Ch 15: Acids and Bases

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1 Ch 15: Acids and Bases A c i d s a n d B a s e s C h 1 5 P a g e 1 Homework: Read Chapter 15 Work out sample/practice exercises in the sections, Bonus problems: 39, 41, 49, 63, 67, 83, 91, 95, 99, 107, 111, 115, 117, 123, 139 Check for the MasteringChemistry.com assignment and complete before due date Acids, Bases, Salts: Acids, bases and salts are very important and perform many essential functions. Digestive juices (0.1 M HCl; needed to kill bacteria, break down food and activate enzymes) ph buffers; Households and Industrial uses Drain cleaner (NaOH) fertilizer (NH 4 NO 3 ) Car battery acid (40% H 2 SO 4 ) Table salt as a food preservative or for flavor (NaCl) Example 1: Come up with more common acids, bases, or salts and their uses.

2 A c i d s a n d B a s e s C h 1 5 P a g e 2 A few common acids, their uses and relative strength: A few common bases, their uses and relative strength:

3 A c i d s a n d B a s e s C h 1 5 P a g e 3 Review Electrolytes and Double Displacement Reactions: Electrolytes: Nonelectrolyte: A molecule or substance that remains whole in aqueous solutions, it cannot split apart. Examples: any insoluble solid, gas (CO 2, O 2, SO 2 ), and molecules (sugar, CH 4, H 2 O) Weak Electrolyte: An ionic substance that will partially ionize into its separate ions in aqueous solution. Examples: Weak acids (HF, HC 2 H 3 O 2 ) and Weak bases (NH 4 OH, CH 3 NH 2 ), and slightly soluble solids (PbCl 2 ). Partial ionization is an equilibrium reaction in which the reactant is favored, K<1 ; HF (aq) H +1 (aq) + F -1 (aq). Strong Electrolyte: An ionic substance that completely dissociates into its ions in aqueous solution. Examples: Strong Acids (HCl), Strong Bases (NaOH), Soluble Salts (KBr); K>>1; HCl (aq) H +1 (aq) +Cl -1 (aq). Titration: Reacting a solution of unknown concentration with a known (standard) concentration, stopping the titration when an indicator (phenolphthalein) changes color. Example 2: Write the dissociation reactions in the solvent water for the following substances. Strong electrolytes will have ( ), weak electrolytes have ( ), and nonelectrolytes are no reaction (NR). FeCl 3 (aq) HNO 3 (aq) Sr(OH) 2 (aq) CH 3 OH (aq) HF(aq)

4 A c i d s a n d B a s e s C h 1 5 P a g e 4 Double Displacement Reactions: Double Displacement reactions have two ionic reactants. Reactants will exchange ions in making products AB + CD AD + CB A reaction occurs if a nonelectrolyte (Solid, Liquid (H 2 O), Gas), or Weak Electrolyte is formed as one or more of the products. If all the reactants and the products are strong electrolytes, then no reaction takes place. Review the Solubility Rules. a) Whole or Molecular equation 2 AgNO 3 (aq) + CaCl 2 (aq) Ca(NO 3 ) 2 (aq) + 2 AgCl (s) b) Total Ionic Equation with Spectator Ions 2 Ag +1 (aq) + 2 NO 3-1 (aq) + Ca +2 (aq) +2 Cl -1 (aq) c) Net Ionic Reaction Ag +1 (aq) + Cl -1 (aq) AgCl (s) Ca +2 (aq) + 2 NO 3-1 (aq) + 2 AgCl (s) Example 3: For the Double Displacement reactions, write the whole, total ionic (circling the spectator ions) and net ionic equations given the reactants. Identify the type of reaction as Precipitation, Neutralization, or No Reaction. a) HCl (aq) + NaOH (aq) b) NH 4 OH, same as NH 3 (aq) + H 2 S (aq) c) Al(NO 3 ) 3 (aq) + KCl (aq) d) K 2 S (aq) + Zn(ClO 3 ) 2 (aq)

5 A c i d s a n d B a s e s C h 1 5 P a g e 5 Review Acids, Bases and Salts : Acids: Properties: Taste Sour Reacts with active metals to liberate H 2 : Zn (s) + 2 HCl(aq) H 2 (g)+ ZnCl 2 (aq) Reacts with carbonates to liberate CO 2 : CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + H 2 O(l) + CO 2 (g) React with bases to form ionic salts (neutralize): NaOH (aq) + HCl (aq) NaCl (aq) + H 2 O (l) Conduct electrical current Certain dyes change color with acids (litmus-red) Acids ionize in water to increase the H +1 ion concentration Nomenclature: a) Binary acids, those anions ending in ide: Hydro root ic acid; (HCl), Hydrochloric Acid; (H 2 S), hydrosulfuric acid b) Ternary oxyacids: if anion ends with ate, root ic acid; (HNO 3 ), nitric acid, if anion ends with ite, root ous acid, (HNO 2 ), nitrous acid Bases (also known as alkalis): Properties: Taste bitter Slimy to the touch Conducts electricity Certain dyes change color with bases (litmus-blue) React with acids to form ionic salts (neutralize): Ionize in water to increase the OH -1 ion concentration Nomenclature: metal name + hydroxide; (NaOH), Sodium Hydroxide; or common name (NH 3 ), ammonia, organic bases with N are amines, (CH 3 NH 2 ), methyl amine Salts: The ionic substances not readily identified as an acid or base, some are very soluble, others only slightly soluble in water. Nomenclature: cation name + anion name; (CuCl 2 ) copper (II) chloride, (NH 4 ) 2 SO 4, ammonium sulfate, (NaCl), sodium chloride.

6 A c i d s a n d B a s e s C h 1 5 P a g e 6 Example 4: Fill in the Table with names or formulas. Name Formula Hydrosulfuric acid Ethyl amine Potassium fluoride MgHCO 3 H 2 C 2 O 4 Ba(OH) 2 Acids and Bases Defined : Arrhenius Definition (1884): Most limited definition requires water. Acid: Substance that will increase the H +1 ion concentration in an aqueous solution. (HF) Base: Substance that will increase the OH -1 ion concentration in an aqueous solution. (KOH) Neutralization is the combination of an acid with a base to form water and a salt. Bronsted Lowry Definitions (1923): broader definition, more base possibilities Acid: Donates one H +1 ion. (HA), NH 4 +1 Base: Accepts one H +1 ion. (A -1 ), NH 3 Conjugate Acid/Base Pairs: These are different by only a single H +1. The acid has one more H +1 compared to the base in a conjugate pair. NH 4 +1 is the conjugate acid for NH 3 the conjugate base. HF is the conjugate acid for F -1 the conjugate base. Amphiprotic Substance: One that can both accept and donate H +1 and can be either acid or base dependent on the environment. H 2 O can accept H +1 and become H 3 O +1 or donate H +1 and become OH -1. HCO 3-1 is also amphiprotic.

7 A c i d s a n d B a s e s C h 1 5 P a g e 7 Amphoteric Substance is another term used. Although all amphiprotic species must be amphoteric, not all amphoteric substances are amphiprotic. Amphoteric substances are able to react with an acid or a base. For example, the amphoteric metal oxide, ZnO, contains no hydrogen and cannot donate a proton. ZnO acts as a Lewis acid or Lewis base which accepts or donates electron pairs. Amphoteric ZnO reacts with both acids and with bases: In acid: ZnO + 2H + Zn 2+ + H 2 O In base: ZnO + H 2 O + 2OH - [Zn(OH) 4 ] 2- Example 5: a) Write the formulas for the conjugate bases given the acids: +1 Acid NH 4 HF HNO 2 H 2 SO 3 H 2 O Conjugate base b) Write the formulas for the conjugate acids given the bases: -1 Base C 2 H 3 O 2 (CH 3 ) 2 NH ClO -1-2 SO 3 H 2 O Conjugate acid c) Which of the species in (a) and (b) above are amphiprotic?

8 A c i d s a n d B a s e s C h 1 5 P a g e 8 Lewis Definitions: This definition does not require water or aqueous reactions. Acid: Accepts a share of a nonbonding electron pair. (BH 3 ) Base: Donates a share of a nonbonding electron pair. (NH 3 ) Generally results in a covalent bond forming, the product is called an adduct Example 6: Identify the Lewis acid and base for the reactants below. a) (CH 3 ) 3 N + BF 3 (CH 3 ) 3 NBF 3 b) FeBr 3 + Br -1 FeBr 4-1 c) Zn NH 3 Zn(NH 3 ) 4 +2 d) SO 2 + H 2 O H 2 SO 3

9 A c i d s a n d B a s e s C h 1 5 P a g e 9 Structure of Acids: Binary acids have the hydrogen attached to a nonmetal atom: HF Oxy acids or ternary acids have the hydrogen attached to an oxygen atom: Carboxylic acids have COOH group the hydrogen attached to the COO group is acidic: *ADD THE LONE PAIR ELECTRONS TO THE PICTURES OF SULFURIC ACID, NITRIC ACID AND CARBOXYLIC ACID GROUP. Structure of Bases: Most ionic bases contain OH -1 ions Some contain CO 3-2 ions Molecular bases contain structures that would like to add an H +1 ion, mostly NH 3 and amine groups. Relative strengths: A stronger acid will have a weaker conjugate base and vice versa. Strong acids have a negligible conjugate base. Stronger bases have weaker conjugate acids. Reactions always favor having a greater amount of the weaker acid and base at equilibrium.

10 A c i d s a n d B a s e s C h 1 5 P a g e 10 Example 7: a) Which acid (HI or HF) has the weaker conjugate base? b) Which base (C 2 H 3 O 2-1 or OH -1 ) has a weaker conjugate acid?

11 A c i d s a n d B a s e s C h 1 5 P a g e 11 Autoionization of Water: Pure water will ionize just slightly; it is generally considered a nonelectrolyte since the amount it ionizes is so small. K w = 1.0 x at 25 C for the reaction H 2 O (l) + H 2 O (l) H 3 O +1 (aq) + OH -1 (aq) The hydronium ion (H 3 O +1 ) is often written as a proton in water, H +1 (aq), even though the H +1 is so reactive it cannot exist alone in water. H +1 is chemically bonded to one or more water molecules in an aqueous solution connected by hydrogen bonding. H(H 2 O) n +1 H 2 O (l) + H 2 O (l) H 3 O +1 (aq) + OH -1 (aq) or H 2 O (l) H +1 (aq) + OH -1 (aq); K w = [H +1 ][OH -1 ] Temperature: As the temperature changes, so will K w 0 C K w = 1.1 x C K w = 1.0 x C K w = 2.5 x C K w = 9.6 x ph Scale: ph is a value that helps in determining the acidity of a solution. ph can be determined using ph meter, ph paper, or indicators (intense colored organic dyes that change color at different ph values). ph = -log[h +1 ] [H +1 ] = 10 -ph K w = 1.0 x = [H +1 ][OH -1 ] at 25 C poh = -log[oh -1 ] [OH -1 ] = 10 -poh pk w = -log[k w ] pk w = ph + poh = 14 at 25 C Acid has a ph < 7 Neutral has ph = 7 Base has a ph > 7 px = -log[x], pk a = -logk a, pk b = -logk b ; K a = 10 -Ka, K b = 10 -Kb

12 A c i d s a n d B a s e s C h 1 5 P a g e 12 Significant Figures and Logs: When you take the log of a number written in scientific notation, the digit(s) before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number log(2.0 x 10 6 ) = log(10 6 ) + log(2.0) = = Because the part of the scientific notation number that determines the significant figures is the decimal part, the sig figs are the digits after the decimal point. Example 8: Given that the ph of a solution is 4.88, solve for the [H +1 ], [OH -1 ], and poh using appropriate significant digits. Strong Acids and Bases: Strong Acids: HCl, HBr, HI, HNO 3, HClO 4, HClO 3, H 2 SO 4 Strong Soluble Bases: LiOH, NaOH, KOH, RbOH, CsOH, FrOH, Ca(OH) 2, Sr(OH) 2, Ba(OH) 2, Ra(OH) 2 Strong Acids and Strong Bases dissociate into ions nearly completely, so the dissociation reaction equilibrium constant K is very, very large.

13 A c i d s a n d B a s e s C h 1 5 P a g e 13 Example 9: Calculate the concentration of hydrogen and hydroxide ions, and the ph and poh. a) M HCl; [H +1 ] = [Acid] for monoprotic strong acids. b) M Ba(OH) 2 ; [OH -1 ] = (number OH -1 )x[base] for strong soluble bases Equilibrium Involving Weak Acids: Weak acids are much more common and numerous compared to strong acids. Weak acids (WA or HA) only will partially ionize. We use the equilibrium constant, K a, in which K a <1, the smaller the K a, the weaker the acid. The book appendix has K a values. The weak acid equilibrium reaction is generally of the form HA H +1 + A -1 or HA + H 2 O H 3 O +1 + A -1 Approximation in Calculations: When using RICE you may avoid the quadratic equation if x is added or subtracted from a relatively large number compared to the value of K a. The addition or subtraction of the x can be considered negligible when K a is 1000 times smaller than the concentration. Approximation values can be used if K a is between 100 to 1000 times smaller than the concentrations.

14 A c i d s a n d B a s e s C h 1 5 P a g e 14

15 A c i d s a n d B a s e s C h 1 5 P a g e 15 Example 10: In M solution, a weak monoprotic acid (HA) is 5.00% ionized. Calculate using the RICE equation the equilibrium concentrations, ph and K a. Example 11: The ph of a M solution of a monoprotic acid (HA) is Using RICE, calculate K a. Example 12: Calculate the equilibrium concentrations and ph for the following solutions. (a) M HC 2 H 3 O 2 (b) M HCN

16 A c i d s a n d B a s e s C h 1 5 P a g e 16 Equilibrium Involving Weak Bases: Weak soluble bases are generally going to be amines, ammonia, and conjugate base ions of acids. Weak bases (B or A -1 ) need to have water as an additional reactant to provide the H +1 and only will partially ionize. We use the equilibrium constant, K b, in which K b <1, the smaller the K b, the weaker the base. Appendix D has K values The weak base equilibrium reaction is generally of the form B + H 2 O BH +1 + OH -1 Or A -1 + H 2 O HA + OH -1 every anion can potentially act as a base and accept an H +1. Example 13: Calculate the % ionization, poh, ph and the equilibrium concentrations in 0.15 M NH 3 (aq). K b = 1.8 x 10-5

17 A c i d s a n d B a s e s C h 1 5 P a g e 17 Example 14: The ph is for a NH 3 (aq) solution. Calculate the Molarity of NH 3. Polyprotic Acid Equilibria: Ionization of polyprotic acids occur stepwise. K a is different for each step, decreasing as each H +1 is lost (K a1 > K a2 > K a3 ). Generally, the difference in K a values is enough so that the second ionization does not happen to a large extent except for extremely dilute solutions, the [H +1 ] can be assumed to come from the first step alone. [A -2 ] = K a2 as long as the second ionization is negligible.

18 A c i d s a n d B a s e s C h 1 5 P a g e 18 Example 15: Calculate the concentrations of all species in 0.25 M H 2 SO 4 solution. Given: K a1 >>1 (assume complete ionization), K a2 = 1.2 x 10-2 Example 16: Calculate the concentrations of all species in 0.40 M H 3 AsO 4 solution. Given: K a1 = 5.5 x 10-3, K a2 = 1.7 x 10-7, K a1 = 5.1 x 10-12

19 A c i d s a n d B a s e s C h 1 5 P a g e 19 Relating Ka and Kb for Conjugate Acid/Base Pairs: K a HF H +1 + F -1 K a = 3.5 x 10-4 K b F -1 + H 2 O HF + OH -1 K b =? K w H 2 O H +1 + OH -1 K w = 1.0 x K w = K a x K b Example 17: Demonstrate how K w = K a x K b and calculate the K b value for F -1 Salt Solutions (Acid-Base Properties): Salts are ionic compounds with cation and anion that may be conjugates of a base and acid. The reaction Acid + Base Water + Soluble Salt has several General Categories forming soluble salts: SA + SB soluble neutral salt + water WA + SB soluble basic salt + water SA + WB soluble acidic salt + water WA + WB salt of unknown acidity + water The reaction Acid + Base Water + Salt Can be reversed to create the reaction for the hydrolysis of a salt Salt + Water Acid + Base Writing the whole, total ionic equation and net ionic equations: Balance all atoms and charges SA, SB and SS (Soluble salts) are written as charged ions WA, WB, gases, solids, liquids are written whole Include phases

20 A c i d s a n d B a s e s C h 1 5 P a g e 20 Example 18: Write the whole, total ionic, and net ionic reactions for the hydrolysis of the following salts. Identify spectator ions. Predict acidity of the salt, (Is it neutral, acidic or basic) a) CaBr 2 b) NaNO 2 c) NH 4 NO 3 d) NH 4 C 2 H 3 O 2 e) CH 3 NH 2 F Example 19: Calculate [OH -1 ], ph, % hydrolysis for 0.10 M NaClO solution found in Clorox bleach. K a of HClO = 3.5 x 10-8

21 A c i d s a n d B a s e s C h 1 5 P a g e 21 Example 20: Write the whole, total and net ionic hydrolysis reactions for NH 4 Br. Calculate [H +1 ], ph, % hydrolysis for 0.20 M NH 4 Br solution. Look up the appropriate equilibrium constant you will require in the calculations. Hydrated Metal Cations can act as Weak Acids: Alkali metal and alkaline earth metal cations are ph neutral, negligible counter-ions of strong bases. They do not act as weak acids. Small highly charged metals can coordinate with water and release H +1 ion from water to reduce the charge. Cu(H 2 O) 6 +2 (aq) H +1 (aq) + Cu(H 2 O) 5 (OH) +1 (aq) K a = 3 x 10-8 Smaller metals with higher positive charges more acidic Acid Strength: The stronger an acid is at donating H +1, the weaker the conjugate base is at accepting H +1 Cation makes a stronger acid than neutral molecule which is more acidic than anion H 3 O +1 > H 2 O > OH -1 or NH + 4 > NH 3 > NH 2 Larger K a = stronger acid

22 A c i d s a n d B a s e s C h 1 5 P a g e 22 Chemical Structure Influences the Acid/Base Strength: BinaryAcids: Bond Strength (Greatest factor in same vertical column), weaker bond (larger anion) more acidic Bond strengths: HF << HCl < HBr < HI Electronegativity difference: (factor when comparing anions in the same period, horizontal row) greater difference more acidic. H 2 S is a weaker acid than HCl F -1 ion (special case) causes an increased ordering in water molecules creating the unfavorable lowering of entropy. This helps to explain why HF is a weak acid. Ternary Oxyacids: Electronegativity of Center Atom, higher electronegativity more acidic H 2 SO 4 > H 2 SeO 4 > H 2 TeO 4 Oxidation State of Center, Larger oxidation number (more oxygens) more acidic. HClO 4 > HClO 3 > HClO 2 > HClO Carboxylic Acids: The ability for the conjugate base to have resonance structures will stabilize the base and it is more likely to have the hydrogen ion lost. R-COOH Polyprotic Acids: The fewer H +1 a species has, the weaker the acid becomes. H 3 PO 4 is a stronger acid than H 2 PO 4-1 and both are stronger than the HPO 4-2 acid since each successive K a gets much smaller. Example 21: Explain the following observations: a) H 3 PO 4 is a stronger acid than H 3 AsO 4 b) H 2 SO 3 is a stronger acid than HSO 3-1 Acid Rain: Over 85% of U.S. fuel is from fossil fuels producing CO 2, SO 2, and NO 2 which are linked to acid rain and damages to ecosystems and structures. Natural processes as volcanoes also add to it. Nonmetal oxides and water create acids.

23 A c i d s a n d B a s e s C h 1 5 P a g e 23 More Practice: 1. Give the conjugate base of the following Bronsted-Lowry acids +1-1 a) NH 4 b) H 2 PO 4 c) HC 7 H 5 O 2 2. Give the conjugate acid of the following Bronsted-Lowry bases a) CN -1 b) H 2 PO 4-1 c) C 2 H 5 NH 2 3. Designate the Bronsted-Lowry acid and base on the reactant side of each equation and the conjugate acid and base on the product side: (all aqueous) +1 a) NH 4 + CN -1 HCN + NH 3 b) (CH 3 ) 3 N + H 2 O (l) (CH 3 ) 3 NH +1 + OH -1 c) HCHO 2 + PO -3 4 CHO HPO 4 4. The hydrogen oxalate ion, HC 2 O 4-1, is amphiprotic. Write the balanced chemical equation showing how it acts as an acid and how it acts as a base in water. 5. Which of the following is the stronger acid, HBrO or HBr? 6. Which is the stronger base, F -1 or Cl -1? 7. Calculate [H +1 ], ph and poh for the following and determine if acidic or basic a) [OH - ] = M b) 2.5 x 10-2 M HCl c) solution where 100x[H + ] =[OH - ] 8. By what factor does [H + ] change for a ph change of a) 2.00 units, b) 0.50 units? 9. Predict the products of the following acid-base reactions, and also predict whether the reactants or the products are preferred when at equilibrium. a) NH 3 (aq) + HBr (aq) b) HCO 3-1 (aq) + F -1 (aq) c) H 2 O (l) + NO 3-1 (aq) 10. The average ph of normal arterial blood is 7.40 at body temperature (37 C), at which K w = 2.4 x Calculate [H +1 ], [OH -1 ] and poh for this temperature. 11. A M solution of lactic acid (HC 3 H 5 O 3 ) has a ph of Calculate K a. 12. A M solution of chloroacetic acid is 11.0% ionized. Solve for the equilibrium concentrations of its ions and the K a.

24 A c i d s a n d B a s e s C h 1 5 P a g e Saccharin (HNC 7 H 4 SO 3 ) has a pk a of 2.32 at 25 C. Solve for the ph of a M solution of saccharin. 14. Phosphoric acid is triprotic (H 3 PO 4 ). Calculate the ph and equilibrium concentrations involved in phosphoric acid for a M solution. Given: K a1 =7.5 x 10-3, K a2 = 6.2 x 10-8, K a3 = 4.2 x Calculate the ph of M ethylamine (C 2 H 5 NH 2 ), K b = 6.4 x Ephedrine (C 10 H 15 ON) is used in nasal sprays as a decongestant. A M solution has a ph of Solve for the equilibrium concentrations and K b. 17. Predict wheter the following aqueous salts will be acidic, basic, or neutral a) NH 4 Br b) NaC 2 H 3 O 2 c) KClO An unknown salt is either NaOCl or NaF. When moles of the salt is dissolved in L the ph of the solution is What is the salt?

25 A c i d s a n d B a s e s C h 1 5 P a g e 25 Answers: 1) a) NH 3 b) HPO c) C 7 H 5 O 2 2) +1 a) HCN b) H 3 PO 4 c) C 2 H 5 NH 3 3) +1 a) NH 4 + CN -1 HCN + NH 3 acid base acid base b) (CH 3 ) 3 N + H 2 O (l) (CH 3 ) 3 NH +1 + OH -1 base acid acid base -3 c) HCHO 2 + PO 4-1 CHO2-2 + HPO 4 acid base base acid 4) Acid: HC 2 O H 2 O C 2 O H 3 O +1 Base: HC 2 O H 2 O H 2 C 2 O 4 + OH -1 5) HBr 6) F -1 7) a) [H +1 ]= 2.5 x 10-11, ph = 10.60, poh = 3.40 b) [H +1 ]= 2.5 x 10-2, ph = 1.60, poh = c) [H +1 ]= 1.0 x 10-8, ph = 8.00, poh = ) a) 100x b) 3.16x 9) a) Products: NH 3 (aq) + HBr (aq) NH +1 4 (aq) + Br -1 (aq) b) Reactants: HCO -1 3 (aq) + F -1 (aq) CO -2 3 (aq) + HF (aq) c) Reactants: H 2 O (l) + NO -1 3 (aq) OH -1 (aq) + HNO 3 (aq) 10) [H +1 ]= 4.0 x 10-8, [OH - ] = 6.0 x 10-7, poh = ) K a = 1.4 x ) [H +1 ] = [ClCH 2 COO -1 ] = M, [ClCH 2 COOH] = M, K a = 1.4 x ) [H +1 ]= 2.0 x 10-2, ph = ) [H 3 PO 4 ] = M, [H +1 ]= M, [H 2 PO -1 4 ] = M, [HPO -2 4 ] = 6.2 x 10-8, [PO -3 4 ] = 3x10-20, [OH - ] = 3.7 x 10-13, ph= ) ph = ) [OH -1 ] = [C 10 H 15 ONH +1 ] = M, [C 10 H 15 ON] = 0.033M, K a = 1.4 x ) a) acid, b) base, c) neutral 18) NaF

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