CHEM Dr. Babb s Sections Exam #3 Review Sheet

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1 CHEM 116 Dr. Babb s Sections Exam #3 Review Sheet Acid/Base Theories and Conjugate AcidBase Pairs 111. Define the following terms: Arrhenius acid, Arrhenius base, Lewis acid, Lewis base, BronstedLowry acid and BronstedLowry base Identify the Lewis acids and bases in the following reactions. A. BCl 3 + NH 3 < > Cl 3 BNH 3 B. H + + H 2 O < > H 3 O + C. H + + OH < > H 2 O D. Ag+ + 2 NH 3 < > [Ag(NH 3 ) 2 ] Identify the BronstedLowry acids and bases in the following reactions. A. HCO 3 + H 2 O < > H 2 CO 3 + OH B. HCO H 2 O < > CO 3 + H 2 O C. 2 HPO 4 + H 2 PO 4 3 < > PO 4 + H 3 PO 4 D. HCO NaOH < > CO 3 + H 2 O + Na Does H + exist alone in solution? How is H + usually written in acid/base equilibrium? 115. What is a conjugate acid/base pair? How does an acid differ from its conjugate base? How does a base differ from its conjugate acid? 116. For the reaction below, identify the conjugate acid/base pairs. HC 2 H 3 O 2 (aq) + H 2 O < > C 2 H 3 O 2 (aq) + H 3 O + (aq) 117. Give the conjugate acid and conjugate base of each of the following species: NH 3, HSO 4, CO 2 3, and H 2 O Does a strong acid have a strong or weak conjugate base? Does a weak acid have a strong or weak conjugate base? Does a strong base have a strong or weak conjugate acid? Does a weak base have a strong or weak conjugate acid? 119. In an acid/base reaction, to tell which way the equilibrium lies we look at the relative strengths of the two bases involved. Which base gets the H +, the stronger or the weaker base? Which way does the equilibrium lie for the following two reactions? A. HCl + H 2 O < > H 3 O + + Cl B. HC 2 H 3 O 2 (aq) + H 2 O < > C 2 H 3 O 2 (aq) + H 3 O + (aq) K W, ph, poh, K A and K B, Calculation of ph for Strong and Weak Acids and Bases 120. Write the reaction for the dissociation of water. Write the form for the ion product constant of water, K w. What does the value of K w depend on? What are the equilibrium concentrations of H 3 O + and OH in pure water? 121. Define an acidic, basic and neutral solution in terms of the concentrations of H 3 O + and OH If [H 3 O + ]= M, what is the [OH ]? If [OH ]= M, what is the [H 3 O + ]? 123. Write the equations used to calculate ph, poh and pk w. Write the equation that relates ph, poh, and pk w A ph>7 indicates what type of solution, acidic, basic or neutral? A ph<7 indicates what type of solution, acidic, basic or neutral? A ph=7 indicates what type of solution? 125. A. A ph=5.17 indicates what type of solution, acidic, basic or neutral? A poh=8.9 indicates what type of solution, acidic, basic or neutral? A ph=7 indicates what type of solution? B. Which of the solutions shown below is basic? A solution with... i. ph=6.5 iii. [H 3 O + ]= ii. poh=5.5 iv. [OH ]= What is K A? Write the equation that goes along with K A for HCl and HC 2 H 3 O 2. What does K A tell us about the strength of an acid? If K A is small (10 3 ) is the acid weak or strong?

2 127. What is K B? Write the equation that goes along with K B for NH 3. What does K B tell us about the strength of a base? If K B is small (10 3 ) is the base weak or strong? 128. Calculate the [H 3 O + ], [OH ], ph and poh for a 0.25 M HClO 4 solution Calculate the [H 3 O + ], [OH ], ph and poh for a M Ba(OH) 2 solution Calculate the ph of a M HCl solution Calculate the ph of a M Ca(OH) 2 solution Calculate the [H 3 O + ], [OH ], ph, poh and percent ionization for a M HF solution. (K A (HF)= ) 133. The weak acid formic acid HCOOH has a ph of 2.14 when the concentration of HCOOH is 0.30 M. What is the value for the aciddissociation constant K A and what is the percent dissociation of the HCOOH? 134. Calculate the [H 3 O + ], [OH ], ph, poh and percent ionization for a 0.75 M NH 3 solution. (K B (NH 3 )= ) 135. A. Write all of the equilibrium equations established in a solution of 0.25 M H 2 SO 3 and calculate the ph of the solution. (K A1 = ; K A2 = ) B. Write all of the equilibrium equations established in a solution of 0.25 M H 3 PO 4 and calculate the ph of the solution. (K A1 = ; K A2 = ; K A3 = ) Hydrolysis of Salts 136. When a salt is dissolved in water (this process is called hydrolysis of the salt) is the ph of the resulting solution always neutral? 137. The general formula for formation of the salt is ACID + BASE > SALT + H 2 O. Where does the anion of the salt come from, the acid or the base? Where does the cation of the salt come from, the acid or the base? 138. A salt that contains the cation from a strong base and the anion from a strong acid will dissolve in water to give a neutral solution. Explain why using NaCl A salt that contains the cation of a weak base and the anion from a strong acid will dissolve in water to give an acidic solution. Explain why using chemical equations for NH 4 Cl A salt that contains the cation of a strong base and the anion from a weak acid will dissolve in water to give a basic solution. Explain why using chemical equations for KNO A salt that contains a small, highly charged metal cation and the anion from a strong acid will dissolve in water to give an acidic solution. Explain why using chemical equations for AlCl When the following salts are dissolved in water, will the resulting solutions be acidic, basic or neutral. A. KC 2 H 3 O 2 D. NaF G. Zn(NO 3 ) 2 B. NH 4 Br E. K 2 CO 3 H. FeCl 2 C. KClO 4 F. AlBr To do problem 144, K B of F (the conjugate base of HF) is needed. Show how K A for HF and K B for F can be related by K w Calculate the ph of a 0.45 M NaF solution. K A (HF)= HINT: First predict whether this solution should be acidic, basic or neutral. Second write the equations for dissociation of the salt and any equilibrium established. Third calculate either K A or K B for the equilibrium Calculate the ph of 0.75 M NH 4 ClO 4. K B (NH 3 )= HINT: First predict whether this solution should be acidic, basic or neutral. Second, write the equations for dissociation of the salt and any equilibrium established. Third, calculate either K A or K B for the equilibrium. Strength of Acids 146. How does the strength of binary acids (consisting of H and some other nonmetal element)

3 vary within a group? Why? Arrange the following in order of increasing acid strength: HCl, HF, HI, and HBr. Arrange the following in order of increasing acid strength: H 2 S, H 2 Te, and H 2 O How does the strength of binary acids vary within a row? Why? Arrange the following in order of increasing acid strength: HF, H 2 O, NH 3, and CH How does the strength of oxyacids (consisting of H, O and some other element) vary within a series? Arrange the following in order of increasing acid strength: HClO. HClO 3, HClO 2, and HClO 4. How does the strength of an oxyacid vary with the electronegativity of the central atom? Arrange the following in order of increasing acid strength: HOBr, HOCl, and HOI Which of the following equilibria lie mostly on the product side (ie. in an equilibrium mixture which one has product concentrations > reactant concentrations) A. HF + Br < > HBr + F D. H 2 O + H 2 O < > OH + H 3 O + B. NH 2 + H 2 O < > NH 3 + OH E. HBrO 3 + IO 3 < > BrO 3 + HIO 3 C. HClO + ClO 2 < > ClO + HClO 2 Neutralization, Common Ion Effect and Buffers 150. When a strong acid is neutralized by a strong base is the neutralization complete? Write the neutralization reaction for HNO 3 and KOH. What is the value of K n (equilibrium constant for neutralization)? If stoichiometric amounts of acid and base are reacted will the resulting solution be acidic, basic or neutral? 151. When a weak acid is neutralized by a strong base is the neutralization complete? Write the neutralization reaction for HF and KOH. What is the value of K n (equilibrium constant for neutralization)? (K A (HF)= ) If stoichiometric amounts of acid and base are reacted will the resulting solution be acidic, basic or neutral? 152. When a strong acid is neutralized by a weak base is the neutralization complete? Write the neutralization reaction for HClO 4 and NH 3. What is the value of K n (equilibrium constant for neutralization)? (K B (NH 3 )= ) If stoichiometric amounts of acid and base are reacted will the resulting solution be acidic, basic or neutral? 153. When a weak acid is neutralized by a weak base is the neutralization complete? Write the neutralization reaction for HF and NH 3. What is the value of K n (equilibrium constant for neutralization)? (K A (HF)= ; K B (NH 3 )= ) If stoichiometric amounts of acid and base are reacted will the resulting solution be acidic, basic or neutral? 154. What happens to the ph of a solution of M HC 2 H 3 O 2 if a strong electrolyte like NaC 2 H 3 O 2 is added. NOTE: Addition of NaC 2 H 3 O 2 adds a common ion. Use Le Chatelier s principle to predict the direction of the reaction. What will happen to the ph of M HC 2 H 3 O 2 if NaCl is added? 155. What happens to the ph of a solution of M NH 3 if a strong electrolyte like NH 4 Cl is added. NOTE: Addition of NH 4 Cl adds a common ion. Use Le Chatelier s principle to predict the direction of the reaction. What will happen to the ph of M NH 3 if KNO 3 is added? 156. Calculate the ph and % ionization of a solution which is M HC 2 H 3 O 2 and M NaC 2 H 3 O 2. How does the ph of this solution compare to that of M HC 2 H 3 O 2? (K A (HC 2 H 3 O 2 )= ) 157. What is a buffer solution and what is its function? If the following substances are mixed in equimolar amounts, will a buffer solution be formed? A. HF and NaF D. NH 3 and NH 4 Cl G. HF and NaOH B. HC 2 H 3 O 2 and NaC 2 H 3 O 2 E. HCl and KCl H. NH 3 and HCl C. HCN and NaCN F. NaOH and NaCl

4 If equal volumes of the following solutions are mixed, will a buffer solution result? A. 0.2 M HF and 0.1 M NaOH E. 0.5 M NH 3 and 1.0 M HCl B. 0.2 M HF and 0.3 M NaOH F. 0.2 M HCN and 0.2 M KOH C. 0.2 M HF and 0.2 M NaOH G. 0.3 M HOCl and 0.15 M Ba(OH) 2 D. 0.2 M NaF and 0.1 M HCl H. 0.3 M HOCl and 0.1 M Ba(OH) Explain using the HC 2 H 3 O 2 /NaC 2 H 3 O 2 buffer system how a buffer maintains a relatively constant ph when small quantity of acid (HCl) or base (NaOH) is added A buffer solution consists of M HC 2 H 3 O 2 and M NaC 2 H 3 O 2. Calculate the ph of the resulting solution if mol HClO 4 is added to 500. ml of the buffer. The ph of the buffer solution before addition of the HClO 4 was A buffer solution consists of M HC 2 H 3 O 2 and M NaC 2 H 3 O 2. Calculate the ph of the resulting solution if mol NaOH is added to 500. ml of the buffer. The ph of the buffer solution before addition of the NaOH was Derive the HendersonHasselbach equation for a buffer solution consisting of weak acid/conjugate base (HF/NaF) and for a buffer solution consisting of weak base/conjugate acid (NH 3 /NH + 4 ) If the concentrations of acid and conjugate base are equal what is the ph of the buffer solution equal to? 163. A buffer solution is 0.25 M HNO 2 and 0.15 M NaNO 2. What is the ph of the buffer? K A (HNO 2 )= A buffer solution with a ph of 4.0 is needed in an experiment. Which of the following buffer systems should be used? For the chosen buffer system, what concentration ratio of conjugate base:acid should be used to attain ph=4.0? A. HNO 2 /NaNO 2 K A = B. HC 2 H 3 O 2 /NaC 2 H 3 O 2 K A = C. HCOOH/NaCOOH D. C 6 H 5 COOH/NaC 6 H 5 COO K A = K A = E. HCN/NaCN F. NH 3 /NH 4 Cl K A = K B = A buffer with a ph of 9.5 is to be prepared from NH 3 and /NH 4 Cl. What ratio of base to salt would you use? K B (NH 3 )= Explain why dilution of a buffer solution does not change the ph. Does a buffer work best when it s solution is dilute or concentrated (i.e. Which has a better buffer capacity a dilute or concentrated solution)? What concentration ratio of acid/salt or base/salt works best for buffers?

5 CHEM 116Dr. Babb s Sections Answer Key to Exam #3 Review Sheet 112. Lewis Acid (e pair acceptor) Lewis Base (e pair donor) A. BCl 3 NH 3 B. H + H 2 O C. H + OH D. Ag + NH BL Acid (H + donor) BL Base (H + acceptor) where BL=BronstedLowry A. H 2 O HCO 3 B. HCO 3 H 2 O C. 2 HPO 4 H 2 PO 4 D. HCO 3 NaOH (or OH ) 114. H + does not exist alone in solution and is actually associated and surrounded by many water molecules. From now on, H + will be represented as H 3 O Conjugate acid/base pairs have chemical formulas that differ by an H +. To obtain the conjugate acid of a base, add an H + to the chemical formula of the base. To obtain the conjugate base of an acid, subtract an H + from the chemical formula of the acid Acid (reactant side): HC 2 H 3 O 2 Conjugate Base (product side): C 2 H 3 O 2 Base (reactant side): H 2 O Conjugate Acid (product side): H 3 O NH 3 : Conjugate acid = NH 4+ ; Conjugate base = NH 2 HSO 4 : Conjugate acid = H 2 SO 4 2 Conjugate base = SO 4 2 CO 3 : Conjugate acid = HCO 3 Conjugate base = none (no H + to remove) H 2 O: Conjugate acid = H 3 O + Conjugate base = OH 118. Strong acids have weak conjugate bases (i.e. the conjugate base has no tendency to react with water to reform the original strong acid). Weak acids have strong conjugate bases (i.e. the conjugate base has a strong tendency to react with water to reform the original weak acid). Strong bases have weak conjugate acids (i.e. the conjugate acid has no tendency to react with water to reform the original strong base). Weak bases have strong conjugate acids (i.e. the conjugate acid has a strong tendency to react with water to reform the original weak base) The stronger base gets the H + and the equilibrium lies on the side opposite that of the strongest base. A. Eq. lies on the product side (i.e. [Products] > [Reactants]) because of the bases H 2 O and Cl, the H 2 O is the stronger base and thus, the eq. lies on the opposite side. B. Eq. lies on the reactant side (i.e. [Reactants] > [Products]) because of the bases H 2 O and C 2 H 3 O 2, the C 2 H 3 O 2 is the stronger base and thus the eq. lies on the opposite side Eq for dissociation of water: 2 H 2 O(l) < > H 3 O + (aq) + OH (aq); IonProduct constant=k w = [H 3 O + ][OH ] and K w has a value of at 25 C. The numerical value of K W depends only on temperature. In pure water, [H 3 O + ] = [OH ] = M Acidic solution occurs when [H 3 O + ] > M or when [OH ] < M or when [H 3 O + ] > [OH ]. Basic solution occurs when [H 3 O + ] < M or when [OH ] > M or when [H 3 O + ] < [OH ]. Neutral solution occurs when [H 3 O + ] = [OH ] = M All of the above definitions are for 25 C.

6 122. When [H 3 O + ] = M, then [OH ] = M. When [OH ] = M, then [H 3 O + ] = M ph = log [H 3 O + ]; poh = log [OH ]; pk W = logk w = log( ) = 14; ph + poh = ph > 7 indicates a basic solution; ph < 7 indicates an acidic solution; ph = 7 indicates a neutral solution A. ph = 5.17 is an acidic solution; poh = 8.9 has a solution ph = = 5.1 and is an acidic solution. ph = 7 is a neutral solution. B. i. ph = 6.5 is an acidic solution. ii. poh = 5.5 has a solution ph of 8.5 and is a basic solution. iii. [H 3 O + ] = is a basic solution since [H 3 O + ] < M and ph = 9.0. iv. [OH ] = is an acidic solution since [OHG] < M and ph = K A = Acid Dissociation Constant; the higher the value of K A, the stronger the acid; the lower the value of K A, the weaker the acid. HCl + H 2 O < > H 3 O + + Cl this eq. specified by K A for HCl HC 2 H 3 O 2 (aq) + H 2 O < > C 2 H 3 O 2 (aq) + H 3 O + (aq) this eq. specified by K A for HC 2 H 3 O K B = Base Dissociation Constant; the higher the value of K B, the stronger the base; the lower the value of K B, the weaker the base. NH 3 (aq) + H 2 O < > NH + 4 (aq) + OH (aq) this eq. specified by K B for NH [H 3 O + ] = 0.25 M; [OH ] = M; ph = 0.60; poh = [H 3 O + ] = G 12 M; [OH ] = M; ph = 12.00; poh = ph = ph = [H 3 O + ] = M; [OH ] = M; ph = 2.53; poh = 11.47; % Ionization HF = 12% 133. K A = ; % Dissociation of HCOOH = 2.4% 134. [H 3 O + ] = M; [OH ] = M; ph = 11.57; poh = 2.43; % Ionization NH 3 = 0.49% 135. A. H 2 SO 3 (aq) + H 2 O < > HSO 3 (aq) + H 3 O + (aq) K A1 HSO 3 (aq) + H 2 O < > SO 2 3 (aq) + H 3 O + (aq) K A2 B. H 3 PO 4 (aq) + H 2 O < > H 2 PO 4 (aq) + H 3 O + (aq) K A1 H 2 PO 4 (aq) + H 2 O < > HPO 2 4 (aq) + H 3 O + (aq) 2 3 HPO 4 (aq) + H 2 O < > PO 4 (aq) + H 3 O + (aq) 136. No, the ph of the resulting solution can be acidic, basic or neutral depending on the chemical identity of the salt The anion for the salt always comes from the acid; while, the cation for the salt always comes from the base NaCl is a neutral salt. NaCl is a strong electrolyte and dissolves and dissociates 100% as shown here: NaCl(s) > Na + (aq) + Cl (aq). The Na + is a weak conjugate acid of the strong base NaOH. Therefore, Na + has no tendency to react with water (hydrolyze) to reform the original strong base. The Cl is the weak conjugate base of the strong acid HCl. Therefore, Cl has no tendency to react with water (hydrolyze) to reform the original strong acid. The ph of an aqueous solution of this type of salt (containing cation from strong base and anion from strong acid) will be the ph of pure water which is NH 4 Cl is an acidic salt. NH 4 Cl is a strong electrolyte and dissolves and dissociates 100% as shown here: NH 4 Cl(s) > NH + 4 (aq) + Cl (aq). The Cl is the weak conjugate base of the strong acid HCl. Therefore, Cl has no tendency to react with water to reform the original K A2 K A3

7 strong acid. The NH is the strong conjugate acid of the weak base NH 3. Therefore, NH 4 has a tendency to react with water (hydrolyze) to reform the original weak base as shown here: NH 4+ (aq) + H 2 O(l) < > NH 3 (aq) + H 3 O + (aq). During this reaction some H 3 O + is produced resulting in a solution with an acidic ph (ph < 7) KNO 2 is a basic salt. KNO 2 is a strong electrolyte and dissolves and dissociates 100% as shown here: KNO 2 (s) > K + (aq) + NO 2 (aq). The K + is a weak conjugate acid of the strong base KOH. Therefore, K + has no tendency to react with water (hydrolyze) to reform the original strong base. The NO 2 is the strong conjugate base of the weak acid HNO 2. Therefore, NO 2 has a tendency to react with water (hydrolyze) to reform the original weak acid as shown here: NO 2 (aq) + H 2 O(l) < > HNO 2 (aq) + OH (aq). During this reaction some OH is produced resulting in a solution with a basic ph (ph > 7) AlCl 3 is an acidic salt. AlCl 3 is a strong electrolyte and dissolves and dissociates 100% as shown here: AlCl 3 (s) > Al +3 (aq) + 3 Cl (aq). The Cl is the weak conjugate base of the strong acid HCl. Therefore, Cl has no tendency to react with water to reform the original strong acid. The Al +3 being a small, highly charged metal cation exists in solution as Al(H 2 O) +3 6 (aq). The association of the H 2 O s with the Al +3 tends to weaken the OH bond in the H 2 O s making the waters surrounding the aluminum more acidic. Therefore, the +3 Al(H 2 O) 6 species has a tendency to react with water (hydrolyze) as an acid as shown here: +3 Al(H 2 O) 6 (aq) + H 2 O(l) < > Al(H 2 O) 5 OH +2 (aq) + H 3 O + (aq). During this reaction some H 3 O + is produced resulting in a solutions with an acidic ph (ph < 7) A. basic D. basic G. acidic B. acidic E. basic H. acidic C. neutral F. acidic 143. K A K B = K W 144. NaF is a basic salt due to hydrolysis of F. ph = NH 4 ClO 4 is an acidic salt due to hydrolysis of NH + 4. ph = In general, any factor that weakens the bond to hydrogen, makes the substance more acidic. For binary acids within the same group such as HCl, HF, HI, and HBr, the atom bonded to hydrogen (F, Cl, Br, I) becomes larger as proceed down the periodic table and this tends to weaken the bond to H. Thus, binary acids become more acidic as proceed from top to bottom with a column. Most acidic: HI > HBr > HCl > HF :Least acidic; Most acidic: H 2 Te > H 2 S > H 2 O :Least acidic For binary acids within the same row such as CH 4, NH 3, H 2 O, and HF, the atom the hydrogen is bonded to (C, N, O and F) becomes more electronegative as proceed across the row and this tends to weaken the bond to H. Thus these binary acids become more acidic as proceed from left to right within a row. Most acidic: HF > H 2 O > NH 3 > CH 4 :Least acidic In oxyacids, the H is always bonded to one of the oxygens. Any factor that weakens the OH bond, makes the substance more acidic. Within an oxyacid series such as HClO, HClO 3, HClO 2 and HClO 4, as the number of oxygens bonded to the central atom increases, the oxidation number of the central atom increases causing a weakening of the OH bond strength and an increase in the acidity. Most acidic: HClO 4 > HClO 3 > HClO 2 > HClO. For a series of oxyacids with the same number of oxygens but with different central atoms such as HOBr, HOCl, HOI, as the electronegativity of the central atom increases, the OH bond strength weakens and the acidity increases. Most acidic: HOCl > HOBr > HOI :Least acidic Compare strengths of acids on reactant and product sides. The stronger acid has a greater

8 tendency to give up its proton; therefore, the equilibrium will lie on the opposite side from the strongest acid. A. HBr is a stronger acid than HF. Eq lies on side opposite the HBr (i.e. on reactant side); thus, [Reactants] > [Products] and K < 1. B. H 2 O is a stronger acid than NH 3. Eq lies on side opposite the H 2 O (i.e. on product side); thus, [Products] > [Reactants] and K > 1. C. HClO 2 is a stronger acid than HClO. Eq lies on side opposite the HClO 2 (i.e. on reactant side); thus, [Reactants] > [Products] and K < 1. D. H 3 O + is a stronger acid than H 2 O. Eq. lies on side opposite the H 3 O + (i.e. on the reactant side); thus, [Reactants] > [Products] and K < 1. E. HBrO 3 is a stronger acid than HIO 3. Eq. lies on side opposite the HBrO 3 (i.e. on the product side); thus, [Products] > [Reactants] and K > As solid NaC 2 H 3 O 2 is added to M HC 2 H 3 O 2, the [H 3 O + ] decreases and the ph increases. Why? Look at effect of addition of NaC 2 H 3 O 2, a strong electrolyte, on the acetic acid eq. HC 2 H 3 O 2 (aq) + H 2 O(l) < > C 2 H 3 O 2 (aq) + H 3 O + (aq) NaC 2 H 3 O 2 (s) > Na + (aq) + C 2 H 3 O 2 (aq) As NaC 2 H 3 O 2 is added to an eq. mixture of M HC 2 H 3 O 2, the common ion C 2 H 3 O 2 is added which shifts the acetic acid eq. to the left decreasing the [H 3 O + ] and increasing the ph. Addition of NaCl to an eq. mixture of M HC 2 H 3 O 2 has no effect on the ph or acetic acid eq. because NaCl has no ion in common with the acetic acid eq As solid NH 4 Cl is added to M NH 3, the [OH ] decreases, the [H 3 O + ] increases and the ph decreases. Why? Look at effect of addition of NH 4 Cl, a strong electrolyte, on the ammonia eq. NH 3 (aq) + H 2 O(l) < > NH + 4 (aq) + OH (aq) NH 4 Cl(s) > NH + 4 (aq) + Cl (aq) + As NH 4 Cl is added to an eq. mixture of M NH 3, the common ion NH 4 is added which shifts the ammonia eq. to the left decreasing the [OH ] and decreasing the ph. Addition of KNO 3 to an eq. mixture of M NH 3 has no effect on the ph or ammonia eq. because KNO 3 has no ion in common with the ammonia eq For a M HC 2 H 3 O 2 /0.250 M NaC 2 H 3 O 2 : ph = 4.44 and % Ionization = %. For a M HC 2 H 3 O 2 : ph = 2.52 and % Ionization = 0.60% Buffer solution: A solution consisting of a weak acid and the conjugate base (or salt) of the weak acid. OR A solution consisting of a weak base and the conjugate acid (or salt) of the weak base. Function of Buffer: Acts to maintain the ph of the solution at a relatively constant value even with the addition of strong base or strong acid. If the following substances are mixed in equimolar amounts, will a buffer solution be formed? A. HF and NaF; Yes D. NH 3 and NH 4 Cl; Yes B. HC 2 H 3 O 2 and NaC 2 H 3 O 2 ; Yes E. HCl and KCl; No C. HCN and NaCN; Yes F. NaOH and NaCl; No G. HF and NaOH; No H. NH 3 and HCl; No If equal volumes of the following solutions are mixed, will a buffer solution result? A. 0.2 M HF and 0.1 M NaOH; Yes B. 0.2 M HF and 0.3 M NaOH; No C. 0.2 M HF and 0.2 M NaOH; No

9 D. 0.2 M NaF and 0.1 M HCl; Yes E. 0.5 M NH 3 and 1.0 M HCl; No F. 0.2 M HCN and 0.2 M KOH; No G. 0.3 M HOCl and 0.15 M Ba(OH) 2 ; No H. 0.3 M HOCl and 0.1 M Ba(OH) 2 ; Yes 159. ph = ph = HendersonHasselbach equation: ph = pk A + log[base]/[acid] 162. When [Base] = [Acid] then ratio [base]/[acid] = 1 and log(1) = 0 so ph = pk A. Thus, the pk A is the midpoint of the buffer range ph = If a buffer with a ph = 4.0 is needed in an experiment, then a buffer system with a pk A value around 4.0 is needed. pk A A. HNO 2 /NaNO 2 K A = B. HC 2 H 3 O 2 /NaC 2 H 3 O 2 K A = C. HCOOH/NaCOOH K A = D. C 6 H 5 COOH/NaC 6 H 5 COO K A = E. HCN/NaCN K A = F. NH 3 /NH 4 Cl K B = The best buffer system is that given in D because its pk A value is closest to 4.0. To obtain a ph=4.0 using this buffer system, a concentration ratio of base:acid of 0.65 should be used. Therefore, if the [C 6 H 5 COOH] is 1.0 M; the [NaC 6 H 5 COO] should be 0.65 M [NH 3 ]/[NH 4+ ] = 1.7