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Chem 222 Study Guide for Exam 2

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by: Leslie Pike

Chem 222 Study Guide for Exam 2 Chem 222

Marketplace > Western Kentucky University > Chemistry > Chem 222 > Chem 222 Study Guide for Exam 2
Leslie Pike
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Acidic and basic salts, equilibrium constants, pH determination, buffers, ICE, Ka and Kb, titrations, solubilities
College Chemistry 2
Darwin Dahl
Study Guide
Chemistry, chem 222, chem222
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This 7 page Study Guide was uploaded by Leslie Pike on Wednesday March 30, 2016. The Study Guide belongs to Chem 222 at Western Kentucky University taught by Darwin Dahl in Spring 2016. Since its upload, it has received 56 views. For similar materials see College Chemistry 2 in Chemistry at Western Kentucky University.


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Date Created: 03/30/16
Solubility and Dissociation A system is at equilibrium when there is no observable change with time. The reaction going from the reactants to the products is equal to the reaction going from the products to the reactants. For some reactions, there is no equilibrium point, the reaction goes until all reactants are used up. Ex. 2H 2 O  2H O 2 For other reactions, they can go both ways: N 2 4 2NO 2 Equilibrium can be physical equilibrium, such as water or mercury in a closed container at equilibrium with its vapor. Equilibrium can be chemical, such as in the reaction shown above. For a reaction aA + bB  cC + dD, the equilibrium constant is determined as: c d K= [C] [D] [ A] [B]b K is unitless because it is a ratio. K can be written as Kc, the concentration equilibrium constant. At a homogenous equilibrium, all reactants are in the same phase. For a heterogeneous equilibrium, the reactants are in different phases. For gases, there is a pressure equilibrium constant (the partial , Kp, in addition to the concentration equilibrium constant. One can convert between the two with the following equation: Δ n Kp=Kc(RT) Here, Δn is the change in the number of moles of gas, or the number of moles of gas on the right minus the number of moles of gas on the left. It can be a negative number. R is 0.0821 L*atm/(mol*K) Q can be found by taking the equation for K and plugging in the initial concentrations of the reactants and products. If Q is gceater than K, the reaction will go to the left because there are too many products. If Q is lcss than K, the reaction will go to the right because there are too many reactants. If Q = K, c equilibrium is reached and nothing happens. To work an equilibrium problem, remember ICE: Initial, Change, and Equilibrium. Initial is your initial concentrations. Change is how they change (multiply by balancing coefficient, and use a positive or negative sign in front depending on whether the species is being produced or consumed). Equilibrium is the final concentrations (initial concentrations plus or minus however much they changed). Le Chatlier’s Principle: The system does the opposite of what you do to it. For example, take the following reversible reaction: N +3H  2NH , delta H > 0 2 2 3 All three of these are in the gas phase. Increasing the concentration of N will cause2 the system to want to decrease the concentration of N , so the reaction will go to 2 the right. Taking N o2t of the system will cause the system to want to produce more N 2 so the reaction will go to the left. The reaction is exothermic, meaning that heat can be considered a product. Adding more heat to the system will cause the reaction to go to the left, absorbing heat. Removing heat will cause the system to go to the right, releasing heat. If the above system is in a closed container, decreasing the size of the container (or increasing the pressure) will mean that the molecules have less room. Therefore, to save space, N an2 H will 2ecome NH . (Four mo3es of gas on the left, two moles on the right.) Increasing the size of the container (or decreasing the pressure) will give the molecules more room, and NH will bre3k apart into N and H . 2 2 EASY MISTAKES TO MAKE: The stressor on the system MUST be something that will actually affect K. Adding more liquid water to a system, for instance, won’t change anything, because water is a solvent instead of a reactant. Increasing or decreasing the pressure or volume of a system which contains no reacting gases won’t change the equilibrium. The presence of a catalyst doesn’t affect K. Temperature affects K differently, depending on whether the reaction is exothermic or endothermic. If endothermic, treat heat as a reactant and solve the problem accordingly. If exothermic, treat heat as a product and solve the problem accordingly. Sample problem: The equilibrium constant for silver carbonate, Ag CO , is K = 2 * 3 sp -12 10 . Find the solubility of silver carbonate. First write the equation for the dissociation. + 2- Ag 2O 32Ag + CO 3 Then, write the expression for K : sp + 2 2- K spAg ] [CO ] 3 Notice that silver carbonate, being a solid, is NOT included in the expression! Notice that the products are raised to their balancing coefficients. Use the ICE principle: Initial: Ag = 0, CO = 3 Change: Ag = +2x, CO = +x 3 Equilibrium: Ag = 0+2x = 2x, CO = 0+x 3 x Plug the values into the K expspssion: 8*10 -12= [2x] [x], x=1.26*10 M -4 -4 Therefore, the solubility for silver carbonate is x=1.26*10 M. Given sufficient silver carbonate in the water, it will dissolve until solubility concentration is reached. (Note: silver carbonate splits into three ions, so the concentration of silver ion will be twice the solubility number and the concentration of carbonate ion will be equal to the solubility number.) Determining whether or not a given concentration of ions will cause a precipitate: Calculate Q for the given concentrations of the two ions. (Remember, the expression for Q is the same as the expression for K, and you plug in your initial concentrations instead of your equilibrium concentrations.) If Q < K, you have a dilute solution and no precipitate will form. If Q = K, you have a saturated solution. If Q > K, you have more compound in the system than can be dissolved, and some will precipitate out. Common ion effect: the presence of an ion decreases the solubility of a compound containing that ion. Ex. An 0.1 M solution of silver nitrate will dissolve less silver chloride than a solution of pure water. When making the ICE table, be sure to write in the initial concentration of whatever ion is already present in the solution. Other than that, the problem is worked no differently than when you are dissolving a compound in pure water. Acids and Bases + An Arrhenius acid produces hydronium ion (H O ) in w3ter. An Arrhenius base produces hydroxide ion (OH ) in water. A Bronsted-Lowry acid is a proton donor. A Bronsted-Lowry base is a proton acceptor. An acid-base conjugate pair differs by one proton, i.e. HF/F , NH /NH . 4+ 3 2- NOT an acid/base conjugate pair: H SO /SO2. T4ese 4wo compounds do not differ by one proton. They differ by two protons. -14 The equilibrium constant for the auto-ionization of water is K =10 w . The expression is Kw=[H ][OH ], where the concentrations of H and OH are equal. The pH is calculated as –log[H ] and is on a scale from 1 to 14, with 1 being the most acidic (highest concentration of hydrogen) and 14 being the most basic (highest concentration of hydroxide). Pure water has a pH of 7. The pOH is - calculated as –log[OH ]. The pH and pOH sum to 14, so it is easy to convert from one to the other. Strong acids dissociate completely in water and have K=infinity. The six strong acids are:  HCl  HBr  HI  HNO 3  HClO 4  H2SO 4 Strong bases dissociate completely in water and have K=infinity. Almost all bases are strong. Ammonia is a weak base. When given a problem, assume a strong base unless you are told otherwise. Sample problem: You mix 20 mL 0.1 M HCl with 15 mL 0.2 M HClO . What i4 the pH of the resulting solution? Since pH only takes into account the hydrogen ion concentration, we will ignore the fact that two different acids are used. We don’t care about chloride and perchlorate ions because they don’t affect the pH. The first step is to calculate the number of moles of hydrogen ion we have. Each acid is a strong acid, so the number of moles of hydrogen will be equal to the number of moles of acid. 0.1 M HCl * 0.02 L = 0.002 mol HCl = 0.002 mol H + 0.2 M HClO *40.015 L= 0.003 mol HClO = 0.043 mol H + + 0.002 + 0.003 = 0.005 mol H The next step is to convert this to molarity. Our total volume of acid is .02 + .015 = . 035 L. .005mol/.035L = .357 M pH=-log[.357]=.845 Is our answer reasonable? We are mixing two strong acids, so our pH value should indicate an acidic solution. A pH value of .845 is very acidic. The pH of a weak acid is calculated using the same method used to determine the + concentration of sparingly soluble salts. The concentration of H is determined, and then the pH is determined by taking the negative log of this value. In using the above method, it is often useful to neglect x to save having to use the quadratic equation. X can be neglected when concentration/K > 100. The bigger Ka is, the stronger the acid is, and the less basic that its anion is. For all strong acids, the anion is neutral, but for weak acids, the anion is basic. Ex. Perchlorate and sulfate are neutral anions, but fluoride and cyanide are basic anions. If a salt contains a basic anion, the salt will form a basic solution when dissolved in water. Ex. Sodium acetate, potassium fluoride. Useful equations for pH problems: −¿ ¿ OH ¿ ¿ +¿¿ H ¿ ¿ pKa+pKb=14 Sample problem: You have a solution of 0.1 M HCl and 0.1 M HF. Calculate the pH. The HCl will dissociate completely, so that the concentration of the free hydrogen ions will be equal to the HCl concentration. The HF does not dissociate completely. It dissociates according to the following equation: HF  H + F - According to Le Chatelier’s Principle, addition of an ion to one side of the equation + will push the reaction to the other side. The additional H caused by the dissociation of HCl will make HF even less likely to dissociate. Thus, the weak acid can be ignored in the calculation of the pH; only the concentration of the strong acid is needed. Take the negative logarithm of the concentration of the strong acid to find the pH. Acidic salts are the result of the neutralization of a strong acid and a weak base. Acidic salts can also result when the salt contains small, highly charged ions, like aluminum. The easier the free hydrogen atom is to remove, the stronger the acid. For a binary (two-atom) acid, strength increases down a column (HI is stronger than HBr which is stronger than HCl which is stronger than HF). The bigger atoms have more electron shells between themselves and the hydrogen proton; therefore, they do not attract it as strongly. For oxoacids (acids containing oxygen atoms), acid strength increases as the number of oxygen atoms involved increases. Acid strength also increases as the non-oxygen non-hydrogen atom increases in electronegativity. HClO is stronger than HBrO which is stronger than HIO. For polyprotic acids, acid strength decreases as the acid loses protons. How to choose a buffer: the pKa of the buffer should match the pH you desire to buffer at. Sample problem: You mix 5 ml of 0.1 M NaOH with 20 ml of 0.1 M of a weak acid, -6 HA, with a Ka of 10 . What is the resulting pH of this solution? This is a limiting reagents problem. The reaction will go to completion because strong bases such as NaOH react until all is used up. Convert everything to mmol by multiplying molarity by mL. You will get 0.5 mmol NaOH and 2 mmol HA. Use an ICE table. Your limiting reagent is NaOH, as there are fewer moles of it present and species react in a 1:1 mole ratio. Your final concentration of HA is 2mmol-0.5mmol=1.5mmol. Your final concentration of NaA (which will dissolve in solution and be present at A ) is 0.5 mmol. Now, ask yourself: Do I have a buffer in solution? In this case, the answer is yes, because both HA and A- are present in the solution. Since you have a buffer, use the Hendersen-Hasselbalch equation. −¿ A HA ¿ ¿ pH=pKa−log¿ Is your answer reasonable? You mixed a base and an acid together at the beginning, but you added more acid than base, so your solution will be acidic, so your pH value should be less than 7. Sample problem 2: You mix 20 ml of 0.1 M NaOH with 5 ml of a weak acid, HA, with the same Ka as before. What is the resulting pH of the solution? This is a limiting reagents problem. This reaction will go to completion because of your strong base. Convert everything to mmol by multiplying M by ml. You should get 2 mmol NaOH and 0.5 mmol HA. Use an ICE table. All of your HA will be used up in the reaction; you should end up with 1.5 mmol NaOH and 0.5 mmol A . - Now, ask yourself: Do I have a buffer in solution? In this case, the answer is no, because no HA is present, only A-. You CANNOT use Hendersen-Hasselbalch when you do not have a buffer in solution. What you DO have in solution are two anions, OH and A . One is the ion of a strong base and the other is the ion of the weak base. Ignore the weak base and calculate pH using the strong base. −¿ ¿ OH ¿ ¿ pOH=−log¿ pH=14−pOH=12.78 Now ask yourself: Is this answer reasonable? You mixed an acid and a base, and you had more base than acid, so your resulting solution will be basic, so your pH should be above 7. Titrations Titration can be used to determine the concentration of an acid or base. A titration curve is a plot of pH versus amount of titrant added. It is shaped like an “S” with the midpoint of the S being the point at which there is neither acid nor base present in solution, only salt and water. If both the acid and base are strong, your pH will be 7 at the equivalence point. If one is strong and the other is weak, your salt will be acidic or basic and will affect the pH. To tell if a salt is acidic or basic: Strong base strong acid = two neutral ions making a neutral salt. Strong base weak acid = a neutral cation and a basic anion (makes sense that an acid containing a basic anion won’t be as strong of an acid, right?) making a basic salt. Weak base strong acid = an acidic cation and a neutral anion (a base containing an acidic cation will be a weaker base—think aqueous ammonia, NH OH, whi4h contains the acidic cation NH )4making an acidic salt. Weak base weak acid = if the acid is stronger, you get an acidic salt and vice versa. For example, strontium carbonate is a salt containing strontium and carbonate (well, duh). The strontium would have come from strontium hydroxide, a strong base. Carbonate would have come from carbonic acid, a weak acid (if they are not strong, acids are weak—again, duh). Neutral cation (strontium) and basic anion (carbonate) make a basic salt. To distinguish between base-acid titrations and acid-base titrations: If you have a base and you are titrating it with an acid, your pH will start out high and then decrease. If you have an acid and you are titrating it with a base, your pH will start out low and then increase.


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