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NOTES – Acids, Bases and Ph, Chapter 14

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NOTES – Acids, Bases and Ph, Chapter 14 NOTES – Acids, Bases and pH, Chapter 14 Terms: Acid dissociation constant, Ka the equilibrium constant for a reaction in which a proton is removed + from an acid by the H2O to form a conjugate base and H30 Acidic oxide a covalent oxide that dissolves in water to give an acidic solution Amine: an organic base derived from ammonia in which one or more of the hydrogen atoms are replaced by organic groups Amphoteric substance a substance that can behave either as an acid or a base Basic oxide an ionic oxide that dissolves in water to produce a basic solution Carboxyl group the –COOH group in an organic acid Conjugate base what remains of an acid molecule after a proton is lost Conjugate acid the species formed when a proton is added to a base Conjugate acid-base pair two species related to each other by the donating and accepting of a single proton Diprotic acid an acid having two acidic protons (ex: sulfuric acid) + Hydronium Ion the H30 ion; a hydrated proton Ion- product constant, (Kw) The equilibrium constant for the auto ionization of water; (Kw)= + - -14 [H ][OH ] at 25 ºC. (Kw)= 1.0x10 Lewis Acid an electron-pair acceptor Lewis base an electron-pair donor Lime-soda process a water-softening method in which lime and soda ash are added to water to remove calcium and magnesium ions by precipitation Monoprotic acid an acid with one acidic proton Oxyacids an acid in which the acidic proton is attached to an oxygen atom Organic acid an acid with a carbon-atom backbone; often contains the carboxyl group pH scale a log scale based on 10 and equal to –log[H+]; a convenient way to represent solution acidity that ranges from 0 - 7(neutral/H2O) - 14 Major species the components present in relatively large amounts in a solution Percent dissociation the ratio of the amount of a substance that is dissociated at equilibrium to the initial concentration of the substance in the solution multiplied by 100. Polyprotic Acid an acid with more than one acidic proton. It dissociates in a stepwise manner, one proton as a time Salt an ionic compound Slaked lime calcium hydroxide Strong acid an acid that completely dissociates to produce an H+ ion and the conjugate base Strong bases a metal hydroxide salt that completely dissociates into its ions in water Triprotic Acid three protons that dissociate (ex: Phosphoric acid) Weak acid an acid that dissociates only slightly in aqueous solutions Weak base a base that reacts with water to produce hydroxide ions to only a slight extent in aqueous solution. Misc Concepts I. Brønsted -Lowery Acid- H+ donor (labile O-H bond) Base- H+ acceptor (has lone e- pairs) NOTES – Acids, Bases and pH, Chapter 14 Ex: Brønsted -Lowery acid: HNO2 2- Brønsted -Lowery base: SO3 II. Arrhenius + Acid- forms H3O ions in solution Base- forms OH- ions in solution III. Lewis Acid- e- pair acceptor Base- e- pair donor Ex: BF3 + NH3 → BF3NH3 L.A. L.B. Acceptor Donor IV. Acids and Acid strength Strong acid- 100% dissociates in water HCl, HBr, HI, H2SO4, HNO3, HIO3, HClO4 Weak acid < 100% dissociates in water Acid strength is based on strength of the bond that includes the “H” that becomes H+ Stronger bond = weaker acid and weaker bond = stronger acid + – Dissociation equation for a strong acid: HA + H2O → H3O + A + – Dissociation equation for a weak acid: HA + H2O H3O + A +- [H3 O ][A ] Acid dissociation constant: Ka = [HA] V. Bases and Base strength Strong bases- release OH- into solution 100% dissociates in aqueous solution Group 1 and some of group 2 (Ca, Ba) hydroxide Ex: NaOH, KOH (generically written as … MtOH) Dissociation equation for a strong base: MtOH (s) → Mt+ + OH– Dissociation equation for a weak base: + 1- 1- 1- if no charge: B + H2O HB + OH or for an anion B + H2O HB + OH [HB+- ][OH ] [HB][OH- ] Base dissociation constant: Kb = Kb = [B] [B- ] 1- Weak bases- are typically molecules with CO2 in them or amines (have NH groups) 1- for example CH3CO2 and C2H5NH2 VI. Water as an acid and a base Amphoteric- can be either an acid or a base Water is most common + - Ex: 2H2O → H + OH + + - -7 Kw = [H ][OH¯] if [H ] = [OH ] = 1.0 x 10 M -7 -7 Kw = ( 1.0 x 10 )(1.0 x 10 ) -14 Kw = 1.0 x 10 VII The pH Scale The pH scale is a scale from 1-14 to represent solution acidity It is a log scale based on 10, where pH = -log [H+] i. pH changes by 1 for every power of 10 change in [H+] NOTES – Acids, Bases and pH, Chapter 14 ii. pH decreases as [H+] increases because pH = -log [H+] + - Consider the log form of the expression: Kw = [H ][OH ] + - Kw = [H ][OH ] + - log Kw = log [H ] + log [OH ] + - - log Kw = -log [H ] - log [OH ]) + - Therefore: pKw = p[H ] - p[OH ]) -14 Since Kw = 1.0 X 10 , -14 pKw = -log (1.0 x 10 ) = 14.00 Thus, for all aqueous solutions at 25°C, pH, and pOH add up to 14.00 pH + pOH = 14.00 ex: Calculate pH and pOH for each of the following solutions at 25°C a. 1.0 X 10 –3 M OH– Kw 1.0 x 10-14 [H+] = = = 1.0 x 10 –11 M [OH] 1.0 x 10-3 pH = -log [H+] = -log (1.0 x 10 –11 ) = 11.00 pOH = -log [OH-] = -log (1.0 x 10 –3 ) = 3.00 b. 1.0 M H+ K 1.0 x 10-14 [OH-] = w = = 1.0 x 10 –14 M [H ] 1.0 pH = -log [H+] = -log (1.0) = 0.00 pOH = -log [OH-] = -log (1.0 x 10 –14 ) = 14.00 VIII Calculating the pH of Acidic Solutions a. Mainly deals with the solution components and their chemistry, so it is important to identify and focus on the major species i. 1.0 M HC is actually H+ and Cl- ions b. Major species are those that are present in large amounts + - i. In the solution 1.0 M HCl, the major species are H , Cl , and H2O c. Strong acids are those that dissolve (nearly) completely in solution EX: Calculate the pH of 0.10 M HNO3 + 1. List major species: H , NO 3 , and H2O 2. Consider the major source of acid: HNO3 3. [H+] = 0.10 M and pH = -log(0.01) = 1.00 d. Strong Acid Equilibrium Problems i. List the major species in solution ii. Choose the species that produce H+ ions, and write balanced equations for those rxns iii. Using the values of the equilibrium constants (K) for the rxns you have written, decide which is the major producer of H+ ions iv. Write the equilibrium expression for the dominant equilibrium v. Make an ICE table to find the change in equilibrium in terms of “x” vi. Solve for x vii. Check to see if approximation is valid (5%) rule viii. Calculate [H+] and pH EX: see pg. 673 for an in-depth example of this type of problem e. The pH of a Mixture of Weak Acids i. Sometimes a solution may contain two weak acids of different strengths NOTES – Acids, Bases and pH, Chapter 14 -10 EX: Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 x 10 ) and 5.00 M -4 HNO2 (Ka = 4.0 x 10 ). (p 676) 1. Major species: HCN, HNO2 , and H2O 2. All rxns produce H+: + - -10 a. HCN H + CN Ka = 6.2 x 10 + -4 b. HNO2 H + NO2 Ka = 4.0 x 10 + - -14 c. H2O H + OH Ka = 1.0 x 10 3. Because of the Ka values, it can be determined that HNO2 is the major producer of H+ ions 4. Write the equilibrium equation for HNO2: -4 [H ][NO2 ] a. Ka = 4.0 x 10 = [HNO2 ] 5. Make and ICE table to find x = [H+] + a. HNO2 H + NO2 Initial: 5.00 0 0 Change: -x +x +x Equilibrium: 5.00 – x x x -4 b. Solve for x {let (5.00 - x = 5.00)} using the Ka = 4.0 X 10 = 6. Therefore [H+] = x = 4.5 X 10-2 M and pH = 1.35 f. Percent Dissociation i. Used to find the amount of weak acid that has dissociated in reaching equilibrium in aqueous solution amount dissociated (mol/L) ii. Percent dissociation = 100% initial concentration (mol/L) + -2 EX: Refer to previous example and find the percent dissociation of HNO2 where [H ] = x = 4.5 x 10 M and pH = 1.35 [H ] The percent dissociation is: 100% = .9% [HNO2 ] IX. Bases According to the Bronsted and Lowry model, and the Arrhenius concept, a base both accepts protons and yields OH- ions. This can be shown by the dissociation of NaOH: NaOH Na+(aq) + OH- (aq) - The dissociation of a base can also be represented by B + H20 BH + OH The neutralization of an acid and base will always result in water and a salt. Acid Base Water Salt HCl + NaOH H2O + NaCl a. Calculating pH of Strong Bases Since strong, assume 100% dissociation. Calculate pOH, then substract from 14 to get the pH EX: Calculate the pH of a 0.050 M solution of KOH. Because [OH-] =0.050 M: pOH = - log (0.050) = 1.30 pH = 14.00 - 1.30 = 12.70 NOTES – Acids, Bases and pH, Chapter 14 b. Calculating pH of Weak Bases Weak bases are only partially ionized in solution while strong bases are completely ionized, so need to determine the hydroxide concentration via equilibrium, then calculate the pOH and then subtract from 14 to get pH -5 EX: Calculate the pH for a 15.0M solution of NH3 (Kb = 1.8 x 10 ) - This reaction includes NH3, H20, NH4 and OH ; H2O is negligible + - Kb= [NH4 ][OH ] [NH3] Use an ice table to find the values to plug into the equation: Kb= [x][x] [15.0-x] x 2 1.8 x 10-5 = 15.0 Therefore: [OH-] = 1.6 x 10-2 K 1.0x1014 [H+] = w = = 6.3 x 10-13 [OH ] 1.6x102 pH= -log(6.3x10-13) = 12.20 X.
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