EoC 9-1 Briefly describe or define and given an example of… (a) a weak electrolyte A weak electrolyte only partially ionizes when dissolved in water. H2CO3 is an example of a weak electrolyte. (b) a BrønstedLowry acid A Brønsted-Lowry acid is a molecule that donates a proton when it encounters a base (proton acceptor). By this definition, NH4+ can be a Brønsted-Lowry acid. (c) the conjugate acid The conjugate acid of a Brønsted-Lowry base is the potential proton donator formed when a Brønstedof a Brønsted-Lowry Lowry base accepts a proton. For example, the NH 4 + is a conjugate acid in the reaction, NH3 + base proton ⇌ NH4 +. (d) neutralization, in terms of the Brønsted-Lowry concept Neutralization, according to the Brønsted-Lowry concept, occurs when a reaction involving an acid and its conjugate base is combined with a second reaction involving a base and its conjugate acid. Thus, NH3 + H2 O ⇌ NH4 + + OHIn the example above, NH 3 acts as a base with NH4 + as its conjugate acid. H2 O acts as an acid with OH– as its conjugate base. (e) and amphiprotic An amphiprotic solvent can act either as an acid or a base depending on the solute. Water is an solvent example of an amphiprotic chemical species. (f) a zwitterion A zwitterion is a chemical species that bears both positive and negative charges. Free amino acids, such as glycine, can exist as zwitterions in solution. NH2 CH3 COOH ⇌ NH 3 +CH2 COO - . (g) autoprotonolysis Autoprotolysis is the act of self-ionization to produce both a conjugate acid and a conjugate base. EoC 9-2 (h) a strong acid A strong acid dissociates completely such that no undissociated molecules are left in aqueous solution. Hydrochloric acid, HCl, is an example of a strong acid. (i) Le Châtelier principal The Le Châtelier principle states that the position of an equilibrium always shifts in such a direction that it relieves the stress. A common ion like sulfate added to a solution containing sparingly soluble BaSO 4 is an example (j) the common-ion effect The common-ion effect is responsible for the reduced solubility of an ionic precipitate when one of the soluble components reacting to form the precipitate is added to the solution in equilibrium with the precipitate. Chloride ion added to a AgCl solution decreases the solubility of Ag + because of the common ion effect. 9-2 Briefly describe or define and given an example of… An amphiprotic solute is a chemical species that possesses both acidic and basic properties. The (a) an amphiprotic solute dihydrogen phosphate ion, H2 PO 4 – , is an example of an amphiprotic solute. A differentiating solvent reveals different strengths of acids. By this definition, anhydrous acetic acid is a differentiating solvent because perchloric acid dissociates 5000 times more than hydrochloric (b) a differentiating solvent acid. (c) a leveling solvent A leveling solvent shows no difference between strong acids. Perchloric acid and hydrochloric acid ionize completely in water; thus, water is a leveling solvent. A mass-action effect arises when a shift in the chemical equilibrium occurs due to the introduction of (d) a mass-action ion effect one of the participating chemical species (i.e., addition of one of the reactants or products. EoC 9-3 9-3 Briefly explain why there is no term in an equilibrium constant expression for water or for a pure solid, even though one (or both) appears in the balanced net ionic equation for the equilibrium. For dilute aqueous solutions, the concentration of water remains constant and is assumed to be independent of the equilibrium. Thus, its concentration is included within the equilibrium constant. For a pure solid, the concentration of the chemical species in the solid phase is constant. As long as some solid exists as a second phase, its effect on the equilibrium is constant and is included within the equilibrium constant. EoC 9-4 9-4 Identify the acid on the left and its conjugate base on the right in the following equations: (a) HOCl + H2 O ⇌ H3 O + + OCl(b) HONH2 + H2 O ⇌ HONH3 + + (c)OH NH- 4 + + H2 O ⇌ NH3 + H3 O + HOCl A HONH2 B NH4 + A H2 O B H2 O A H2 O B NH3 B H3 O + A HONH3 + A OCl B OH- B H3 O + A (d) 2 HCO 3 - ⇌ H2 PO 4 - + CO 3 2- (e) PO 4 3- + H2 PO 4 - ⇌ 2 HPO 4 22 HCO 3 - B and A PO 4 3- B H2 PO 4 A H2 PO 4 - A 2CO 3 B 2HPO 4 2- B and A EoC9 P5 9-5 Identify the base on the left and its conjugate acid on the right from equations the equations for problem 9-4 (a) HOCl + H2 O ⇌ H3 O + + OCl(b) HONH2 + H2 O ⇌ HONH3 + + (c)OH NH- 4 + + H2 O ⇌ NH3 + H3 O + HOCl BL acid HONH2 BL base NH4 + BL acid H2 O BL base H2 O BL acid H2 O BL base NH3 conj. Base H3 O + conj. Acid HONH3 + conj. Acid OCl conj. Base OH- conj. Base H3 O + conj. Acid (d) 2 HCO 3 - ⇌ H2 PO 4 - + CO 3 2- (e) PO 4 3- + H2 PO 4 - ⇌ 2 HPO 4 22 HCO 3 - amphiprotic species PO 4 3- BL base H2 PO 4 conj. Acid H2 PO 4 - BL acid 2conj. Base CO 3 2HPO 4 2- amphiprotic species, both conj. acid and conj. base EoC9 P6 9-6 Write the expressions for the autoprotonolysis of (a) H2 O 2 H2 O ⇌ H3 O + + OH(b) CH3 COOH 2 CH3 COOH ⇌ CH 3 COOH 2 + (c) CHNH2 2 CHNH2 ⇌ CHNH3 + + (d) CH3 OH 2 CH3 OH ⇌ CH3 OH2 + + EoC9 P7 9-7 Write the equilibrium-constant expressions and obtain numerical values for each constant in (a) the basic dissociation of aniline, C 6 H4 NH2 . K eq = [C 6 H4 NH3 +][OH - ] = [C 6 H4 NH2 ] (b) the acidic dissociation of hypochlorous acid, HClO. K eq = [ClO - ][H3 O +] [HClO] = (c) the acidic dissociaton of methyl ammonium hydrochloride, CH3 NH3 Cl. (d) the basic dissociation of NaNO3 . EoC9 P8 K eq = [CH3 NH2 - ][H3 O +] = [CH3 NH3 +] K eq = [HNO 3 ][OH - ] [NO 3 - ] = (e) the dissociation of H3 AsO 3 to H3 O + and AsO3 3- . K eq = [H3 O +] 3 [AsO 3 3- ] [H3 AsO 3 ] = (f) the reaction C2 O 4 2- with H2 O to give H2 CO 4 and OH K eq = [H2 CO 4 ][OH - ] 2 [C 2 O 4 2- ] = 9-8 Generate the solubility-product expression for (A) CuBr (B) HgClI (C) PbCl 2 (D) La(IO 3 )3 (E) Ag3 AsO 4 3+ + 3CuBr (s ) ⇌ Cu- + Br- HgClI (s ) ⇌ Hg2+ + ClPbCl + I-2 (s ) ⇌ Pb 2+ + 2Cl La(IO + Ag 3IO 3 )3 (s ) ⇌ La 3 AsO 3 4 (s ) ⇌ Ag + AsO4 + 2+ 2+ - 2 3+ - 3 + 3 3K sp = [Cu ][Br ] K sp = [Hg ][Cl ][I ] K sp = [Pb ][Cl ] K sp = [La ][IO 3 ] K sp = [Ag ] [AsO 4 ] EoC9 P9 9-9 Express the solubility-product constant for each substance in Problem 9-8 in terms of its molar solubility S. (A) CuBr (B) HgClI (C) PbCl 2 (D) La(IO 3 )3 (E) Ag3 AsO 4 S = [Cu+]=[Br - ] S = [Hg2+]=[Cl - ]=[I - ]S = [Pb 2+]=½[Cl - ] S = [La 3+]=⅓[IO 3 - ] S = [AsO 4 3- ]=⅓[Ag +] K sp = S × S = S 2 K sp = S × S × S = S 3 K sp = S × (2S )2 = 4S 3 K sp = S × (3S )3 = 27S K4 sp = 3S × S = 3S 2 EoC9 P10 9-10 Calculate the solubility-product constant for each of the following substances, given that the molar concentrations of their saturated solutions as indicated (A) CuBr (B) HgClI (C) PbCl 2 (D) La(IO 3 )3 (E) Ag3 AsO 4 2+ 22+ 3+ AgSeCN ⇌ Ag+ + SeCN RaSO ⇌ Ra + SO Pb(BrO ) ⇌ Pb + 2BrO Ce(IO ) ⇌ Ce + 3IO 4 4 3 2 3 3 3 3 + [Ag ] = [SeCN ] EoC9 P11 9-11 Calculate the solubility of the solutes in Problem 9-10 for solutions in which the cation concentration is 0.030 M. (A) CuBr (B) HgClI (C) PbCl 2 (D) La(IO 3 )3 (E) Ag3 AsO 4 2+ 22+ 3+ AgSeCN ⇌ Ag+ + SeCN RaSO + SO4Pb(BrO + 2BrO Ce(IO + 3IO 3 4 ⇌ Ra 3 )2 ⇌ Pb 3 3 )3 ⇌ Ce [Ag+] = [SeCN - ] EoC9 P12 9-12 Calculate the solubility of the solutes in Problem 9-10 for solutions in which the anion concentration is 0.030 M. (A) (B) (C) (D) 2+ 22+ 3+ AgSeCN ⇌ Ag+ + SeCN RaSO + SO4Pb(BrO + 2BrO Ce(IO + 3IO 3 4 ⇌ Ra 3 )2 ⇌ Pb 3 3 )3 ⇌ Ce + [Ag ] = [SeCN ] EoC9 P13 9-13 What CrO 4 2- concentration is required to… Ag2 CrO 4 (s ) ⇌ 2Ag+ + CrO 4 2K sp = 1.20E-12 [CrO 4 2- ] = K sp / [Ag+] 2 (a) initiate a precipitate of Ag2 CrO 4 from a solution that is [Ag+] = 0.00413 [CrO 4 2- ] =7.04E-08 ###### EoC9 P14 (b) lower the concentration of Ag+ in a solution to [Ag+] = 9E-07 [CrO 4 2- ] =1.48E+00 ###### 9-14 What hydroxide concentration is required to (a) initiate (a) lower EoC9 P15 9-15 The solubility-product constant for Ce(IO 3 )3 is 3.2×10-10 . What is the Ce 3+ concentration in a solution prepared by mixing 50.00 mL of 0.0450 M Ce 3+ with 50.00 mL of… (a) (b) (c) (d) water 0.0450 M 0.250 M 0.0500 M EoC9 P16 9-16 The solubility-product constant for K2 PdCl 6 is 6.0×10-6 (K2 PdCl 6 ⇌ 2 K+ + PdCl 6 2- ). Wjhat is the K+ concentration of a solution prepared by mixing 50.0 mL of a 0.200 M KCl with 50.0 mL of… (a) (b) (c) 0.160 M PdCl 6 2- ? 0.0800 M PdCl 6 2- ? 0.240 M PdCl 6 2- ? EoC9 P17 9-17 The solubility product for a series of iodides are… list these four compounds in decreasing order of molar solubility in… CuI K sp AgI K sp = PbI 2 K sp BiI 3 K sp EoC9 P18 1 × 10-12 8.3 × 107.1 × 10-9 8.1 × 10- (a) water (b) 0.20 M (c) a 0.020 9-18 The solubility products for a series of hydroxides are + Which hydroxide… BiOOH K sp = 4.0 × 10-10 = [BiO ][OH ] -22 (a) has the lowest molar solubility in H 2 O Be(OH) 2 K sp = 7.0 × 10 (b) the lowest molar solubility in a solution that is 0.30 M in NaOH? Tm(OH)3 K sp = 3.0 × 10-24 Hf(OH)4 K sp = 4.0 × 10-26 EoC9 P19 9-19 Calculate the pH of water at 25°C and 75°C. The values for p K w at these temperatures are 13.99 and 12.70, respectively. EoC9 P20 9-20 At 25°C, what are the molar H3 O + and OH- concentrations in… (a) (b) (c) (d) (e) (f) (g) (h) 0.0300 M C6 H5 COOH 0.600 M HN3 0.100 M ethylamine 0.200 M trimethylamine 0.200 M C6 H5 COONa (sodium benzoate) 0.0860 M CH3 CH2COONa 0.250 M hydroxulamine hydrochloride? 0.0250 M ethyl ammonium chloride EoC9 P21 9-21 at 25° C, what is the hydronium ion concentration? (a) (b) (c) (d) (e) (f) 0.200 M chloroacetic acid 0.200 M sodium chloroacetate 0.0200 M methylamine 0.0200 methylamine hydrochloride 2.00 × 10-3 M aniline hydrochloride 0.300 M HIO3 EoC9 P22 9-22 define buffer solution a buffer solution resists change in pH with dilution or with addition of acids or bases. A buffer is composed EoC9 P23 9-23 define buffer capacity Buffer capacity of a solution is defined as the number of moles of a strong acid (or a strong base) that causes 1.00 L of a buffer to undergo a 1.00-unit change in pH EoC9 P24 9-24 Which has the greater buffer capacity? (Kb NH3 = 1.75 × 10-5 ) pH = pKa + log([NaA]/[HA]) (b) NH3 NH4 Cl a mixture containing 0.100 mol NH 4.756962 0.026 mol NH 0.065 3 and 0.200 4 Cl or 4.356962 0.13 0.325 a mixture containing 0.050 mol NH 4.356962 3 and 0.100 mol NH4 Cl EoC9 P25 9-25 Consider solutions prepared by (a), (b), and (c). In what respects do these solutions resemble one another? How do they differ? (a) (b) (c) dissolving 8.00 mmol of NaOAc in 200 mL of 0.100 M HOAc adding 100 mL of 0.0500 M NaOH to 100 mL of 0.175 M HOAc adding 40.0 mL of 0.1200 M HCl to 160.0 mL of 0.0420 M NaOAc a) EoC9 P26 (a) (b) (c) (d) <<< this one has more because of higher concentration of HA (weak acid) and NaA (conjugate base 9-26 Consult Appendix 3, and pick out a suitable acid/base pair to prepare a buffer with a pH of 4.5 8.1 10.3 6.1 look at the K a of each and then use 3.98E-10 9.09E-09 4.33E-04 9.52E-07 EoC9 9-27 What mass of sodium formalate should be added to 500.0 mL of 1.00 M formalic acid to P27 produce a buffer solution with a pH of 3.50? NaF = sodium formalate pH = pKa + log [NaF]/[HF] HF = formic acid HF volume0.500 HF molarity 1.00 moles HF 0.5 HF Ka 0.00018 HF pKa 3.744727 pH = 3.50 pH - pKa =-0.24473 10(pH - pKa) =0.57 moles NaF0.284605 molar mass68.0069 NaF 1.00E-14 mass NF =19.3551 1.70E-01 9.400117 8.041436 3.363512 6.021363 2.51E-05 1.1E-06 2.31E-11 1.05E-08 4.599883 5.958564 10.63649 7.978637 anilium hydroxyl ethyl hydrazine 5.88E-14 EoC P9 28 9-28 What mass of sodium glycolate should be added to 400.0 mL of 1.00 M glycolic acid to produce a buffer solution with a pH of 4.00? pH = pKa + log [NaG]/[HG] NaG = sodium glycolate HG = glycolic acid HG volume0.400 HG molarity 1.00 moles HG 0.4 HG Ka 0.000147 HG pKa 3.832683 pH = 4.00 pH - pKa =0.167317 10(pH - pKa) =1.47 moles NaG0.588 molar mass98.0327 NaG mass NG =57.64323 EoC P9 29 0.500 1.10 0.55 0.00018 3.744727 3.40 -0.34473 0.45 0.248677 69.0148 17.16238 9-29 What volume of 0.200 M HCl must be added to 500.0 mL of 0.300 M sodium mandelate to produce a buffer solution with a pH of 3.37? HCl molarity 0.2 vol NaA 0.5 NaA molarity 0.3 moles NaA 0.15 [HA (conjugate 0.0004 of NaA)] Ka [HA]'s pKa 3.39794 pH 3.37 pH - pKa =-0.02794 10(pH-pKa) 0.937692 moles HCl 0.077412 volume HCl387.0585 = EoC P9 30 9-30 What volume of 2.00 M NaOH must be added to 150.0 mL of 1.40 M glycolic acid to produce a buffer solution having a pH of 3.80? NaOH molarity 2 vol HA 0.2 HA molarity 1 moles HA 0.2 Ka 0.000147 pKa 3.832683 pH = 4 pH - pKa =0.167317 [A-]/[HA] = 1.47 moles NaOH 0.119028 volume NaOH 59.51417 = EoC P9 31 (This problem requires values in your textbook's specific appendic es, which you can access through (This problem requires values in your textbook's specific appendic es, which you can access through 9-31 Is the following statement true or false, or both? Define your answer with equations, examples, or graphs. "a buffer maintains the pH of a solution constant."