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## Chapter 16. Acid –Base Equilibria

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**Equilibria in Solutions of Weak Acids**• The dissociation of a weak acid is an equilibrium situation with an equilibrium constant, called the acid dissociation constant, Ka based on the equation • HA (aq) + H2O (l)H3O+ (aq) + A- (aq)**Equilibria in Solutions of Weak Acids**• The acid dissociation constant, Ka is always based on the reaction of one mole of the weak acid with water. • If you see the symbol Ka, it always refers to a balanced equation of the form • HA (aq) + H2O (l)H3O+ (aq) + A- (aq)**Problem**• The pH of 0.10 mol/L HOClis 4.23. Calculate Kafor hypochlorous acid. • HOCl (aq) + H2O (l)H3O+ (aq) + ClO- (aq)**Calculating Equilibrium Concentrations in Solutions of Weak**Acids • We can calculate equilibrium concentrations of reactants and products in weak aciddissociationreactions with known values for Ka. • To do this, we will often use the ICE table technique we saw in the last chapter on equilibrium.**Calculating Equilibrium Concentrations in Solutions of Weak**Acids • We need to figure out what is an acid and what is a base in our system. • For example, if we start with 0.10 mol/L HCN, then HCN is an acid, and water is a base. • HCN (aq) + H2O (l)H3O+ (aq) + CN- (aq) • Ka = 4.9 x 10-10**Like our previous equilibrium problems, we then create a**table of the initial concentrations of all chemicals, the change in their concentration, and their equilibrium concentrationsin terms of known and unknown values.**We canALWAYS solve this equation using thequadratic formula**and get the right answer, but itmight be possibleto do it more simply.**Every time we do an weak acidequilibrium problem, divide the**initial concentration of the acid by Ka. • For this example • 0.10 / 4.9 x 10-10 = 2 x 108**0.10 / 4.9 x 10-10 = 2 x 108**• Since this value isgreater than 100, we can assumethat theinitial concentration of the acid and the equilibrium concentration of the acidare the same. • This assumption will lead to answers with less than 5% error since this pre-check is greater than 100.**The assumption we will make is that**• x << [HCN]i so [HCN]eqm[HCN]I • 4.9 x 10-10 = x2 / 0.10 • x2 = (4.9 x 10-10)(0.10) • x = 4.9 x 10-11 • x = 7.0 x 10-6 mol/L**Based on the assumption we’ve made, at equilibrium**• x = [H3O+]eqm=[CN-]eqm • =7.0 x 10-6 mol/L • (-ve value isn’t physically possible) • [HCN]eqm = 0.10 mol/L.**.**• Any time we make an assumption, • we MUST check it. • We assumed x << [HCN]i • To check the assumption, we divide x by [HCN]i and express it as a percentage**As long as the assumption check is less than 5%, then**• the assumption is valid! • If the assumption was not valid, we would have to go back and use the quadratic formula!**Remember!**• H2O (l)+H2O (l) H3O+ (aq) + OH- (aq) • is alwaystaking place inwaterwhether or not we have added an acid or base. • This reaction also contributes • H3O+ (aq) and OH- (aq) • to our systemat equilibrium**Remember!**• Since at 25 C • Kw=[H3O+][OH-]=1.0 x 10-14 • it turns out that if ouracid-base equilibrium we’re interested ingives a pH value between about6.8 and 7.2 then the auto-dissociation of water contributes a significant amount of [H3O+] and [OH-]to our system and the real pH would not be what we calculated in the problem.**Problem**• Acetic acid CH3COOH (or HAc) is the solute that gives vinegar its characteristic odour and sour taste. Calculatethe pHand the concentration of all species present in: • a) 1.00 mol/L CH3COOH • b) 0.00100 mol/L CH3COOH**Problem a)**Let’s check the initial acid concentration / Ka ratio. 1.00 / 1.8 x 10-555000 is larger than 100.**Problem a)**We can probably assume that x << [HAc]i so [HAc]eqm[HAc]i 1.8 x 10-5 = x2 / 1.00 x2 = (1.8 x 10-5)(1.00) x = 1.8 x 10-5 x = 4.2 x 10-3 mol/L (but must be + value sincex =[H3O+])**Problem a)**• So at equilibrium, • [H3O+] = [CH3COO-] = 4.2 x 10-3 mol/L • [CH3COOH] = 1.00 mol/L. • The assumption was valid and so • pH = - log [H3O+] • pH = - log 4.2 x 10-3 • pH = 2.38**0.00100**0.00100 Problem b) Let’s check the initial acid concentration / Ka ratio. 0.00100 / 1.8 x 10-556 is smaller than 100.**0.00100**0.00100 Problem b) We can probably CAN NOT assume that x << [HAc]i so [HAc]eqm[HAc]i**Problem b)**Since [H3O+] = x we must use the positive value, so [H3O+] = [CH3COO-]= 1.3 x 10-4 mol/L [CH3COOH] = 0.00100 mol/L – 1.3 x 10-4 mol/L = 0.00087 mol/L.**Problem b)**Let’s confirm that x << [HAc]i IS NOT TRUE pH = - log [H3O+] pH = - log 4.2 x 10-4 pH = 3.38**Problem**• A vitamin C tablet containing 250 mg of ascorbic acid (C6H8O6; Ka = 8.0 x 10-5 is dissolved in a 250 mL glass of water to give a solution where [C6H8O6] = 5.68 x 10-3 mol/L. • What is the pH of the solution?**Problem**• Check the initial acid concentration / Ka ratio. • 5.68 x 10-3/ 8.0 x 10-5 71 • which is not larger than 100 so**Problem**• Since [H3O+] = x the answer must be the positive value • [H3O+] =[C6H7O6-]= 6.3 x 10-4 mol/L • [C6H8O6] = (5.68 x 10-3 - 6.3 x 10-4)mol/L • = 5.05 x 10-3 mol/L • pH = - log [H3O+] • pH = - log 6.3 x 10-4 • pH = 3.20**Degree of ionization**• The pH of a solution of a weak acid like acetic acid will depend on the initial concentration of the weak acid andKa. Therefore, we can define a second measure of the strength of a weak acid by looking of the • degree (or percent) ionization of theacid. • %ionization = [HA]ionized / [HA]initial x 100%**Percent ionization**• In part a) of an earlier problem an acetic acid solution with initial concentration of 1.00 mol/Lat equilibrium had • [H3O+]eqm=[HA]ionized=4.2 x 10-3 mol/L • %ionization= [HA]ionized / [HA]initial x 100% • %ionization = 4.2 x 10-3 mol/L / 1.00 mol/L x 100% • %ionized= 0.42%**Percent ionization**• In part b) of an earlier problem an acetic acid solution with initial concentration of 0.00100 mol/Lat equilibrium had • [H3O+] = [HA]ionized = 1.3 x 10-4 mol/L • %ionization= [HA]ionized / [HA]initial x 100% • %ionization = 1.3 x 10-4 mol/L / 0.00100 mol/L x 100% • %ionization= 13%**Equilibria in Solutions of Weak Bases**• The dissociation of a weak base is an equilibrium situation with an equilibrium constant, called the base dissociation constant, Kb based on the equation • B (aq) + H2O (l)BH+ (aq) + OH- (aq)**Equilibria in Solutions of Weak Bases**• The base dissociation constant, Kb is always based on the reaction of one mole of the weak base with water. • If you see the symbol Kb, it always refers to a balanced equation of the form • B (aq) + H2O (l)BH+ (aq) + OH- (aq)**Equilibria in Solutions of Weak Bases**• Our approach to solving equilibria problems involving bases is exactly the same as for acids. • 1. Set up the ICE table • 2. Establish the equilibrium constant expression • 3. Make a simplifying assumption when possible • 4. Solve for x, and then for eq’m amounts**Problem**• Strychnine (C21H22N2O2),a deadly poison used for killing rodents,is aweak base having Kb = 1.8 x 10-6. Calculate the pH if • [C21H22N2O2]initial = 4.8 x 10-4 mol/L**Problem**Check the initial base concentration / Kb ratio 4.8 x 10-4 / 1.8 x 10-6267 which is greater than 100 We are probably good to make a simplifying assumption that x << [C21H22N2O2]i**The assumption we will make is that**• x << [C21H22N2O2]i so • [C21H22N2O2]eqm[C21H22N2O2]I • 1.8 x 10-6 = x2 / 4.8 x 10-4 • x2 = (1.8 x 10-6)(4.8 x 10-4) • x = 8.64 x 10-10 • x = 2.94 x 10-5 mol/L**Problem**• Since x = [OH-], the answer must be the positive value, • x =[C21H23N2O2+]= [OH-] =2.9 x 10-5 mol/L • [C21H22N2O2] = 4.8 x 10-4 mol/L – 2.9 x 10-5 mol/L • = 4.5 x 10-4 mol/L. • We should check the assumption!**Problem**• In this case, the error is more than 5%. • I will leave it to you to go back and use the quadratic formula. • Compare the two answers**Problem**• To continue towards the answer of the problem AS IF the assumption WERE VALID • pOH = - log [OH-] • pOH = - log 2.9 x 10-5 • pOH = 4.54 • pH + pOH = 14.00 • pH = 14.00 - pOH • pH = 14.00 - (4.54) • pH = 9.46**Relation Between Ka and Kb**• The strength of an acid in water is expressed throughKa, while the strength of a base can be expressed through Kb • Since Brønsted-Lowryacid-base reactions involve conjugateacid-basepairsthereshould be a connectionbetween the • Ka value and the Kb value of a • conjugateacid-basepair.**Relation Between Ka and Kb**• HA (aq) + H2O (l)H3O+ (aq) + A- (aq) • A- (aq) + H2O (l)OH- (aq) + HA (aq)**Since these reactions take place in the same beaker at the**same time let’s • add them together**The sum of the reactions is the dissociation of water**reaction, which has the ion-product constant for water • Kw=[H3O+][OH-]=1.0 x 10-14 at 25 °C • Closer inspection shows us that**As the strength of an acid increases(larger Ka) the strength**of the conjugate base must decrease(smaller Kb) because their product must always be thedissociation constant for water Kw.**Strong acidsalways havevery weak conjugate bases. Strong**basesalways havevery weak conjugate acids. • Since Ka x Kb = Kw • then Ka = Kw / Kb • and Kb = Kw / Ka**Problem**• a) – Piperidine (C5H11N) is an amine found in black pepper. Find Kb for piperidine in Appendix C, and then calculate Ka for the C5H11NH+ cation. • Kb = 1.3 x 10-3 • b) Find Ka for HOCl in Appendix C, and then calculate Kbfor OCl-. • Ka = 3.5 x 10-8**Acid-Base Properties of Salts**• When acids and bases react with each other, • they form ionic compounds called salts. • Salts, when dissolved in water, can lead to acidic, basic, or neutral solutions, depending on the relative strengths of the acid and base we derive them from. • Strong acid + Strong baseNeutral salt solution • Strong acid + Weak baseAcidic salt solution • Weak acid + Strong baseBasic salt solution