Study Guide for CHM 116 Exam
Study Guide for CHM 116 Exam CHM116
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This 3 page Study Guide was uploaded by Margaret Frazier on Friday October 14, 2016. The Study Guide belongs to CHM116 at Purdue University taught by in Spring 2015. Since its upload, it has received 71 views. For similar materials see General Chemistry 2 Lecture in Chemistry at Purdue University.
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Date Created: 10/14/16
Study Guide for CHM 116 Exam II 1. Finding which way the reaction will move a. If Q is greater than Keq then the reaction will move towards reactants b. If Q is less than Keq then there is too much on the reactant side and reaction will move towards the products 2. Finding Knet a. Look at the two equations and see if you can cancel anything. If you can, multiply by a constant in order to cancel more. i. If you multiply by a constant, then the Keq of that equation will be raised to the power of the constant. ii. Ex. 1. I2 2I K1 = 10^8 2. H2 + I H2IK2 = 10^4 3. H2I + I 2HI K3 = 10^9 a. Knet = K1xK2xK3 iii. Ex. 1. (NO3 NO + O2) x2 K1 = ? a. Multiply this row by two so that 2NO and 2NO will cancel 2. 2NO N2+ O2 K2 = 2e12 3. Net reaction: 2NO3 N2 + 3O2 Knet = 1x10^3 a. Finding K1: Knet = K1^2 x K2 which equals 1x10^3 iv. If in order to cancel you must flip an equation, take Keq^1 of that equation. 3. Kp a. Kp is the pressure at equilibrium b. An expanded version of this equation is: i. Kp = [C]^c[D]^d/([A]^a[B]^b) * (RT)^(c+dab) 1. c+dab is the chance in n, or the change in the number of moles 2. Kc is the first part of that equation 4. Finding pH and pOH a. To find pH and pOH, determine which species is the acid and which is the base i. Equations will be modeled this way 1. HA + H2O Ka H3O^+ + A^ Acid reaction donates a H 2. A^ + H2O Kb OH^ + HA Base reaction receives a H 3. 2H2O Kw OH^ + H3O^+ Reaction of water ii. In the first equation the acid is HA because it has a hydrogen. The conjugate base of this is A^ because in the base reaction it is in place of HA b. Using the concentration of [OH^] we can find pOH by taking the –log[OH^]. The pH of a concentration of [H3O^+] can be found this same way. i. If given a concentration of [OH^] and asked to find pH, find pOH and then take 14 – pOH = pH. The same can be said for finding the pOH given an initial concentration of [H3O^+]. c. If we are given pH or pOH we can find [H3O^+] or [OH^] by taking 10^pH or 10^pOH. 5. ICE table a. ICE tables are used to find the concentration of species in a solution at equilibrium given an initial concentration and Keq. aA + bB cC Initial concentration of A was 0.05, initial (the concentrations of B and C are origionally 0) The Ka of the solution is 1e5 A B C Initial 0.05 0 0 Change ax +bx +cx Equilibrium 0.05ax bx cx b. We can solve for x and then find the concentrations at equilibrium. i. [cx][bx]/(0.05ax) = Ka = 1e5 1. This is where it can get kind of confusing. If x is a very small number (< 5%) then we can ignore the x on the bottom in order to not get a quadratic formula. X is small when it is smaller than 5% which can be found by doing x/initial concentration. 2. So therefore we have cbx^2 = (1e5) x (0.05a) and we solve for x. c. We can also use ICE tables to find the pH, or pOH of a solution. i. It will be solved the same way as we solved above, however when you find the concentration of the acid or base then you will convert the concentration to pH or pOH depending on what you need. d. If a certain amount of solution is added to a solution we can use a different formula in order to not have to deal with the ICE table. Again, we can only use this is x is very small. i. pH = pKa + log(acid/base) ii. This is called the HendersonHasselbach Equation or HH equation iii. We can use this to solve for either pH or pKa. 6. Solving for pKa/pKb a. If we are given a solution with an [OH^] concentration, then K at equilibrium will be Kb. This can get tricky when we need to find the pH of a solution. When solving for pH using an ICE table, use this formula: i. pKw = pKa x pKb 1. pKw always equals 1x10^14. So we can easily solve for pKa or pKb. Make sure that if you are solving for the pH that your Keq is Ka. Once you have pKa you can use the equation 10^pKa to find the Ka. 2. Ka or Kb = [cx][bx]/(Initial concentrationax) a. If asked to solve for pH use Ka, Kb for pOH. 7. Titrations a. A buffer solution minimizes the large changes in pH during a titration b. The buffer region on a titration graph is the region where the pH is changing slowly before the equivalence point. i. The equivalence point is where the pH changes very fast c. For a weak acid or a weak base, the pH range is a result of Ka d. In order for a buffer solution to regulate the pH the buffer concentration must be higher than the strong acid of base, otherwise the acid of base will drive the the pH. e. A buffer solution can consist of a weak acid and its conjugate base or a strong base and its conjugate acid. f. At the halfequivalence point (which is in the center of the buffer region) the moles of Acid exactly equal the moles of the base i. At this point [A] approximately equals [HA] so pH = pKa g. The equivalence point will tell you what the initial concentration of something was before the addition of the acid or base (in Liters) 8. Diprotic Acids a. These acids contain multiple H’s that they can give away, so they undergo two reactions
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