Answer Key, Problem Set 5 (With explanations)--complete

Chemistry 122 Mines, Spring 2016 Answer Key, Problem Set 5 (With explanations)complete 1. NT1; 2. NT2; 3. MP; 4. MP (15.38); 5. MP (15.40); 6. MP (15.42); 7. NT3; 8. NT4; 9. MP; 10. NT5; 11. NT6; 12. MP; 13. NT7; 14. NT8; 15. NT9; 16. NT10; 17. MP; 18. MP; 19. NT11; 20. NT12; 21. NT13; 22. NT14; 23. NT15; 24. NT16; 25. MP; 26. NT17; 27. MP; 28. NT18; 29. MP; 30. NT19; 31. MP; 32. MP BrønstedLowry Definition of Acids and Bases and AcidBase Conjugates 1. NT1 15.36. For each reaction [equation], identify the BrønstedLowry (BL) acid, the BrønstedLowry base, the conjugate acid, and the conjugate base. (a) HI(aq) + H 2 O(l) H 3 O + (aq) + I (aq) (b) CH 3 NH 2 (aq) + H 2 O(l) CH 3 NH 3 + (aq) + OH (aq) (c) CO 3 2 (aq) + H 2 O(l) HCO 3 (aq) + OH (aq) (d) HI(aq) + H 2 O(l) Answers: H 3 O + (aq) + I (aq) BL Acid BL Base Conjugate Base Conjugate Acid (a) HI H2O I H3O + (b) H2O CH3NH2 OH CH3NH3 + (c) H2O CO3 2 OH HCO3 (d) HBr H2O Br H3O + Reasoning: (1) A (BrønstedLowry) acid is a proton donor: a base is a proton acceptor. So you must look at what a given species ends up as (i.e., what it does during the reaction) in order to properly assess whether it is an acid or a base. (2) Conjugates are pairs of species that differ in structure (and thus formula) by exactly one H +. (Look closely at the relationship between the species in the two rectangles and the relationship between the species in the two ovals.) NOTE: The conjugate acidbase pair species are NOT the acid and base that react with one another! They are on opposite sides of the equation (not the same side). A conjugate is formed when an acid or a base acts like an acid or base (i.e., it gives away an H + or accepts one). 2. NT2 15.37 (add (e): NH2 ) Write the formula for the conjugate base of each acid. (a) HCl CB is Cl (d) HF CB is F (b) H 2 SO 3 CB is HSO3 (e) NH 2 CB is NH 2 (c) HCHO 2 CB is CHO2 Reasoning: To determine the formula of the conjugate base of an acid, you must remove exactly one H +, which means to remove one H from the formula, and also subtract +1 from whatever the charge on the acid formula was that is, make the charge one unit more negative. 3. MP No answer in this key. 4. MP (15.38) Write the formula for the conjugate acid of each base. Mastering problem, so answers not shown. However, to determine the formula of the conjugate acid of an base, you must do exactly the opposite of what was done in the prior problem. That is, you must add exactly one H +, which means to add one H from the formula, and also add +1 to whatever the charge on the acid formula was that is, make the charge one unit more positive. 5. MP (15.40) Both HCO 3 and HS are amphoteric. Write an equation to show how each substance [species!!] can act as an acid, and another equation to show how it can act as a base. **NOTE: The question 15.40 in Mastering is actually a bit different than the question asked in the physical text 15.40. The answer below is therefore not the answer to the Mastering question, although the concepts are obviously analogous. So I ve left my answer here in case it might PS51

help, but please make sure to realize these are not the answers to the Mastering problem, as it is worded. Always read every question carefully! Don t assume anything. NOTE: There are many possible answers to this problem because they did not specifically ask you to write out the acid ionization and base ionization reaction equations for these species (even though those equations would represent one set of correct answers to this question). As such, I will provide two correct answers for each, one of which is the acid (or base) ionization equation and one of which is not. Please know that a species can act as an acid by giving its proton away to any species not just water. Acid ionization, however, is the (very specific) acidbase reaction in which an acid gives its proton to H2O. HCO 3 acting as an acid: HCO3 (aq) + OH (aq) CO3 2 (aq) + H2O(l) HCO 3 needs to be a reactant and CO 3 2 must be a product. HCO3 (aq) + H2O(l) OR CO3 2 (aq) + H3O + (aq) HCO 3 acting as a base: HCO3 (aq) + HSO4 (aq) H2CO3(aq) + SO4 2 (aq) HCO 3 needs to be a reactant and H 2 CO 3 must be a product. HCO3 (aq) + H2O(l) OR H2CO3(aq) + OH (aq) HS acting as an acid: HS (aq) + F (aq) S 2 (aq) + HF(aq) HS needs to be a reactant and S 2 must be a product. HS (aq) + H2O(l) OR S 2 (aq) + H3O + (aq) HS acting as an base: HS (aq) + HCN(aq) H2S(aq) + CN (aq) HS needs to be a reactant and H 2 S must be a product. HS (aq) + H2O(l) OR H2S(aq) + OH (aq) Acid Ionization Definition, Ka, and Relating Relative Acid Strength to Ka Values and to Nanoscopic Pictures of Solutions of Acids 6. MP (15.42) Classify each acid as strong or weak. If the acid is weak, write an expression for the acid ionization constant (K a ) No answer in key. However, learn the 6 common strong acids so that a) You won t need to worry about their Ka values (because they re basically 100% ionized), and b) You will know that any acid other than one of these is a weak acid. Separately, know that acid ionization means an acid reacting with water (using the BronstedLowry definition of acid) so that you can generate the acid ionization reaction equation and Ka expression (Law of Mass Action) for any acid of interest. 7. NT3. The following pictures represent aqueous solutions of three acids HA (A = X, Y, or Z); water molecules have been omitted for clarity: NOTE: The black circles represent oxygen atoms. The open circles represent hydrogen atoms (charges are not indicated here). Answer the following, with reasoning: (a) What is the conjugate base of each acid? Answer: X, Y, and Z Reasoning: A CB is a species with exactly one H + removed from the (C)A (b) Arrange the three acids in the order of increasing acid strength. Answer: HX < HZ < HY Reasoning: Ignoring the concentration dependence on % ionization of a weak acid (which is a very minor issue in the scenario in this problem), the stronger the acid, the greater the fraction of dissolved formula units that will exist as separated ions. HX has the smallest fraction of formula units existing as separated ions (1 out of 10), HZ has the next smallest (2 out of 10), and HY has the greatest (6 out of 6). (c) Which acid, if any, is a strong acid? Answer: only HY Reasoning: A strong acid has nearly 100% of its dissolved formula units existing as separated ions. HY has 100% of its formula units existing as separated ions. PS52

(d) Which acid has the smallest value of K a? Answer: HX Reasoning: Ignoring the concentration dependence on % ionization of a weak acid (which is a very minor issue in the scenario in this problem), the smaller the Ka value, the smaller the fraction of dissolved formula units that will exist as separated ions. HX has the smallest fraction of formula units existing as separated ions (see (b) above), so it has the smallest Ka. 8. NT4 15.44. Rank the solutions in order of decreasing [H 3 O + ]: 0.10 M HCl; 0.10 M HF; 0.10 M HClO; 0.10 M HC 6 H 5 O. Answer: 0.10 M HCl >> 0.10 M HF > 0.10 M HClO > 0.10 M HC6H5O Reasoning: For solutions of acids with the same (prepared) concentration (same [HA]0), the percentage of formula units that will ionize will be greater for the stronger the acid (i.e., the greater the Ka). This is because a bigger Ka means the acid ionization reaction lies farther to the right (towards products) at equilibrium, and H3O + is always a product of acid ionization (by definition). In this problem, HCl is the only strong acid, and so it is the strongest (it is essentially 100% ionized) and thus has the greatest [H3O + ]eq. From Table 15.5, the Ka s for HF, HClO, and HC6H5O at 25 C are 3.5 x 10 4, 2.9 x 10 8, and 1.3 x 10 10 respectively. Hence the order given. Note: It is unfortunate that the correct order for this problem happens to be the exact order in which the acids appear in the original question. Somebody wasn t looking very carefully 9. MP. No answer in key. This a visual (nanoscopic pictures) question about solutions of acids. 10. NT5. Use Table 15.5 (and 15.3, which you should memorize) to order the following from the strongest to the weakest acid. HClO 2, H 2 O, HC 7 H 5 O 2 (benzoic acid), HClO 4 Answer: HClO4 (strong) > HClO2 (weak) > HC7H5O2 (weak) > H2O ( reference ) Reasoning: HClO2 and HC7H5O2 are listed in the table with HClO2 having a larger Ka value than HC7H5O2 (1.1 x 10 2 vs 6.5 x 10 5 ). So clearly HClO2 is the stronger acid. ( stronger means better at giving away its H + ; a larger K means more product favored [as long as stoichiometry is identical between the two equations being compared], so in the case of acid ionization, a larger K means a greater tendency to give away its H + ). Both acids are considered weak acids because they are not 100% ionized [K values are not >> 1], but are better acids than water [Ka values are larger than Kw, or 1.0 x 10 14 at 25 C]. Thus, HClO4, one of the six common strong acids, must be stronger than both of these, and H2O, which is the reference to whom everyone is compared (and so it defines the border between weak and negligible), must be weaker than both of these. (Note: All weak acids are stronger than water, by definition, and all negligible acids are weaker than water, by definition.) Base Ionization Definition, Kb 11. NT6 15.86 [15.88 in etext or 3/e] Write equations showing how each weak base ionizes water to form OH. Also write the corresponding expression for K b. NOTE: This question is basically asking you to write out the equation for base ionization of a given base, along with the expression for Kb (i.e., the Law of Mass Action) (a) CO 3 2 : CO3 2 (aq) + H2O(l) HCO3 (aq) + OH (aq) (b) C 6 H 5 NH 2 C6H5NH2 (aq) + H2O(l) C6H5NH3 + (aq) + OH (aq) (c) C 2 H 5 NH 2 : C2H5NH2(aq) + H2O(l) C2H5NH3 + (aq) + OH (aq) PS53

Strategy: Remember that base ionization is a base reacting with water (using the BronstedLowry definition of a base). Namely, just write the equation in which the base accepts (takes) a proton (H + ) from a water molecule. The products will therefore be the conjugate acid of the base, and OH (the conjugate base of water, which acts as the acid in this reaction). Relating Relative Base Strength to Strengths of Conjugate Acids (Applying The stronger the acid, the weaker its conjugate base, etc.) 12. MP. No answer in key. The first few parts of this Item are related to conjugate acidbase pairs, but the last part involves the concept that The stronger the acid, the weaker its conjugate base, so I put the problem here. 13. NT7 15.46. Pick the stronger base from each pair. (a) ClO 4 or ClO 2 ClO2 HClO2 is a weak acid whereas HClO4 is a strong acid. The conjugate base of the weaker acid will be the stronger base. (In this case, since HClO4 is a strong acid, its conjugate, ClO4 is a negligible base, which is obviously weaker than any weak base (like ClO2 ). (b) Cl or H 2 O H2O (c) CN or ClO HCl is a strong acid, so its conjugate (Cl ) is a negligible base. A negligible base, by definition, is a poorer base than H2O. (Water is the reference to which all acids and bases are compared. If you are worse at accepting protons than water is, you are, by definition, a negligible base (in water).) CN From Table 15.5, it is clear that HCN (Ka = 4.9 x 10 10 at 25 C) is a weaker acid than HClO (Ka = 2.9 x 10 8 at 25 C). Thus its conjugate (CN ) must be a stronger base than HClO s conjugate (ClO ). NOTE: I used the conceptual reasoning here that the stronger an acid is, the weaker will be its conjugate base. I describe the same kind of reasoning below (in NT3). The idea is that being really good at giving away an H +, means that what s left behind after the giving (i.e., the conjugate) must not be very good at taking an H +. This is important and solid reasoning. However, one could also answer certain kinds of these questions (i.e., ones in which Ka s are known) by utilizing the relationship: KaKb = Kw. That is, one can either solve directly for each K Kb w and then compare the Kb s (whomever s Kb is higher is the stronger base), or just K a use that equation to show that the larger an acid s Ka, the smaller will be its Kb. This is essentially proof of the qualitative statement made at the beginning of this paragraph. 14. NT8. Use Table 15.5 (and appropriate reasoning!) to order the following from the strongest to weakest base. ClO 2, H 2 O, C 7 H 5 O 2, ClO 4, OH Answer: OH (strong) > C7H5O2 (weak) > ClO2 (weak) > H2O ( reference ) > ClO4 (negligible) Reasoning: (very short): Applying the idea that the stronger the acid, the weaker is its conjugate base will yield the order shown above for the four bases other than H2O based on the ordering of the acids in the prior problem. Then H2O is placed between the weakest weak base (ClO2 ) and the negligible base (ClO4 ). (One must be able to identify ClO4 as negligible [see below]). Reasoning (more complete): From the prior problem, we know that HClO2 and HC7H5O2 are both weak acids, but HClO2 is stronger. That means HClO2 gives away its H + better (i.e., with a greater tendency) than does HC7H5O2, which means that ClO2 (the conjugate base of HClO 2 ) must be poorer at taking an H + than C7H5O2 (the conjugate base of HC 7 H 5 O 2 ). PS54

Since both HClO2 and HC7H5O2 are weak acids, their conjugates must also be weak bases ( weak gives weak strong gives negligible! Strong and negligible are the extremes ; weak is in the middle.) Thus both are stronger bases than water, the reference. ClO4, being the conjugate to a strong acid (HClO4), must be even poorer at accepting an H + than water, so it is a negligible base (in water). (At equilibrium, there is essentially no HClO4 in a solution of HClO4, which means that ClO4 cannot take H + away from water to essentially any extent!). OH does not appear in any table. However, since OH is the conjugate base to H2O, and H2O is, by definition, a weaker acid than any weak acid, its conjugate base must be stronger than the (weak) conjugates of any weak acid. NOTE: I have labeled OH a strong base above for two reasons: 1) Soluble hydroxides are considered strong bases (e.g., NaOH, KOH), and 2) The equilibrium constant for the reaction OH + H2O H2O + OH is effectively 1 (since the products [numerator] are the same as the reactants [denominator]. Technically, it is neither reactant favored nor product favored, but since all weak bases have Kb values less than 1, it is not unreasonable to lump OH in with the strong bases. Autoionization of Water, Kw, and the Definition of Acidic, Basic, and Neutral 15. NT9. 15.13 Write an equation for the autoionization of water and an expression for the ion product constant for water (K w ). what is the value of K w at 25 C? Answer: H2O(l) + H2O(l) H3O + (aq) + OH (aq); Kw [H3O + ][OH ] You should not have to memorize anything other than water autoionization is water reacting with water (using the BronstedLowry definitions of acid and base). In a sense, this reaction is like the acid ionization reaction for water and the base ionization reaction for water. This is where the relationship 1.0 x 10 14 [H3O + ][OH ] (at 25 C only) comes from! 16. NT10. What is the definition of a neutral solution (in an acidbase sense, not electrically )? Relate this to the chemical equation for the autoionization of water (see prior problem). Is a sample of pure water at any temperature neutral, even though the concentration of [H3O + ] will vary with temperature? Explain. Answer: A neutral solution is one in which the concentration of hydronium ion equals the concentration of hydroxide ion: [H3O + ] [OH ] It s relation to the equation for autoionization is this: If all you have is water (i.e., a sample of pure water), then the only reaction that occurs in solution is autoionization. In autoionization, for every mole of H3O + that forms, one mole of OH will form also, because the coefficients are both one. Thus, in a sample of pure water, no matter what the value of Kw is, the concentration of hydronium ion will equal the concentration of hydroxide, and thus the solution is neutral. When the temperature varies, changing the value of Kw, the amount of water that undergoes this reaction (net) will change, thus changing the concentration of hydronium ion at equilibrium, but whether is a greater or lesser amount of reaction that occurs, the concentration of hydronium ion will still equal that of hydroxide ion (i.e., if more autoionization occurs, [H3O + ] will increase, but the [OH ] will increase just as much!). Thus a sample of pure water will always necessarily be neutral. **NOTE: When more autoionization takes place and [H3O + ] increases, the ph will decrease. But that does not mean the solution will be more acidic, because the solution will still be neutral (with [H3O + ] still equalling [OH ])!** 17. MP. No answer in key. NOTES: 1) Please note that the authors of this question treat H + as the equivalent of H3O +. As mentioned in class, I don t care for this approach because it makes it look like the acid molecule is dissociating rather than giving a proton to a water molecule. Physically PS55

speaking, the H + ion does not exist to any appreciable extent in aqueous solution it attaches to a water molecule to create a hydronium ion. 2) I m not a big fan of poh, but I was not about to skip this problem just because it has poh in it. Just remember that poh = log[oh ]. I would not try to learn any special relationships between ph and poh, although if you do use such a relationship correctly, I will not take off points, obviously. Definition of ph, and Interconversions Between ph, [H3O + ], and [OH ] (and poh); Revisit Acidic, Basic, Neutral at T s other than 25 C. 18. MP. No answer in key. This problem is similar to the following problem. 19. NT11. 15.52. Complete the table. (All solutions are at 25 C.) NOTE: Underlined values were given in the problem. (See below for strategy / reasoning.) [H3O + ] (M) [OH ] (M) ph Acidic or Basic? 14 1.0 x 10 2.857.. log(3.5 x 10 (a) 3 ) 2.455.. 3.5 x 10 3 M 3 Acidic 3.5 x 10 2.46 2.9 x 10 12 ([H 3 O + ] > [OH ]) M 14 1.0 x 10 2.63.. log(2.63 x 10 7 (b) 8 ) = 7.579.. 3.8 x 10 3.8 x 10 7 M Basic 7.58 2.6 x 10 8 ([OH ] > [H 3 O + ]) M 14 1.0 x 10 5.55.. log(1.8 x 10 (c) 9 ) 8.744.. 1.8 x 10 9 M 9 Basic 1.8 x 10 8.74 5.6 x 10 6 ([OH ] > [H 3 O + ]) M 14 1.0 x 10 10 (d) 7.15 7.07.. 1.41.. 7.1 x 10 8 8 7.07.. x 10 7.15 Basic M 1.4 x 10 7 ([OH ] > [H 3 O + ]) M Strategy / Reasoning: Utilize the two relationships below in order to get between ph and [H3O + ] and between [H3O + ] and [OH ]. Then apply the definition of acidic ([H3O + ] > [OH ]) or basic ([OH ] > [H3O + ]) to make that conclusion. ph = log[h3o + ] and [H3O + ][OH ] Kw ( 1.0 x 10 14 at 25 C) NOTE: Although not independent of the first equation above, I will include the following (3 rd ) ph relationship for convenience: [H3O ] 10 (which helps you get ph from [H3O + ]. as in part (d) of this problem) 20. NT12. If T = 25 C, what is the ph of a solution having [OH ] = 1.3 x 10 15 M? State the concentration of H 3 O + in this solution explicitly somewhere as well. Answers: [H3O + ] = 7.7 M; ph = 0.89 Reasoning: At 25 C, [H 3 O + ][OH ] = 1.0 x 10 14 14 14 + 1.0 x 10 1.0 x 10 [H3O ] 7.69 M 15 [OH ] 1.3 x 10 (NOTE: The above result is 7.69 M, a concentration of H +. This is NOT a ph value!!) PS56

Apply the definition of ph: ph = + log[h3o ] log(7.69) 0.89 NOTE: If the concentration of H 3 O + is greater than 1 M, the ph will be less than zero! This is no big deal (just quite acidic!), but since many people have learned that the ph scale goes from 0 to 14, they find this troubling or problematic. 21. NT13. 15.53. Like all equilibrium constants, the value of K w depends on temperature. At body temperature (37 C), K w = 2.4 x 10 14. What is the [H 3 O + ] and ph of pure water at body temperature? Is the sample in this problem acidic, basic, or neutral? Answers: [H3O + ] 1.5 x 10 7 M; ph 6.81; sample is neutral Strategy / Work: 1) Recognize that in a sample of pure water, the only reaction occurring (for all practical purposes) is autoionization: H2O + H2O H3O + + OH Kw To get concentrations at equilibrium, allow some water to autoionize (like an ICE problem in which you let x [H3O + ] that forms as equilibrium is established). 2) Because of the stoichiometry, the system at equilibrium must have [H3O + ] [OH ] (If you were to create an ICE table, you d have initial concentrations of H3O + and OH equal to zero, and then the change row would have + x for both species since the stoichiometry is 1 : 1.) Thus, for pure water: x 2 Kw (where x [H3O + ]eq [OH ]eq) 3) In this problem, since Kw 2.4 x 10 14, x 2.4 x 10 14 7 1.549... x 10 [H3O + ] 4) ph log[h3o + ] log(1.55 x 10 7 ) 6.809 6.81 5) Recognize that despite the ph not being 7.0, the solution MUST be neutral (since [H3O + ] = [OH ], right?)! Do not forget that the definition of neutral is [H3O + ] [OH ] and that neutral only ends up being ph of 7.00 if the temperature is 25 C! 22. NT14. A solution has a ph of 7.0 but is basic. Explain clearly what this means and how it can be so! (Hint: Apply the definition of basic and the definition of ph!) ADD: What can you say about the value of [H 3 O + ][OH ] in this solution? And what about the value of K w? Finally, what can you conclude about the T of the solution? Answers: [H3O + ] 1.0 x 10 7 M and [OH ] > [H3O + ] Kw [H3O + ][OH ] > 1.0 x 10 14 T 25C (i.e., this can be so as long as T is not 25C; ph of 7.0 means neutral only at 25C!) Reasoning: The ph is 7.0. That must mean (based on our definition of ph) that [H3O + ] 1.0 x 10 7 M. (It turns out that this definition is only approximately true; it applies for extremely dilute solutions only. But I will not hold you accountable for anything but this definition. Ask me if you want to know more.) The solution is basic. That must mean that [OH ] > [H3O + ], period. Why? Because THAT is the definition of basic! THUS: If [H3O + ] 1.0 x 10 7 M and [OH ] > [H3O + ], [OH ] MUST be greater than 1.0 x 10 7 M! This means that [H3O + ] x [OH ] {= (1.0 x 10 7 ) x (a number bigger than 1.0 x 10 7 )} is NOT equal to 1.0 x 10 14 (it must be greater than that). PS57

If [H3O + ] x [OH ] is bigger than 1.0 x 10 14, then Kw must be bigger than 1.0 x 10 14. Since Kw = 1.0 x 10 14 at 25 C, that must mean the temperature of the given solution is NOT 25C! It turns out that the T must be greater than 25C, because the autoionization of water is an endothermic reaction. From our study of equilibrium properties of systems, if a reaction is endothermic, then a T increase will lead to an increase in the value of K. Remember? From another text*, I found the following data (see right) describing how K w depends on T. Note that K w increases as T increases. *Quantitative Chemical Analysis, 2 nd Ed. (1987); Harris, D. C.; W. H. Freeman and Company. T (C) K w pk w 0 0.114 x 10 14 14.9435 20 0.681 x 10 14 14.167 24 1.00 x 10 14 14.000 25 1.01 x 10 14 13.9965 30 1.47 x 10 14 13.833 40 2.92 x 10 14 13.535 60 9.61 x 10 14 13.017 Definition of Ksp and its Associated Chemical Equation; Relation to Molar Solubility (in Pure Water) 23. NT15; 16.85 Write balanced equations and expressions for K sp for the dissociation of each ionic compound: 16.85 (a) BaSO4(s) Ba 2+ (aq) + SO4 2 (aq); Ksp [Ba 2+ ][SO4 2 ] (b) PbBr2(s) Pb 2+ (aq) + 2 Br (aq); Ksp = [Pb 2+ ][Br ] 2 (c) Ag2CrO4(s) 2 Ag + (aq) + CrO4 2 (aq); Ksp [Ag + ] 2 [CrO4 2 ] Added by me: (d) Sr3(PO4)2 (s) 3 Sr 2+ (aq) + 2 PO4 3 (aq); Ksp [Sr 2+ ] 3 [CO3 2 ] 2 (e) Fe(OH)3 (s) Fe 3+ (aq) + 3 OH (aq); Ksp = [Fe 3+ ][OH ] 3 (f) Ag3PO4 (s) 3 Ag + (aq) + PO4 3 (aq); Ksp [Ag + ] 3 [PO4 3 ] NOTE: I have left off the eq subscript designations from these K expressions merely for convenience. Please be aware that these are equilibrium concentrations nonetheless! 24. NT16. 16.88(b). Use the K sp values in Table 16.2 to calculate the molar solubility of each compound in pure water. (b) Ag 2 CrO 4 Strategy Answer: 6.54 x 10 5 M 1) Recognize that this is a Find the concentrations at equilibrium kind of problem in which you are given (or can figure out) the initial situation as well as K. In such problems, it is useful to create an ICE table and define an x to represent a species change. **One difference here is that you are not given the balanced chemical equation you must create that yourself (as in the prior problem). Be very careful to write the proper charges of the ions made as well as the correct coefficients. 2) In a Ksp context, x is essentially always defined as the moles of solid that dissolve in exactly 1 L of solution and as such, x is the molar solubility of the solid (what is asked for in this problem) 3) Procees as in any ICE / x type of problem. That is, use stoichiometry to figure out expressions for the equilibrium concentrations (i.e., fill out the ICE table correctly), substitute into the K expression, and solve for x. Unlike in other problem types, if the only thing asked for is the molar solubility, then you are essentially done when you find x. However, it is always a good idea to solve for all concentrations and check your answer by plugging back into the K expression to see if you get a value of K close to the given value. Execution of Strategy: (a) Ag2CrO4(s) 2 Ag + (aq) + CrO4 2 (aq); Ksp [Ag + ] 2 [CrO4 2 ] 1.12 x 10 12 PS58

"Ag2CrO4(s)" [Ag + ] (M) [CrO4 2 ] (M) I excess 0 0 C x + 2x + x E leftover 2x x Ksp [Ag + ] 2 [CrO4 2 ] (2x) 2 (x) 4x 3 1.12 x 10 12 3 1.12 x 10 4 12 6.542 x 10 5 The (molar) solubility of Ag2CrO4(s) (in pure water) is 6.54 x 10 5 M Check: [Ag + ]eq = 2(6.54 x 10 5 ) = 1.31 x 10 4 M 2 [CrO 4 ]eq = 6.54 x 10 5 M K sp = (1.31 x 10 4 ) 2 (6.54 x 10 5 ) = 1.118..x 10 12 (v. close) 25. MP. No answer in key. NOTE: This problem asks you to calculate the molar solubility of Fe(OH)3 in pure water at a temperature at which the Ksp is given/known. I assigned this before NT17 because this has an interactive worked example video that I thought would be useful. However, since this problem has no hints, if you are struggling with it, I invite you to look at my Strategy for this kind of problem shown in NT17 below. 26. NT17. 16.90(b,c).Use the given molar solubilities in pure water to calculate K sp for each compound. (b) Ag 2 SO 3 ; molar solubility = 1.55 x 10 5 M (c) Pd(SCN) 2 ; molar solubility = 2.22 x 10 8 M Answers: (b) 1.49 x 10 14 ; (c) 4.38 x 10 23 Strategy: 1) Recognize that writing out the balanced equation for dissolution is needed (so that the Ksp expression can be properly written, and so that the proper stoichiometry can be clearly seen.) 2) Although this could be done later, I would just go ahead and write out the Ksp expression right next to the balanced equation. 3) Recognize that the molar solubility represents the moles of solid that dissolve per liter of solution. This means that the concentration of each dissolved ion at equilibrium can be determined using the balanced equation (i.e., stoichiometry). 4) Plug the equilibrium concentration values into the Ksp expression and calculate Ksp. Execution of Strategy (for Ag2SO3): Ag2SO3(s) 2 Ag + (aq) + SO3 2 (aq); Ksp [Ag + ] 2 [SO3 2 ] If 1.55 x 10 5 moles of solid dissolve in a liter, then 2 x (1.55 x 10 5 ) 3.10 x 10 5 moles of Ag + must form (per liter) [Ag + ]eq = 3.10 x 10 5 M ; [SO3 2 ]eq = 1.55 x 10 5 M (because one SO 3 2 forms per formula unit of Ag 2 SO 3 that dissolves) Thus, Ksp (3.10 x 10 5 ) 2 (1.55 x 10 5 ) 1.489.. 1.49 x 10 14 Execution of Strategy (for Pd(SCN)2): Pd(SCN)2(s) Pd 2+ (aq) + 2 SCN (aq); Ksp = [Pd 2+ ][SCN ] 2 [Pd 2+ ]eq = 2.22 x 10 8 M (because one Pd 2+ forms per FU of Pd(SCN) 2 dissolved) [SCN ]eq = 2(2.22 x 10 8 ) M = 4.44 x 10 8 M (because two SCN ions form per FU of Pd(SCN) 2 dissolved) Thus, Ksp (2.22 x 10 8 )(4.44 x 10 8 ) 2 4.376... 4.38 x 10 23 PS59

27. MP. No answer in key. More practice with molar solubility from Ksp and Ksp from molar solubility kinds of problem. Please note that the solubility product constant is the same thing as Ksp. Molar Solubilities In Solutions with Some of a Product Ion Present 28. NT18. 16.95. Calculate the molar solubility of barium fluoride in each liquid or solution [at 25 C]. [Use K sp from Table 16.2.] (a) pure water (b) 0.10 M Ba(NO 3 ) 2 (c) 0.15 M NaF Strategy: Answers: (a) 1.83 x 10 2 M; (b) 7.5 x 10 3 M; (c) 1.1 x 10 3 M Same as in Problem 13, except that in (b) and (c), the initial concentration of both ions is not zero. Be careful to take this into account properly in the ICE table, and then use the small x approximation to simplify the algebra once the Ksp expression is substituted into and you are solving for x. NOTE: The molar solubility is still x in (b) and (c) (It is not, for example, the equilibrium concentration of [Ba 2+ ] in (b)). Execution of Strategy (a): BaF2(s) Ba 2+ (aq) + 2 F (aq); Ksp [Ba 2+ ][F ] 2 2.45 x 10 5 " BaF2(s)" [Ba 2+ ] (M) [F ] (M) I excess 0 0 C x + x + 2x E leftover x 2x 5 2.45 x 10 Ksp [Ba 2+ ][F ] 2 (x)(2x) 2 4x 3 2.45 x 10 5 3 1.829 x 10 4 The (molar) solubility of BaF2(s) (in pure water) is 1.83 x 10 2 M 2 Check: [Ba 2+ ]eq = 1.83 x 10 2 M [F ]eq = 2(1.83 x 10 2 ) M = 3.66 x 10 2 M Execution of Strategy (b): K sp = (1.83 x 10 2 )(3.66 x 10 2 ) 2 = 2.451..x 10 5 (v. close) BaF2(s) Ba 2+ (aq) + 2 F (aq); Ksp [Ba 2+ ][F ] 2 2.45 x 10 5 " BaF2(s)" [Ba 2+ ] (M) [F ] (M) I excess 0.10 0 C x + x + 2x E leftover 0.10 + x 2x Ksp [Ba 2+ ][F ] 2 (0.10 + x)(2x) 2 small x approx;0.10 x 0.10 (0.10)(2x) 2 2.45 x 10 5 5 5 2 2.45 x 10 2.45 x 10 4x x 0.007826 0.10 4(0.10) 0.0078 Check approx: x 100 7.8% Oops! 0.10 The approximation isn t quite good based on the 5% rule. See below for how to address. It s very rare for the small x approximation not to work for situations such as this (a common ion situation where one of the products is present to begin with). It only occurred here because Ksp was not really all that small. If that happens, one can do one more iteration as follows to get a more accurate value for x: PS510

Use the approximate value for x just determined in place of the small x approximation. That is, use 0.10 + x 0.10 + 0.0078 = 0.1078 (Note: only do this for the 0.10 + x term, and don t round to the proper # of SFs [yet]). Now we have: 5 5 (0.1078)(2x) 2 2.45 x 10 5 2 2.45 x 10 2.45 x 10 4x x 0.007537.. 0.1078 4(0.1078) In this case, I d do the check first and then make the final conclusion: Check: [Ba 2+ ]eq = 0.10 + 0.007537 M = 0.107 M [F ]eq = 2(0.007537) M = 1.507 x 10 2 M The (molar) solubility of BaF2(s) in 0.10 M Ba(NO3)2 is 7.5 x 10 3 M **The molar solubility of BaF2 here is not equal to the final (equilibrium) concentration of Ba 2+ ion because most of that Ba 2+ ion was present before the dissolution reaction occurred. If this confuses you, recall that x is defined to be the moles of solid that dissolve (to reach equilibrium), not the moles of the cation present at equilibrium.** Execution of Strategy (c): BaF2(s) Ba 2+ (aq) + 2 F (aq); Ksp [Ba 2+ ][F ] 2 2.45 x 10 5 " BaF2(s)" [Ba 2+ ] (M) [F ] (M) I excess 0 0.15 C x + x + 2x E leftover x 0.15 + 2x K sp = (0.107)(1.507 x 10 2 ) 2 = 2.43..x 10 5 (v. close) Ksp [Ba 2+ ][F ] 2 (x)(0.15 + 2x) 2 small x approx; 0.15 2x 0.15 (x)(0.15) 2 2.45 x 10 5 x 5 2.45 x 10 2 0.15 x 0.001088 Check approx: 2(0.001088) x 100 1. 5% 0.15 The (molar) solubility of BaF2(s) in 0.15 M NaF is 1.1 x 10 3 M Check: [Ba 2+ ]eq = 1.09 x 10 3 M [F ]eq = 0.15 + 2(1.09 x 10 3 ) M = 0.152 M K sp = (1.09 x 10 3 )(0.152) 2 = 2.52..x 10 5 (close) 29. MP. No answer in key. Another Find the molar solubility from Ksp kind of problem (one in pure water and another in a solution of a common ion ). Will a Precipitate Form? Kinds of Problems 30. NT19. 16.104. Predict whether or not a precipitate will form if you mix 175.0 ml of a 0.0055 M KCl solution with 145.0 ml of a 0.0015 M AgNO 3 solution [at 25 C]. Identify the precipitate, if any. [Use K sp from Table 16.2.]. Answer: A precipitate of AgCl(s) should form Strategy: 1) Recognize that if a precipitate would form, it would have to be AgCl(s), because all nitrate salts are soluble, and all compounds with K + as the cation are soluble (1 st semester). 2) Recognize that since Ksp is what is given (in 16.2), you should create the balanced chemical equation (and Ksp expression) for AgCl dissolution. PS511

3) Once this equation is set (see #2), realize that precipitate forms means reverse reaction occurs. Thus, you need to determine Q (initially, but after mixing) and see if it is greater than K. If so, a precipitate should form. If not, the solution is unsaturated (and no precipitate will form.) 4) Use the dilution factor approach (or M1V1 = M2V2) to determine the initial concentrations of Ag + and Cl (also recognizing that KCl ionizes 100% into K + and Cl and that AgNO3 ionizes into Ag + and NO3 ) and then substitute into the Qsp expression to find Q. Then compare to Ksp as noted in 3) to make the final conclusion. Execution of Strategy: AgCl(s) Ag + (aq) + Cl (aq); Ksp [Ag + ][Cl ] 1.77 x 10 10 (from Table 16.2) Reverse reaction occurs a precipitate will form check for Q > K 175.0 ml of 0.0055 M KCl(aq) + 145.0 ml of 0.0015 M AgNO3(aq) Vtotal = 175.0 + 145.0 = 320.0 ml 175.0 ml 320.0 ml Qsp (3.01 x 10 3 )(6.80 x 10 4 ) 2.05..x 10 6 145.0 ml 0.0015 M Ag 0.000680 M Ag 320.0 ml Since 2.0 x 10 6 > 1.77 x 10 10, Qsp > Ksp and thus a precipitate (of AgCl) will form [Cl ]0 0.0055 M Cl 0.00301M Cl [Ag + ]0 31. MP. No answer in key. Another precipitation kind of problem (multiple parts). 32. MP. No answer in key. This is a nice LeChatelier s type of equilibrium problem related to solubility equilibria. Be very careful with these kinds of problems. For some reason, some student tend to get the thinking backwards on this! PS512