Exam 2 Review - THE BEST CHEM REVIEW FOR EXAM 2...

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Unformatted text preview: THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander Bronsted Lowry Theory: Acid is proton donor Base is proton acceptor Lewis: Acid = electron acceptor Base = electron donor Ex. Based on its dot structure, the best classification for O2 ­ is a:Lewis base. The oxygen anion has a complete octet as its dot structure shows. Since it can only make available a share in an electron pair, it must be a Lewis base, because it needs to donate electrons. Arrhenius: only applies to aq. Solutions and only one kind of base OH ­ Acid = H+ generated Base = OH ­ generated, must contain the hydroxide ion or hydroxyl group and produce hydroxide ions in solution General Equations: Kw=[H+][OH ­ pH =  ­log[H+] pOH =  ­log[OH ­ pKa=  ­log[Ka] **make sure you understand that pH CAN be negative **** Dissolve/Soluble: Group 1A metals: Li, Na, K , Rb, Cs Nitrate (NO3 ­), Chlorate (ClO3 ­), Perchlorate (ClO4 ­), Acetate (CH3COO ­) NH4+ Don’t Dissolve/Insoluble: Mg (OH)2 CO32 ­ S2 ­ 1 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander OH ­ PO43 ­ Salts of form AB:[this is how to set up your problem] AB ­> A+ B ­ CAB  ­x CAB – x 0 +x X 0 +x X Salts of form A2B or AB2: AB2 ­> A2+ CAB2  ­x 0 +x 2B ­ 0 +2x X CAB – x X Lets do an example with this: What is the pH of a 0.13 M Ba(OH)2 aqueous solution? 2 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander NOTICE: our tables look exactly the same for this type of problem, just follow the structure of the table and look for the form of the salt to identify which table to use! It’s that easy!! Strong Acids or bases in water: [H+] = Ca for strong acids and [OH ­ = Cb Weak acids or bases in water: [H+]=(KaCa)0.5 for weak acids and [OH ­ = (Kb Ca)0.5 for weak bases Strong Acids: HCl, HBr, HI, HClO4, HClO3, H2SO4, HNO3 H2CO3 Strong Bases: OH ­ of alkali and earth metals LiOH, NaOH, CsOH, KOH, Ba(OH)2, Ca(OH)2, Sr(OH)2 NaCH3COO Weak Acids: HSO4 ­, H3PO4, HNO2, HOCl, acetic acid, benzoic acid Weak Bases: NH4, CH3NH2, C2H5NH2, C6H5NH2, C5H5N Cation – positively charged species Anion – negatively charged species “Strong” electrolytes – dissociate completely “weak” electrolytes – dissociate little “non” electrolytes ­ don’t dissociate at all 3 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander spectator ions – when thrown into solution do not affect the pH ex. Na+, Ca++, Rb+, Mg++, alkali metals and earth cations anions of strong acids: Cl ­, Br ­, I ­, NO3 ­, SO42 ­, ClO3 ­, ClO4 ­ ** If you see a SPECTATOR ION, THROW IT OUT, it has no effect on pH*** ex. A water solution of NaI will exhibit a pH value of about 7. This is because we throw out our spectator ions from affecting the solution and we can say that water usually has a pH of around 7 anyways, even though this isnʼt true. FOR THE PH OF WEAK ACIDS: We assume that the amount of H+ in solution is negligible because it is a weak acid, and it doesn’t dissociate nearly at all. Hence: [H+]=(KaCa)0.5 FOR THE PH OF STRONG DILUTE ACIDS: Ex. What would the pH of a 10 ­9 M solution of HCl be? Neutralization Reactions: When you add H+ ions to a solution they look for: 1. hydroxide ions 2. if no OH ­, look for a weak base 3. if no A ­ is present, attach to H20 to make hydronium ions 4 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander When you add OH ­ ions they: 1. look for protons,H+ to make H20 2. if no H+, look for weak acid, HA to make A ­ + H20 Rules to follow: Rule 1: no H+ added, only A ­ in solution, weak base case, use [OH ­= (KbCb)0.5 Rule 2: H+ added, now there is HA and A ­ buffer so use H+ = Ka(Ca/Cb) Rule 3: same amount of H+ as A ­, so neutralization occurs This is called equivalence pt. with only HA left. Solve using weak acid equation H+ = (KaCa)0.5 Rule 4: More H+ than A ­, we assume H+ concentration in excess is more important, ignore weak acid, perform strong acid equation Monoprotic acids: K<1, ∆G > 0 non ­spontaneous K<1, ∆G < 0 spontaneous Amphoterism: H20 “swings” both ways…but seriously… The auto ­ionization of water occurs naturally, so water can be a base and an acid. Amphoteric species can either accept or donate a proton. Ex. H30+, H20, OH- 5 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander **If the pH of a solution is 1.0, that means that the H+ concentration is 0.1 M. If the pH is 2.0, the H+ concentration is 0.01 M. The difference is one order of magnitude. When Assumptions are not okay: [FRUSTRATING!!] You must assume the amount of neutralized protons is negligible, but in some cases, it isn’t. To check this we use the calculated [H+] concentration we get and divide it by the [HA] we used in the denominator. If it’s under 10% then it’s okay to use! Polyprotic Acids: Some acids contain multiple H+ ions that will eventually fall off causing them to have more dissociation constants and that much more annoying to solve for pH. Ex. H2CO3 Ex. What is the pH of a 0.020 M solution of hydrosulfuric acid, a diprotic acid? 6 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander *** for problems like this make sure to check out the ionization constants for both K1 and K2 values. For this problem they give you a Ka1 of 1.1 E-7 and Ka2 = 1.0 E-14 NOTICE: Ka2 is EXTREMELY small, almost nothing. So we can ignore it and only use ka1 to solve this. [H+][HA-]/ 0.02 = 1.1E-7 Amphiprotism in polyprotic acids: Use the equation pH = 0.5(pK1 + pK2) Ex. What would be the pH of a 1 M solution of Na2HPO4? Assume H3PO4 has a pKa1 of 2 and a pKa2 of 7 and a pKa3 of13. Pk1 = [H2PO4][H+]/[H3PO4] Pk2 = [HPO4][H+]/[H2PO4] Pk1 = [PO4][H+]/[HPO4] Since we are finding the amount of [H] found in the HPO4, we can look at our equations and find which ones have the HPO4 if this makes it easier for you, and weʼll see all we need to use is the Pk2 and pk3. So we pop them into the equation, ph=0.5[pk1+pk2] substituting the pk values we need for our problem, we get 10! 7 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander What QUEST says: Na2HPO4 is the salt of the monoprotic species which is the dominate species between pHʼs of 7.2 and 12.7 which is the orange or shortest dashed curve. The best estimate of the pH is the dead center of that range which is a pH of 10 which is the average of pK2 and pK3 for the acid. What I say: so above the line I wrote out the dominant species above each color. For the red the dominant species is H3PO4 because that’s what it starts out as. For blue I wrote H2PO4, orange HPO4, and finally green is PO4. Since we know HPO4 is dominant in the orange region, we know that the equations for the Pka values surrounding this region will determine the pH. So we use the Pk1 and Pk2 surrounding this area, which are found at the intersections. Since one is about 7 and the other is about 13, we can estimate the equation. Ph = 0.5(pk1+pk2) = 10 8 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander Percent dissociation: For a given weak acid, the percent dissociation increases as the acid becomes more dilute.*** Ex1. A 7.2 × 10−3 M solution of acetic acid is 5.0% dissociated. In a 7.2 × 10−4 M solution, the percent dissociation would be > 5.0% Ex2. What is the percent ionization for a weak acid HX, that is 0.40M? Ka =4.0×10−7 To solve this we first need to recognize that this acid is of type HX, so we use the equation [H+] = (Ka*Ca)0.5 and we plug in our values to get 4E-4. This is the amount dissociated in our solution. Next, we know the initial amount is 0.4 M, so we divide 4E-4/0.4 X 100% and we get 0.1% as our percent ionization/dissociation. Ex3. What is the H+ ion concentration in a 0.50 mol/L solution of a weak base that has an ionization constant (Kb) of 2.0 × 10−8? To solve this we look at what we have. A Kb value usually means we’re going to have to set up an equation for Kb. In this instance, we are given the concentration of the base and asked to find the H+ concentration, but in the Kb equation, it gives us the [OH ­. Take a look: Kb=[OH ­[K]/[KOH] So pretend the values we have are the Kb, and the [KOH] initial value of a weak acid. Plug them in and solve for your [OH ­. 9 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander Then we know all we need to do is divide 1E ­14/[OH] to get [H+]. Then you’re done! Ex4. A diprotic acid H2A has values of Ka1 = 1.0 × 10−6 and Ka2 = 1.0 × 10−10. What is the [A2−] in a 0.10 M solution of H2A? For this type of problem we recognize that they give us two Ka values so lets set up our equations and see what we know. Ka1 = [HA-][H+]/[H2A] = 1.0 x 10-6 Ka2 = [A-][H+]/[HA-] = 1.0 x 10-10 We know the Ka values and the original amount of [H2A]. 1.0 x 10-6 = [HA-][H+]/[0.1] so lets solve for the H+ in this dissociation. We get 3.16227755E-4 for the [H+] from the first equation. This changes our second equation though. [A-][3.2E-4 +H+]/[HA- - 3.2E-4] = 1.0 x 10-10 we know that our new H+ concentration for the second dissociation is going to be added to the amount of [H+] from the first because they are in the same solution. The amount of HAchanges because the HA- will take the concentration of [H+] from the first dissociation and combine to form more of the [HA] again, so we subtract this from the denominator. The [A-] will stay the same. In essence, we can say that this equation will be about equal to the dissociation constant. The answer is 1.0 E -10 M [A-] Ex5. Would the value of Kw change if the temperature were raised from 25◦C to 50◦C? Yes. Kw is only 1E-14 at 25 degrees C. 10 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander Ex6: What is the concentration of SO4 in 2.0 M H2SO4? Ka1 is strong and Ka2 = 1.2 × 10−2. We can write our two equations: [HSO4][H+]/[H2SO4] = Ka1 [SO4][H+]/[HSO4] = Ka2 since we know our values lets plug them in…but before we do, we need to recognize something important in the problem. The fact that Ka1 is strong tells us that it dissociates almost completely. So our concentration of 2.0 H2SO4 will affect our values in the Ka2 equation. Youʼll see… [SO4][H+ +x]/[HSO4-x] = Ka2 where x is the amount dissociated added from the ka1 value. From here we can see this general equation that always yields the conjugate base concentration of the ka value, therefore, the [SO4] = ka2 Ex7: 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 [H3O+] concentration in a saturated carbonic acid solution that is 0.037 molar? Ka1 = [HCO3][H+]/[H2CO3] Ka2 = [CO3][H+]/[HCO3] Since we are solving for the [H+] and we know that the ka2 is REALLY small, and that this diprotic acid is actually a strong acid, we can just assume that the Ka1 value will determine the pH. Plug in values and get your answer to be 1.24E-4. 11 THE BEST CHEM REVIEW FOR EXAM 2 – Holcombe Spring 2010 By: Paul Alexander Mass balance: Which of the following is a correct mass balance expression for the addition of H2CO3 to water? [H2CO3]= [H2CO3]+[HCO3]+[CO3] The mass balance equation consists of the major components of the dissolution. Conjugate bases/acids: Conjugate bases – just take away a H+ from whatever it is Conjugate acids – just add an H+ to whatever it is 12 ...
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This note was uploaded on 04/22/2011 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.

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