AP Chemistry- Acids and Bases General Properties of Acids and Bases. Bases- originally defined as any substance that neutralized an acid

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1 AP Chemistry Acids and Bases General Properties of Acids and Bases Acids Electrolyte Taste Litmus Phenolphthalein React with metals to give off H 2 gas H 2 SO 4 (aq) + Mg (s) MgSO 4 (aq) + H 2 (g) Ionize in aqueous solutions to form Acids react with metal sulfides and metal carbonates to form gases Name: Bases originally defined as any substance that neutralized an acid Electrolyte Taste Aqueous solutions feel Litmus Phenolphthalein Substances that accept H + ions (OH is considered a base because it accepts H + ) HCl (aq) + ZnCO 3 (aq) CO 2 (g) + H 2 O (l) + ZnCl 2 (ag) HNO 3 (aq) + MgS(s) MgNO 3 (aq) + H 2 S (g) Common bases = oxides, hydroxides, ammonia, amines Acids and bases will react with each other give produce water and a salt HNO 3 (aq) + NaOH (aq) Arrhenius Definition of Acids and Bases Arrhenius Acids = increase the concentration of in water Arrhenius Bases = increase the concentration of in water (definition restricted to aqueous solutions) Bronsted Lowry Acids and Bases Acid = any substance that can transfer a proton to another substance Base = any substance that can accept a proton A substance can only function as an acid only if another substance simultaneously behaves as a base Bronsted Lowry Conjugate Acid Base Pairs HCl (g) + H 2 O(l) H 3 O + + Cl NH 3 (aq) + H 2 O(l) > NH OH Strength of Conjugate AcidBase Pairs If a conjugate acid is strong, then its conjugate base is If a conjugate acid is weak, then its conjugate base is 1

2 Amphiprotic (Amphoteric) Substances Substances that can act as both an acid or a base H 3 O + gain H + H 2 O loss of H + OH H 2 SO 4 gain H + HSO 4 loss of H + SO 4 2 Strong Acids Careful! This does not indicate concentration!! Strong electrolytes ionize completely in solution HNO 3 (aq) + H 2 O (l) H 3 O + + NO 3 Generally speaking, in strong acids the H is bonded to either a very electronegative element (eg HBr and HI) or to an oxygen bonded to a non metal (eg H 2 SO 4 ). In these oxyacids the strength of the acid increases with the electronegativity of the non metal (H 2 SO 4 is a strong acid, H 3 PO 4 is a weak acid) and the number of oxygens present (HNO 3 is a strong acid, HNO 2 a weak acid). Memorize: HI, HBr, HCl, HClO 3,HClO 4, HNO 3, H 2 SO 4 The strong acid is the only source of H + in solution. Therefore, the ph of a solution is calculated directly from the initial molarity of the acid [H + ] in 0.01M HCl(aq) =? [H + ] in 0.5M H 2 SO 4 (aq) =? Weak Acids Careful! This does not indicate concentration!! Weak Acids only partially ionize in aqueous solutions HA (aq) + H 2 O (l) H 3 O + + A 99% 1% Almost all organic acids are weak acids (ethanoic acid = acetic acid CH 3 COOH) Aqueous carbon dioxide can act as a weak acid biological function CO 2 (aq) + H 2 O(l) H + + HCO 3 HA (aq) + H 2 O (l) H 3 O + + A 99% 1% There is an equilibrium between the ions and the unionized acid in solution We can therefore write an equilibrium expression for the dissociation: Keq = Concentration of water is constant, therefore, it is incorporated into a new constant; Ka Ka = The larger the Ka, the stronger the acid If Ka >> 1, then the acid is completely ionized = strong acid Strong acids ionize almost completely in water HNO 3 (aq) + H 2 O (l) H 3 O + + NO 3 The concentration of unionized HNO 3 (aq) at equilibrium is nearly zero 2

3 Differentiating between strong a weak acids Strong and weak acids can be differentiated by comparing solutions of equal concentrations. The concentration of H + in the solution of the weak acid will be considerably lower, given rise to a number of differences that may be tested experimentally: 1. a weak acid has a higher ph 2. weak acids do not conduct electricity as well as strong acids 3. weak acids react more slowly in typical acid reactions Problem: First determine if the following acids are strong or weak, then if strong acids, determine the [H + ] M HF M HClO M HC 2 H 3 O M HI Strong Bases Like strong acids, strong bases are strong electrolytes The stronger the base is in water, the more OH will be produced Memorize: Group 1 metal hydroxides (LiOH, NaOH, KOH, RbOH, CsOH); Heavy group 2A metal hydroxides [Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 ] Ionization Constants for Bases Generic form of a base equation: B(aq) + H 2 O (l) BH + + OH Kb measures the relative strength of the base in water solution Kb = The larger the value of Kb, the stronger the base A strong base (NaOH or KOH) does not ionize in solution, it dissociates almost completely NaOH(s) + H 2 O(l) Na + (aq) + OH (aq) The Ionization Constant for Water Selfionization of water water does form a few ions on its own H 2 O (l) + H 2 O (l) H 3 O + + OH Kw= [H 3 O + ] = 1.0 X 10 7 [OH ] = 1.0 X

4 Problem: What are the [H 3 O + ] and [OH ] in 0.10M HCl? Acidic : [H 3 O + ] > [OH ] Basic: [H 3 O + ] < [OH ] Neutral: [H 3 O + ] = [OH ] In every case: [H 3 O + ] X [OH ] = 1.0 X The ph Scale In pure water at 25 o C, the concentration of both H + and OH from the dissociation of water is 1X10 7 M less than one molecule in 10 million is dissociated. The ph (power of hydrogen) of a solution depends on the concentration of the H + and is equal to power of 10 with the sign reversed. The ph of water is 7 as [H + ] = 1X10 7 M. H 2 O (l) + H 2 O (l) H 3 O + + OH According to Le Chat s principle, as we increase the H + concentration, the water equilibrium will shift to the left decreasing the OH concentration. If the concentration of H + is increased by a factor of 10 (1X10 4 to 1X10 3 ), then the ph will decrease by one unit (from 4 to 3). ph = power of hydronium ion Runs from 014 ph = log[h 3 O + ] poh = log[oh ] ph + poh = 14 Problems: 1. A sample of freshly pressed apple cider has a ph of Calculate [H 3 O + ] ml of an aqueous solution of a monoprotic strong acid is added to 90 ml of water. This will cause the ph of the solution to a. increase by 10 b. increase by 1 c. decrease by 1 d. decrease by A solution formed by dissolving an antacid tablet has a ph of Calculate [H + ]. 4. Approximately what ph would you expect for a 0.1M solution of ethanoic acid? a. 1 c. 10 b. 3 d Calculate the ph of a 0.15M solution of NaOH. 4

5 6. The ph of a solution is Calculate the [OH ]. 7. Calculate the ph of a 0.23 M solution of HClO 4. Weak Acid ph calculations Calculating Ka form ph HA (aq) + H 2 O (l) H 3 O + + A In order to find Ka, we need to know all of the equilibrium concentrations o Write a balanced equation showing equilibrium o Write an equilibrium expression for the equation o Use the ph calculation to determine the equilibrium concentration of H + o The equilibrium concentrations of the other species are determined using stoichiometry Problem: A student prepared a 0.10M solution of formic acid, HCHO 2, and measured its ph using a ph meter. The ph at 25 o C was found to be Calculate Ka for formic acid at this temperature. Using Ka to Calculate ph Write a balanced equation showing equilibrium Write an equilibrium expression. Look up value for Ka if it is not given Write down the initial equilibrium concentrations Substitute into the equilibrium constant expression and solve for H + Problem: Calculate the ph of a 0.20M solution of HCN. Ka = 4.9 X What do we do if we are faced with having to solve a quadratic equation? Solve it! If the Ka value is small, make the simplifying assumption Assume X is negligible compared wit the initial concentration of that species Check the validity of your assumption o If X < 10% of the initial concentration, then assumption is probably good o If X > 10% of the initial concentration, then it may be best to solve using the quadratic equation. Problem: Calculate the percentage of HF molecules ionized in 0.1M HF solution. Ka HF = 6.8 X

6 % Ionization is another way to assess acid strength. HA (aq) + H 2 O (l) H 3 O + + A % ionization = [H + ] equilibrium X 100 [HA] initial Problem: Calculate the % ionization for the previous formic acid problem. Buffered Solutions If a small volume of a strong acid is added to water, then the ph of the water will change significantly, for example 0.1 ml of 1M HCl added to a liter of water will change the ph from 7 to 4. If the acid were added to a mixture of a weak acid and it conjugate base rather than water, then the change in the ph would be much less. Buffer A solution consisting of weak acid and its conjugate base or a weak base and its conjugate acid. Buffers are able to resist changes in ph when small amounts of acid or base are added to them. Composition and Action of Buffered Solutions A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X ) HX(aq) > H + (aq) + X (aq) Ka = [H + ][X ] [HX] [H + ] = Ka [HX] [X] Thus a buffer contains both: An acidic species (to neutralize the OH ) and A basic species (to neutralize the H + ) When a small amount of OH is added to the buffer, the OH reacts with HX to produce X and water. OH (aq) + HX (aq) H 2 O(l) + X (aq) As long as the amounts of HX and X in the buffer are large compared to the amount of OH added, the [HX]/[X ] ratio remains more or less constant, so the ph is not significantly changed When a small amount of H + is added to the buffer, X is consumed to produce HX. H + (aq) + X (aq) HX (aq) Once again, the [HX]/[X ] ratio remains more or less constant, so the ph does not change significantly 6

7 Buffer Capacity and ph Buffer capacity Buffer capacity depends on the concentrations of the components of the buffer. The greater the concentrations of the conjugate acidbase pair, the greater the buffer capacity. The ph of the buffer is related to the Ka and on the relative concentrations of the acid and base We can derive an equation that shows the relationship between conjugate acidbase concentrations, ph, and the Ka. Ka = [H + ][A ] [HA] Rearrange [H + ] = Ka [HA] [A ] Take the negative log of each side log[h + ] = log Ka log [HA] [A ] By definition: ph = pka + log [A ] [HA] ph = pka + log [base] [acid] HendersonHasselbalch equation Note that this equation uses equilibrium concentrations of acid and conjugate base However, if Ka is sufficiently small (i.e., if the equilibrium concentration of undissociated acid is close to the initial concentration), then we can use the initial values of the acid and base concentrations in order to get a good estimate of ph. Problem: Calculate the ph of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. Addition of Strong Acids or Bases to Buffers We break the calculation into two parts A stoichiometric calculation An equilibrium calculation 7

8 The addition of strong acid or base results in a neutralization reaction X + H 3 O + HX + H 2 O HX + OH X + H 2 O By knowing how much H 3 O + or OH was added, we know how much HX or X is formed. This is the stoichiometric calculation From the concentrations of HX and X (note the change in volume of the solution) we can calculate the ph from the HendersonHasselbalch equation This is the equilibrium calculation. Problem: Consider a 1.00 L buffer made by adding mol cyanic acid, HCNO, and mol potassium cyanate, KCNO, to sufficient water. Calculate the ph of the buffer a) before any acid or base is added; b) after the addition of mol of HNO 3 ; c) after the addition of mol KOH. In parts b) and c) assume the volume does not change. Acid Base Titrations Titration curve = the plot of ph versus volume during a titration Acid Base Titration Suppose we find a bottled labeled dilute HCl in the laboratory. A logical question would be: How dilute is the HCl? We are really asking for the molar concentration of the acid. Neutralization Total moles of H + ions = Total moles of OH ions from the acid from the base HCl (aq) + NaOH (aq) H 2 O (l) + NaCl (aq) Or Net Ionic Equation: H + + OH H 2 O Materials for a titration: Acid or base standard solution = known concentration Acid or base of unknown concentration Indicator designed to change color at or near the point at which all the acid has reacted with all the base in solution Burette Flask Terms: Equivalence point End point Goal choose an indicator that has an end point equal to the equivalence point Steps of a Titration: 1. The volume of the unknown (acid or base) is measured 2. Acid base indicator added to unknown 3. A standard solution (base or acid) of known concentration is slowly and accurately added to the unknown (burette) 8

9 4. When the indicator changes color, the signal indicates that neutralization has occurred. This is the end point of the titration 5. The molarity of the unknown is calculated N a X V a = N B X V B N= normality equivalents / liter Or (M a X V a ) X # of H + ions = (M B X V B ) X # of OH ions Problems: 1. In an acid base titration, standard 0.10M KOH is used to determine the concentration of an unknown HNO 3 solution. a. Write the equation for the neutralization b. If 50 ml of KOH is needed to neutralize 200 ml of HNO 3, calculate the concentration of the acid. 2. In an acidbase titration, standard 0.1M H 2 SO 4 is used to determine the concentration of an unknown NaOH solution. a. Write the equation for the neutralization b. If 40 ml of H 2 SO 4 is needed to neutralize 100 ml of NaOH, calculate the concentration of the base. Acid Base Titration Solve the problems below. 1. A 25.0 ml sample of HCl was titrated to the endpoint with 15.0 ml of 2.0 N NaOH. What is the normality of the HCl? What was its molarity? 2. A 10.0 ml sample of H 2 SO 4 was exactly neutralized by 13.5 ml of 1.0 M KOH. What is the molarity of the sulfuric acid? What is the normality? 9

10 3. How much 1.5 M sodium hydroxide is necessary to exactly neutralize 20.0 ml of 2.5 M phosphoric acid? 4. How much of 0.5 M nitric acid is necessary to titrate 25.0 ml of 0.05 M calcium hydroxide solution to the endpoint? 5. What is the molarity of a sodium hydroxide solution if 15.0 ml is exactly neutralized by 7.5 m of a 0.02 M acetic acid solution? Strong Acid Strong Base Titrations Consider the titration of 50 ml 0.1 M HCl with 0.1 M NaOH NaOH(aq) + HCl(aq) Na + (aq) + Cl (aq) + HOH(l) Four Regions of Titration Curve 1. Initial ph (before any base is added) The ph is determined by the concentration of the strong acid solution Therefore, ph < 7 ph = log of [H + ] ph = 1 2. Between the initial ph and the equivalence point Equivalence point = The moment in a titration when the number of equivalents of the reactant added from a buret equals the number of equivalents of another reactant in the receiving flask. Moles of base added = moles of acid When base is added, before the equivalence point, the ph is given by the amount of strong acid in excess Therefore, ph < 7 Calculate the ph of the resulting solution when 25 ml of 0.1 M NaOH solution is added to 50 ml of 0.1 M HCl (This is a neutralization reaction use moles in calculations) H + (aq) + OH (aq) HOH(l) Before rxn Change After rxn Note total volume of the solution has changed! 10

11 [H + ] = ph = log[h + ] ph = 3. At the equivalence point The amount of base added is stoichiometrically equivalent to the amount of acid originally present Therefore, the ph is determined by the hydrolysis of the salt in solution Therefore, the ph = 7 (hydrolysis of a salt from a strong acid and strong base is neutral) 4. After the equivalence point The ph is determined by the excess base in solution Therefore, ph > 7 Calculate the ph of the resulting solution when 85 ml of 0.1 M NaOH solution is added to 50 ml of 0.1 M HCl (Again, this is a neutralization reaction, use moles in calculations) H + (aq) + OH (aq) HOH(l) Before rxn Change After rxn Note again total volume of the solution has changed! [OH ] = poh = log[oh ] poh = Indicators The role of an indicator is to signal the endpoint of a titration through a color change The chemist must select an indicator whose color changes at the appropriate end point ( equal to the equivalence point) The difference between the equivalence point and the end point of the titration (indicator color change) is called the titration error Common strong acid strong base titration indicator = Phenolphthalein changes color between ph 8.3 and 10.0 Weak Acid Strong Base Titrations Consider the titration of 50 ml of 0.1 M HC 2 H 3 O 2, with 0.1 M NaOH HC 2 H 3 O 2 (aq) + OH (aq) HOH(l) + C 2 H 3 O 2 (aq) 1. Before any base is added: The solution contains only the weak acid Therefore, the ph is given by the equilibrium calculation (since this is an equilibrium calculation, use Molarity for calculations) Initial Change Final HC 2 H 3 O 2 (aq) < H + (aq) + C 2 H 3 O 2 (aq) Ka = 1.8 X 10 5 = ph = 11

12 2. Between the initial ph and the equivalence point As strong base is added it consumes a stoichiometric quantity of weak acid HC 2 H 3 O 2 (aq) + OH (aq) C 2 H 3 O 2 (aq) + H 2 O(l) However, there is an excess of acetic acid Therefore, we have a mixture of weak acid and its conjugate base Thus the composition of the mixture is that of a buffer The ph is given by the buffer calculation First, the amount of C 2 H 3 O 2 generated is calculated, as well as the amount of HC 2 H 3 O 2 consumed. (neutralization reaction) Then the ph is calculated using equilibrium conditions. (HendersonHasselbalch equation) Calculate the ph of the resulting solution when 25 ml of 0.1 M NaOH is added to the initial solution, 50 ml of 0.1 M HC 2 H 3 O 2 (calculate neutralization reaction first using moles, then the equilibrium calculation using Molarity) Neutralization reaction: HC 2 H 3 O 2 (aq) + OH (aq) C 2 H 3 O 2 (aq) + HOH(l) Before rxn Change After rxn Equilibrium Equation: Henderson Hasselbalch equation ph = pka + log [Base] [Acid] ph = 3. At the equivalence point, all the acetic acid has been consumed and all the NaOH has been consumed. However, C 2 H 3 O 2 has been generated Therefore, the ph is given by the C 2 H 3 O 2 solution This means ph > 7 More importantly, the ph of the equivalence point does not equal 7 for a weak acid strong base titration Calculate the equivalence point in the titration of 50ml of 0.1 M HC 2 H 3 O 2 with 0.1 M NaOH 50 ml of 0.1M NaOH needed to react with 50 ml of 0.1 M HC 2 H 3 O 2 total volume of solution 100 ml Neutralization reaction: HC 2 H 3 O 2 (aq) + OH (aq) C 2 H 3 O 2 (aq) + HOH(l) For every mol of HC 2 H 3 O 2 neutralized, 1 mol of C 2 H 3 O 2 is produced Therefore, 5.00 X 10 3 mol of C 2 H 3 O 2 are present at the equivalence point Equilibrium reaction: [C 2 H 3 O 2 ] = 5.00 X 10 3 mol /.100 L = 5.00 X 10 2 M C 2 H 3 O 2 is an anion of a basic salt solve ph using equilibrium equation for the anion Initial C 2 H 3 O 2 (aq) + HOH(l) > HC 2 H 3 O 2 (aq) + OH (aq) 12

13 Change Final Kb = [OH ] = poh = ph = 4. After the equivalence point: The ph is given by the concentration of the excess strong base Calculate the resulting ph when 51 ml of 0.1M NaOH is added to 50 ml of 0.1M HC 2 H 3 O 2 The ph curve for a weak acidstrong base differs significantly from that of a strong acidstrong base titration For a strong acid strong base titration: The ph begins at less than 7 and gradually increases as base is added Near the equivalence point, the ph increases dramatically For a weak acid strong base titration: The initial ph rise is steeper that that for a strong acid strong base titration However, then there is a leveling off due to buffer effects The inflection point is not as steep for a weak acid strong base titration The shape of the two curves after the equivalence point is the same because ph is determined by the strong base is excess The ph values at the equivalence points differ also: The ph = 7.00 for the strong acid strong base equivalence point The ph > 7.00 for the weak acid strong base equivalence point The ph curve for the titration of a weak base with a strong acid also differs significantly from that of a strong base strong acid titration Consider the titration of NH 3 with HCl The equivalence point occurs at ph 5.28 so phenolphthalein should not be used for this titration The color change for methyl red occurs in the ph range from 4.2 to 6.0 so its is a good indicator for this titration Titration of Polyprotic Acids In polyprotic acids, the ionizable protons dissociate in a series of steps Therefore, in a titration there are n equivalence points corresponding to the number of ionizable protons In the titration of H 2 CO 3 with NaOH there are two equivalence points: One for the formation of HCO 3 One for the formation of CO

14 Additional Aspects of Aqueous Equilibria The Common Ion Effect What happens to the ionization of acetic acid when a strong electrolyte such as sodium acetate is added? HC 2 H 3 O 2 (aq) H + + C 2 H 3 O 2 If NaC 2 H 3 O 2 is added, how does this change our equilibrium. 1 st consider if the substance being added is soluble 2 nd is there a common ion? 3 rd how does this ion shift the equilibrium? 4 th does this shift have an effect on the ph? Does it change the concentration of H + or OH? Acid Base Properties of Salt Solutions Nearly all salts are strong electrolytes therefore they exist as ions in aqueous solutions Acid Base properties of salt solutions are a consequence of the reaction of their ions in solution. Many salts can react with water to form OH or H + This reaction with water is called hydrolysis What happens when sodium acetate is placed in water? NaC 2 H 3 O 2 Na + + C 2 H 3 O 2 The anion C 2 H 3 O 2 can react with water to form a weak acid. Remember weak acids are weak electrolytes kind of like saying they are insoluble. Also recall that the formation of insoluble compounds drive reactions. C 2 H 3 O 2 + HOH H C 2 H 3 O 2 + OH Since this reaction forms OH, the salt produces a basic solution. What happens when ammonium chloride is placed in water? + NH 4 Cl NH 4 + Cl NH 4 + will react with the water the following way NH HOH H 3 O + + NH 3 This reaction drops the ph due to the formation of H 3 O + Summary: 1. Salts derived from a strong acid and strong base are neutral 2. Salts derived from a strong base and a weak acid are basic 3. Salts derived from a strong acid and a weak base are acidic 4. Salts derived from a weak acid and a weak base can be either acidic or basic depending on the values for Ka and Kb. 14

15 Common Ion Effect Practice Questions mole of NaC 2 H 3 O 2 is added to 1.0 liter of 0.10 M HC 2 H 3 O 2 solution. What is the ph of the acetic acid solution before sodium acetate is added? What is the ph after sodium acetate is added? ( Ka for acetic acid = 1.8 X 10 5 ) ml of 1.0 M HC 2 H 3 O 2 is mixed with ml of 0.25M NaOH. Calculate the ph of the resulting solution. Hydrolysis of Salts and ph of Salt Solutions 1. Does potassium acetate hydrolyze? If so, is its aqueous solution acidic or basic? 2. Does potassium bromide hydrolyze? If so, is its aqueous solution acidic or basic? 3. What is the ph of a 0.15M solution of potassium acetate? 4. Calculate the ph of a 0.15M solution of sodium lactate, NaC 3 H 5 O 3. Lactic acid, HC 3 H 5 O 3, the acid responsible for sore muscles, is a monoprotic acid with Ka = 1.4 X

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