Test Review #3 - THE BEST REVIEW FOR EXAM 3 ...

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Unformatted text preview: THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER Buffer: mixture of a weak acid and conjugate base or a weak base and conjugate acid. Buffers are solutions, which contain comparative amounts of a conjugate weak acid/base part and allow for only small changes in the pH of a chemical system, compared to non- buffered solutions, when an acid or a base is introduced. Buffer Capacity: amount of protons or hydroxide ions it can absorb without a significant change in pH. Mmol = 1 mole /1000 Henderson Hasselbalch equation: pH = pKa + log (A ­/HA) for a buffer – this is what you’ll use for the majority of problems that look like this… Ex1. What is the pH of a solution made to be 0.5 M in NH3 and 0.3 M in NH4Br? Kb for NH3 is 1.8 × 10−5. We first need to solve for Ka. Then we use the concentrations of NH3 and NH4+ in the equation. We always put the concentration of the molecule with more Hʼs in it on the bottom – easiest way to remember how to do it. So pH = pKa + log (NH3/NH4). Simple as that. Ans: 9.48 Ex2. What is the pH of an aqueous buffer solution that is 0.1 M HF and 0.3 M KF? HF, pKa = 3.45 Same type of problem. More Hʼs goes on bottom. pH= pka + log(KF/HF). Answer: 3.93 1 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER Ex. Recognizing a buffer problem: What is the pH of an aqueous solution that is 0.10 M HCOOH (Ka = 1.8 × 10−4) and 0.10M NaHCO2? The key to this question is to realize that the solution contains a weak acid and its' conjugate base in equal concentrations, i.e., a buffer. For this type of solution, pH = pKa. Salts of form AB: We solve just like for a weak acid. Molar solubility is x = Ksp1/2 for salts of type AB Salts of form A2B or AB2: Ksp = [A+][B ­2 = [x][2x]2= 4x3 Molar solubility of type A2B and AB2 = (Ksp/4)1/3 Ex. Determine the molar solubility of calcium chloride (CaCl2) if Ksp = 2.5 × 102. Salts of form AB3 or A3B: 4 ions so take the 4th root A saturated solution of MX4 (of MW 145 g/mol) contains 0.0175 grams per liter. What is the solubility product constant of MX4? I couldnʼt believe I missed this on the test. But I knew how to do it. I just plugged it in the wrong way. I know the Ksp = [x][4x]4 but if you try to multiply the xʼs out and get 16X^5, then plug X in, you get the wrong answer. You have to plug in 4X and then put that to the 4th power, then multiply that by x. x being the concentration you find from 0.0175/145. Donʼt miss this!!! Ksp = Solubility Constant 2 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER When k is low or K is really small (10 ­28) everything “stays left”, which means the reaction does not dissociate and is insoluble. Ex. CaF2  ­> Ca2+ + 2F ­ Ksp=10 ­12 so the reaction stays left, doesn’t diss. NaCl ­> Na+ + Cl ­ K = infinity so reaction goes to completion, to right Precipitations: Does a precipitate form when 100 mL of 0.00250 M AgNO3 and 100 mL of 0.00200 M NaBr solutions are mixed? To solve problems like these you can follow a few simple steps: identify what’s going to break apart. You know that when NaNO3 forms they both dissolve. So your AgBr is going to be used to find if it precipitates. Now you look for the respective moles of each from their given volumes and molarities. Then calculate the new total volume and find the molarity of each in solution. Multiply them based upon their Ksp equation and find if this value is greater than the Ksp given. If it is, it precipitates!! Not too bad, but not the easiest if you just come to class and don’t read the book. :P 3 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER If Q > Ksp a precipitate forms!!! Titrations: M1V1 = M2V2 you’ll need this to solve for basic probs. When you titrate bases with acids, they give you a pH of 7 when the number of moles of each is equal. This is a weak acid with a strong base titration. To tell this I always look where the line starts and ends. Since it starts on the left at about 5 or 6, we know this is a weak acid, right? If it had started at like 2 ­4 then it could be a strong acid. Look where it ends: pH of 13. This makes for a strong base, no? If it ended between 7 ­10 I’d say it could be a weak base, but this is not the case for this problem. 4 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER The pKa of this acid can be found by looking at half of the volume where the line is vertical, or half of the equivalence point volume. For this, the equivalence point is at a volume about 70 mL and pH of 10. The pKa is about 35 mL then, and a pH of about 7.5. That’s so easy to do!! Now: what type of salt is formed from this acid/base titration? The salt is formed from the neutralization of a weak acid and a strong base, so it is a basic salt. Just look where the graph ends. If it ends in a base area, then its basic. Of those listed, which indicator would be best for this titration? For this titration we need an indicator that is best around the buffer region for the titration. This means where the line is vertical, we need a buffer with a pKin around this value. Since the closest to our equivalence point pH of 10 is phenolphthalein at 9.4, we must use this one. A TOUGHER titration problem: Actually not that hard, but here it is: Calculate the ratio of the molarities of HPO2−4 and H2PO−4 ions required to achieve buffering at pH = 7.00. For H3PO4, pKa1 = 2.12, pKa2 = 7.21, and pKa3 = 12.68. For this problem we could draw out that diagram with the polyprotic acids and their relative amounts in solution. If you even remember what Iʼm talking about. looks like this. Anyways, itʼs easier if you write H3PO4 above the red, H2PO4 above the blue, HPO4 above the orange, and PO4 above the green. Then you know that to find the ratio of H2PO4 and HPO4 in solution you need to use the Pk value that falls between those two. Which would be between blue and orange. The pKa2! Plug into our equation what we have. 7 = 7.21 + log X, where X is the ratio. Solve! Should get 5 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER 0.62. Indicators: An appropriate indicator has color change range pH = pKa ± 1. For Weak acid titration with strong base: Type of Solution: Calculation Weak Acid Buffer Weak Base Strong Base H+ = (KaCa)0.5 pH=pKa + logCa/Cb OH ­=(Kw/KaCb)0.5 OH ­=Cb For Weak base titration with strong acid: Type of Solution: Calculation Weak Base Buffer Weak Acid Strong Acid OH ­= (KbCb)0.5 OH ­=KbCb/Ca H+=(Kw/KbCa)0.5 H+=Ca Weak Base example: A 1 M solution of NaOH is used to titrate a 1 M solution of C6H5COOH (benzoic acid). If the Ka of C6H5COOH is 5 × 10−5, what is the pH of the solution at the equivalence point? 6 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER Quest says: Equal volumes of the titrant and analyte will be used to reach the equivalance point. Regardless of the starting volume, at the equivalance point the volume will be double the starting value and all of the benzoic acid will have been converted to benzoate. The solution will be 0.5 M C6H5COO−. I say: while this is true, it doesnʼt help us solve the problem or really understand. So you should recognize that weʼre titrating a weak base with an acid. Weʼre going to need to solve for Kb in this case. Once you find Kb, we need to figure out the appropriate concentration. Since we know that at the equivalence point the two acid/base pairs are titrated until they are equal in number of moles, which means that they have to meet at 0.5 and 0.5, respectively. This gives us a better understanding of what quest meant when they said “double the starting value” yadah, yadah yadah… So solve [OH-] = (Kb*Cb)0.5 Strong base example: For the titration of 50.0 mL of 0.020 M aque- ous salicylic acid with 0.020 M KOH(aq), calculate the pH after the addition of 55.0 mL of KOH(aq). For salycylic acid, pKa = 2.97. For this example we recognize the titration of a strong base with an acid. If we look at our table the equation for a strong base is OH ­=Cb From the information given we know our moles of salicylic acid is .001 and the moles of KOH is .0011. Therefore when we titrate, our moles of base will be in excess, determining the pH, which is why we chose the equation for a strong base. We subtract our moles .0011 ­.001 to get .0001 moles KOH, but need to find the new molarity of the solution in total volume. Total volume is .105 L. Our new molarity is 9.52 E  ­4. This is all we need to find the [OH ­, just do –log (9.52E ­4) and then subtract from 14 to get your pH! answer: 10.98 7 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER Coulomb – unit for charge I mole of charge = a Faraday = 9.6 E 4 C/mole Make sure you can properly assign oxidation numbers*** Not hard, just look at the periodic table. Balancing redox reactions: 1. assign oxidation numbers to each atom 2. determine number of e’s for each red/oxid. Half reaction 3. determine least common multiples for two half reactions to make e ­ cancel out on both sides 4. if in acid, add one H20 to opposite side for each deficient O. Add 2H+ to excess O side for each added H20. 5. If in base, add 2 OH ­ to opposite side for each deficient O. Add one H20 to excess O side for every 2 OH ­ added 8 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER Type of batt: Sponta E neity: Batt/volt/ga lv Electrolytic/ electrolysis Sponta neous Non ­ sponta neous +  ­ ∆G K  ­ + Red Oxid E Sign Sign flow @ @ cat an >1 Cath Anod To +  ­ e cath <1 Cath anod To  ­ + e cath Ex. For an electrochemical reduction process, electrons are added to solution species at the surface of the cathode.**** In this picture, we need to recognize that the oxidation of Ag is occurring and Ag+ is being plated onto the cathode. By convention, anode is always written on the left side. Remember it like this: AN ­OX, RED ­CAT  RTlnK = ∆G =  nFE =  qE If E is positive, ∆G must be negative and K must be greater than 1. 9 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER Ecell=Ecathode Eanode Standard potential: E0 = keep more positive, then flip other, add. Reduction potential: E Ecell > 0 SPONTANEOUS!!! Ex1. Mn2+ + 2 e− → Mn E0 = −1.029 V Ga3++3e−→Ga E0=−0.560V If we were to build a galvanic cell based on the half reactions above, the E0 would be: 0.469 V The reduction half reaction with the more positive E0 will remain as reduction; the other will flip to oxidation. For the oxidation half reaction, change the sign on E0, then sum the E values: +1.029 V + (-0.560 V) to get E0 is 0.469 V. Ex2. The equilibrium constant for the reaction 2 Hg(l) + 2 Cl−(aq) + Ni2+(aq) → Ni(s) + Hg2Cl2(s) is 5.6 × 10−20 at 25◦C. Calculate the value of E◦ for a cell utilizing this reaction. Log K = nE0/0.0591 at 25o So plug in values log(5.6E-20) = (2)(E)/0.0591 Solve. -0.56889 Strong oxidizing agents: are the things being reduced Strong reducing agents: are the things being oxidized 1 Ampere = 1 Coulomb/second E = Ecell ­ 0.0591/n (ln([oxid]/[reduc])) N= number of e ­ Ex. This problem confused me at first, but you just need to put some though into it…. 10 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER Solution: Set up your equation given what you have. Ecell = .430 E0 = .222 by taking more positive .222 adding to flipped 0.00 N=2 electrons overall transferred .430 = .222 – (.0591/2) log [oxid/red] Whats being oxidized? From the equation given, we see an ­ox side is our H2. And our reduced? Red ­cat side is the Cl ­. Since our Cl ­ concentration is given we can substitute 1 for the [red]. For our H2 we have to be careful. There are 2 H molecules. So our oxidized concentration in log[oxid/red] is actually [H]2. Where most people mess up. Don’t mess up!! Solve .430 = .222 – (.0591/2) log( [H]2/[1] ) for your H concentration to find pH. Easy. Dead Batteries: Q  ­ ­ K And Ecell ­ ­ 0 This means that the battery comes to equilibrium where the reactants have decreased and turned into products A battery formed from the two half re- actions below dies (reaches equilibrium). If [Fe2+] was 0.24 M in the dead battery, what would [Cd2+] be in the dead battery? 11 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER For this dead battery problem we know that Ecell is 0. E0 = 0.04 from keeping more positive -0.4, flipping -0.44, and adding. Number of etransferred is 2. And since we kept -0.4, for Cd2+, this means this is the reaction that is reduced. The Fe is oxidized then. So we can plug our values into the equation. Ecell = E0 – 0.0591/2 Log (oxid/red) 0 = 0.04–(0.0591/2) log [.24]/[Cd] Solve! Ex. Carbonic acid (H2CO3) is a diprotic acid with Ka1=4.2×10−7 and Ka2 = 4.8×10−11.The ion product for water is Kw =1.0 × 10−14. What is the [HCO3 ] concentration in a 0.037 M H2CO3 solution? Ka1 = [HCO3][H+]/[H2CO3] 4.2E-7 = [HCO3][H+]/0.037 solve for [HCO3] = 1.217E-4 and thatʼs it. Ex. Partially from last exam, but still good to look over for exam 3 review What would be the pH of a 6×10−9 M solution of NaOH? ex. Calculate the CO3 concentration in 0.070 M H2CO3 solution. [HCO3][H+]/[0.070] = 4.2×10−7 [CO3][H+]/[HCO3] = 4.8×10−11 12 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER for diprotic acids that have a really small pk2 value in comparison to the pk1 value, we can assume that the [base] produced by the dissociation constant of the pk2 value is equal to the concentration of the pk2. So we assume that [CO3] = 4.8x10-11 problems with pH: When ammonium chloride is added to NH3(aq): Unless you already know the equation for NH3 in aqueous solution, you would know that NH3 + H2O = NH4+ + OHTherefore, when NH4Cl is added to this solution, it goes to the left, combining with OH- molecules and decreasing the pH (or making it lower/closer to 0) confusing question and answer. Youʼd think decreasing the pH would mean it would go up, afterall, decreasing the percent of hydrogen ions means itʼs going up…right? Confusing. Just ignore this and recognize what it really meant. When adding more base to a buffer: What is the final pH of a solution containing 100mL of 0.2 M HX and 300 mL of 0.1M NaX after 0.01 mol of NaOH is added? The pKa is 3.00. For this problem we can set up our RICE diagrams like usual. R HX-> H+ XI 20 0 30 C -10 +10 E 10 0 40 As you can see we started out with 100 mL X .2 M HX = 20 mmol HX And we had: 300 mL X 0.1 M NaX = 30 mmol NaX But we also added 0.01 mol NaOH, which is just 10 mmol if we mult. By 1000. Since we know that this base is going to react with the H in HX we can subtract it from the HX concentration, and that means its going to force the reaction to the right, producing more X- since itʼs going in the forward direction, so we add it to the the Xconcentration. Finally use pH = pka + log (X-/HX) 13 THE BEST REVIEW FOR EXAM 3  ­ HOLCOMBE SPRING 2010 BY: PAUL ALEXANDER For buffers: Choose the effective pH range of an aniline-anilinium chloride buffer. The value of the Kb for aniline is 4.3 × 10−10. Find Ka = 2.32 E -5 pKa = pH for buffer = -log(pka) = 4.63 Effective buffer range: 3.6 to 5.6 Ex. from Exam 3: What is the pH when 100 mL of 0.1 M NH3 is titrated with 50 mL of 0.2 M HCl? The pKa of NH4Br is 9.25. 10^-9.25 = Ka = 5.62341 E-10 [H30+][NH3]/[NH4] = Ka = 5.62 E-10 we know NH4 to be the amount of NH3 0.1/ total volume 0.15 L = 0.0667 M NH4. We don’t care about the HCl because it’s being titrated with our weak base, and since we are only given pKa of the base, we have to use it to find our pH. [X]2/[0.0667-X] = Ka = 5.62 E-10 so we plug this in and get our [H+]. –log[H+] = pH = 5.21 14 ...
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This note was uploaded on 09/27/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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