]. The substance is neutral e.g., saline solution, dilute NaCl(aq), is neutral. CHEM139: Zumdahl Chapter 16 page 5 of 10

Ex. 1: Calculate the hydrogen or hydroxide ion concentration for each of the following, and indicate if the solution is acidic, basic, or neutral. a. milk, [H + ]=3.0 10-7 M [OH ]= acidic basic neutral b. carrots, [H + ]=7.9 10-6 M [OH ]= acidic basic neutral c. eggs, [OH ]=6.3 10-7 M [H + ]= acidic basic neutral d. urine, [H + ]=1.8 10-5 M [OH ]= acidic basic neutral Reviewing Logarithms Logarithms provides a more convenient way to deal with very large and very small numbers. e.g., using log (or log 10 = log base-10) the number 2.50 10 6 can be expressed as log 2.50 10 6 = 6.39794 or 10 6.39794 =2.50 10 6 The components of a logarithm are called the character and the mantissa. In the example above, the character is 6, corresponding to the exponent of the number when it s expressed in scientific notation, and the mantissa is 39794, which When a number is expressed in scientific notation, The exponent in scientific notation = the character of the logarithm, and the digit term in scientific notation = the mantissa of the logarithm. Consider the following examples, log 2.50 10 3 = 3.39794 log 2.50 10-1 = 0.39794 1 log 2.50 10 2 = 2.39794 log 2.50 10-2 = 0.39794 2 log 2.50 10 1 = 1.39794 log 2.50 10-3 = 0.39794 3 log 2.50 10 0 = 0.39794 log 2.50 10-4 = 0.39794 4 log 2.50 10-5 = 0.39794 5 log 2.50 10-6 = 0.39794 6 Note that the digit term of 2.50 in all of these examples always has the same mantissa, 0.39794 for non-negative exponents, and for negative exponents, the mantissa is the difference between the negative of the exponent and 0.39794. Thus, when dealing with logarithms, the number of significant figures is determined by the number of digits in the mantissa. For example, log 2.50 10 6 = 6.398 and 10 3.18 = 1.5 10 3 CHEM139: Zumdahl Chapter 16 page 6 of 10

16.4 The ph Scale ph Scale: a measure of the concentration of hydrogen ions, [H + ] The value of K w at 25 C (1.0 10-14 ) is so small, indicating very few water molecules ionize to form H + and OH. The hydrogen ion concentration, [H + ], is typically quite small, so the ph scale provides a convenient measure of a solution s acidity. The ph is a log 10 scale and is defined as ph = log [H + ] so [H + ] = 10 -ph Similarly, poh = log [OH ] and [OH ] = 10 -poh ph Scale: ph < 7: acidic ph = 7: neutral ph > 7: basic [H + ] scale 10-1 10-3 10-5 10-7 10-9 10-11 10-13 ph scale 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 strongly weakly acidic weakly basic strongly acidic basic contains H + neutral contains OH 1M HCl lemon juice pure eggs drain cleaner stomach acid vinegar water baking soda 1M NaOH coffee NaCl(aq) liquid bleach Calculate ph or [H + ]: Recognize that ph = -log [H + ] and [H + ] = 10 ph To get ph if [H 3 O + ] = 0.1M for a given solution, change [H + ] given to 10 ph form. so [H + ] = 0.1M = 10 1 M ph = 1 and [H + ]=0.0001M=10 4 M ph = 4 Calculate poh or [OH ]: Recognize that poh = -log [OH ] and [OH ] = 10 poh To get poh, change [OH ] given to 10 poh form. so [OH ] = 0.01M = 10 2 M poh = 2 Converting between ph and poh: ph + poh = 14.00 Converting between [H + ] and [OH ]: K w = [H + ][OH ] = 1.0 10-14 CHEM139: Zumdahl Chapter 16 page 7 of 10

Complete the following table: [H + ] ph [OH ] poh Classify 3.1 10-8 M 2.6 10-9 M 2.65 3.18 1.5 10-2 M 16.5 CALCULATING THE ph of STRONG ACID SOLUTIONS Ex. 1: Calculate the ph for the following solutions: a. [HNO 3 ]= 2.5 10-4 M [H + ]= ph = b. [HCl]= 1.5 10-3 M [H + ]= ph = c. [H 2 SO 4 ]= 3.5 10-5 M [H + ]= ph = d. [NaOH]= 1.5 10-3 M [OH ]= poh = ph = e. [Ca(OH) 2 ]= 2.0 10-5 M ph = f. [Sr(OH) 2 ]= 8.0 10-4 M ph = CHEM139: Zumdahl Chapter 16 page 8 of 10

15.7 NEUTRALIZATION REACTIONS Let s consider three different kinds of acid-base neutralization reactions: A strong acid reacts completely with a strong base to produce water and a salt. A strong acid reacts completely with a weak base to produce water and a salt. A strong base reacts completely with a weak acid to produce water and a salt. In all three of these examples, we can determine the amount of acid (or base) present in a solution if we know the balanced chemical equation and how much base (or acid) is needed to completely neutralize it. Ex 1: Find the molarity of a hydrochloric acid solution if 25.50 ml of HCl is required to neutralize 0.375 g of Na 2 CO 3 as shown in the following equation: 2 HCl(aq) + Na 2 CO 3 (s) 2 NaCl(aq) + H 2 O(l) + CO 2 (g) Ex 2: A 25.0 ml sample of hydrochloric acid requires 37.55 ml of a 0.250M NaOH solution for complete neutralization. Calculate the molarity of the hydrochloric solution if the balanced equation for the reaction is: HCl(aq) + NaOH(aq) H 2 O(l) + NaCl(aq) CHEM139: Zumdahl Chapter 16 page 9 of 10

Ex. 3: A 10.00 ml sample of calcium hydroxide requires 26.18 ml of a 0.5565M nitric acid solution for complete neutralization. Calculate the molarity of the calcium hydroxide. a. Write the balanced equation for the reaction below. b. Calculate the molarity of the calcium hydroxide solution. Ex. 4: Citric acid (abbreviated H 3 X) is a triprotic acid i.e. it has three H + ions that can react to produce water that contains a combination of carbons, hydrogens, and oxygens. If 36.10 ml of 0.223M NaOH is used to neutralize a 0.515 g sample of citric acid, calculate the molar mass of the acid. H 3 X(aq) + 3 NaOH(aq) Na 3 X(aq) + 3 H 2 O(l) CHEM139: Zumdahl Chapter 16 page 10 of 10