Final Exam Review Questions - General Chemistry | CHEM 142, Study notes of Chemistry

Download Final Exam Review Questions - General Chemistry | CHEM 142 and more Study notes Chemistry in PDF only on Docsity! Final Exam: Tues. at 8:30 here Same rule on calculators. Check seating assignments Review for Ch. 1-5 posted on web within lecture notes: Test 1 Review Test 2 Review Now: Chapter 6-8 Review (also posted on web, as Final Exam Review) Characteristics of Chemical Equilibrium States • Reaching equilibrium requires reactions to occur. • Once reached, they show no macroscopic evidence of further change. • Reached through dynamic balance of forward and reverse reaction rates. Equilibrium Expressions Involving Pressures For a reaction of the type a A + b B = c C + d D It is sometimes convenient to write the equilibrium expression in terms of partial pressures, e.g. P indicates the partial pressures of the species in equilibrium and KP is a constant called the equilibrium constant in terms of partial pressures. KP depends only on T , and not on pressure. ( ) ( ) ( ) ( ) c d C D P a b A B P P K P P = How is KP related to K? Answer: Through the use of the ideal gas law. RTCRTRTnVP AAA =⎟ ⎠ ⎞ ⎜ ⎝ ⎛== V nor P AA ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) c d c d C D C D P a b a b A BA B c d c d C D c d a b n a b a b A B P P C RT C RT K C RT C RTP P C C RT K RT K RT RTC C + + − + Δ + × × = = × × = × = = Where Δn = c+d-(a+b) Note that K =KP when Δn = 0 Equilibrium involving pure solids or liquids: set their activity = 1 Example: NH4NO2(s) = N2(g) + 2 H2O(g) The equilibrium constant for this reaction would normally be expressed as: [ ][ ] [ ] 2 2 2 4 2 N H O ' NH NO K = 4 2 However, a pure solid or liquid retains the same activity during the reaction. Thus we set the activity of NH NO ( ) to one. s [ ][ ] ( )( )2 2 22 2 2 N H ON H O and P PpK K= = 1.0000 1.0000 Example: Calculating Equilibrium pressures and concentrations from K and initial conditions. Consider the equilibrium: CO(g) + H2O(g) = CO2(g) + H2(g) 0.250 mol CO and 0.250 mol H2O are placed in a 125 mL flask at 900 K. What is the composition of the equilibrium mixture if K = 1.56? The original reactant concentrations are: [CO]0 = [H2O]0 = 0.250 mol/ 0.125 L = 2.00 M Q = 0.Therefore, Q < K, so reactants are consumed and products made. 0+x = x0+x = x2.00 - x2.00 - xEquil. [i] = Init.+Change +x+x-x-xChange = x times coefficient = neg. for reactants 002.002.00Init. [i] H2(g)CO2(g)H2O(g)CO(g)Conc. (M) CO(g) + H2O(g) = CO2(g) + H2(g) Le Chatelier’s Principle If a change in conditions (a ‘stress’ such as change in P, T, or concentration) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions. Henri Le Chatelier, 1884 Ways of Expressing the Reaction Quotient, Q Form of Chemical Equation Form of Q Value of K Reference reaction: A B Q(ref) = K(ref) = Reverse reaction: B A Q = = K = Reaction as sum of two steps: [B] [A] [B]eq [A]eq 1 [A] Q(ref) [B] 1 K(ref) (1) A C (2) C B Qoverall = Q1 x Q2 = Q(ref) Koverall = K1 x K2 = x = = K(ref) [C] [B] [B] [A] [C] [A] Coefficients multiplied by n Q = Qn(ref) K = Kn(ref) [A] [C] [C] [B]Q1 = ; Q2 = Reaction with pure solid or Q’ = Q(ref)[A] = [B] K’ = K(ref)[A] = [B] liquid component, such as A(s) Brønsted-Lowry Acid-Base Definition An acid is a proton donor, any species that donates an H+ ion. An acid must contain H in its formula; HNO3 and H2PO4- are two examples, all Arrhenius acids are Brønsted-Lowry acids. A base is a proton acceptor, any species that accepts an H+ ion. A base must contain a lone pair of electrons to bind the H+ ion; a few examples are NH3, CO32-, F -, as well as OH -. Brønsted-Lowry bases are not Arrhenius bases, but all Arrhenius bases produce the Brønsted-Lowry base OH-. Therefore in the Brønsted-Lowry perspective, one species donates a proton and another species accepts it: an acid-base reaction is a proton transfer process. Acids donate a proton to water Bases accept a proton from water The Acid-Dissociation Constant (Ka) Strong acids dissociate completely into ions in water: HA(g or l) + H2O(l) H3O+(aq) + A-(aq) In a dilute solution of a strong acid, almost no HA molecules exist: [H3O+] = [HA]init or [HA]eq = 0 K = [H3O +][A-] [HA][H2O] at equilibrium, and Ka >> 1 Nitric acid is an example: HNO3 (l) + H2O(l) H3O+(aq) + NO3-(aq) Weak acids dissociate very slightly into ions in water: HA(aq) + H2O(aq) H3O+(aq) + A-(aq) In a dilute solution of a weak acid, the great majority of HA molecules are undissociated: [H3O+] << [HA]init or [HA]eq = [HA]init K = [H3O +][A-] [HA][H2O] at equilibrium, and Ka << 1 Ka = Ka = Classifying the Relative Strengths of Acids and Bases–II Strong bases. Soluble compounds containing O2- or OH- ions are strong bases. The cations are usually those of the most active metals: 1) M2O or MOH, where M= Group 1A(1) metals (Li, Na, K, Rb, Cs) 2) MO or M(OH)2, where M = Group 2A(2) metals (Ca, Sr, Ba) [MgO and Mg(OH)2 are only slightly soluble, but the soluble portion dissociates completely.] Weak bases. Many compounds with an electron-rich nitrogen are weak bases (none are Arrhenius bases). The common structural feature is an N atom that has a lone electron pair in its Lewis structure. 1) Ammonia (:NH3) 2) Amines (general formula RNH2, R2NH, R3N), such as CH3CH2NH2, (CH3)2NH, (C3H7)3N, and C5H5N : : : : ::: The Conjugate Pairs in Some Acid-Base Reactions Acid + Base Conjugate Base + Conjugate Acid Conjugate Pair Conjugate Pair Reaction 1 HF + H2O F– + H3O+ Reaction 2 HCOOH + CN– HCOO– + HCN Reaction 3 NH4+ + CO32– NH3 + HCO3– Reaction 4 H2PO4– + OH– HPO42– + H2O Reaction 5 H2SO4 + N2H5+ HSO4– + N2H62+ • • • • • • • • • • • • • • • • • • • • The Meaning of Ka, the Acid Dissociation Constant For the ionization of an acid, HA: HA(aq) + H2O(l) H3O+(aq) + A-(aq) Eqbm. constant: Ka = [H+] [A-] [HA] The stronger the acid, the higher the [H+] at equilibrium, and the larger the Ka Water itself is a weak acid and weak base: Two water molecules react to form H3O+ and OH- (but only slightly) K = Ka = Kw = 1.0 x 10-14 Autoionization of Water H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Ka = = [H+][OH-] = Kw [H3O+][OH-] [H2O]2 The ion-product for water, Kw: Ka = Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25°C) For pure water the concentration of hydroxyl and hydronium ions must be equal: [H3O+] = [OH-] = √1.0 x 10-14 = 1.0 x 10 -7 M (at 25°C) The molarity of pure water is: = 55.5 M1000g/L 18.02 g/mol Set = 1.0 since nearly pure liquid water even when salts, acids or bases present. The Definition of pH pH = -log10[H3O+] = -log10[H+] Significant Figures for Logarithms The number of decimal places in the log is equal to the number of significant figures in the original number. Determining Concentrations from Ka and Initial [HA] Problem: Hypochlorous acid is a weak acid formed in laundry bleach. What is the [H3O+] of a 0.125 M HClO solution? Ka = 3.5 x 10-8 Plan: We need to find [H3O+]. First we write the balanced equation and the expression for Ka and solve for the hydronium ion concentration. Solution: HClO(aq) + H2O(l) H3O+(aq) + ClO-(aq) Ka = = 3.5 x 10-8 [H3O+] [ClO-] [HClO] Concentration (M) HClO H2O H3O+ ClO- Initial 0.125 ---- 0 0 Change -x ---- +x +x Equilibrium 0.125 - x ---- x x Ka = = 3.5 x 10-8 (x)(x) 0.125-x Use 0.125 - x = 0.125 since x must be tiny. x2 = 0.4375 x 10-8 x = 0.661 x 10-4 Finding the Ka of a Weak Acid from the pH of its Solution–I Problem: The weak acid hypochlorous acid is formed in bleach solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the value of the Ka of this weak acid. Plan: We are given [HClO]initial and the pH which will allow us to find [H3O+] and, hence, the hypochlorite anion concentration, so we can write the reaction and expression for Ka and solve directly. Solution: Calculating [H3O+] : [H3O+] = 10-pH = 10-4.19 = 6.46 x 10-5 M = x Concentration (M) HClO(aq) + H2O(l) H3O+(aq) + ClO -(aq) Initial 0.12 ---- 0 0 Change -x ---- +x +x Equilibrium 0.12 -x ---- +x +x Assumptions: [H3O+] = [H3O+]HClO since it’s >> 10-7 M H+ from water The Meaning of Kb, the Acid Dissociation Constant For the generalized reaction between a base, B(aq), and water: B(aq) + H2O(l) BH+(aq) + OH-(aq) Eqbm. constant: Kb = [BH+] [OH-] [B] The stronger the base, the higher the [OH-] at equilibrium, and the larger the Kb K = TABLE 7.3. Values of K,, for Some Common Weak Bases Conjugate Name Formula Acid Ky Ammonia NH; NH,” 138 <10— Methylamine CH3NH, CH;3NH;3~ 438 Ome Ethylamine C,H;NH> C,H;NH;* 5.6 x 10-* Aniline C.HsNH> C.H;NH;* oreo Ome Pyridine C;H;N C.H;-NH" 7 1082 The Relation Between Ka and Kb of a Conjugate Acid-Base Pair Acid HA + H2O H3O+ + A- Base A- + H2O HA + OH- 2 H2O H3O+ + OH- [H3O+] [OH-] = x [H3O+] [A-] [HA] [HA] [OH-] [A-] Kw = Ka x Kb For HNO2 Ka = 4.5 x 10-4 Kb = 2.2 x 10-11 Ka x Kb = (4.5 x 10-4)(2.2 x 10-11) = 9.9 x 10-15 or ~ 10 x 10-15 = 1 x 10 -14 = Kw Like Example 7.7 (P247-8)-I Calculate the pH of a 5.0 M H3PO4 solution and determine equilibrium concentrations of the species: H3PO4 , H2PO4-, HPO4-2, and PO4-3. Solution: H3PO4 (aq) H+(aq) + H2PO4-(aq) Ka = 7.5 x 10-3 = [H+][H2PO4-] [H3PO4] Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [H3PO4]0 = 5.0 [H3PO4] = 5.0 - x [H2PO4-]0 = 0 [H2PO4-] =x [H+]0 = 0 [H+] = x [H+][H2PO4-] [H3PO4] Ka1 = 7.5 x 10-3 = = = (x)(x) 5.0-x x2 5.0 ~ Like Example 8.1 (P 278-9) - I Nitrous acid, a weak acid (Ka = 4.0 x 10-4), is only 2.0% ionized in a 1.0 M solution. Calculate the [H+], the pH, and the percent dissociation of HNO2 in a 1.0 M solution that is also 1.0 M in NaNO2! HNO2(aq) H+(aq) + NO2-(aq) Ka = = 4.0 x 10-4 [H+] [NO2 -] [HNO2] Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HNO2]0 = 1.0 M [HNO2] = 1.0 – x (from dissolved HNO2) [NO2-]0 = 1.0 M [NO2-] = 1.0 + x (from dissolved NaNO2) [H+]0 = 0 [H+] = x (neglect the contribution from water) Added OH- ions are not allowed to accumulate but are replaced by A- : OH- + HA → H2O + A- HA H+ + A- Ka = ← original [HA] and [A-] >> [H+] = Ka [HA]/[A-] added [OH-] [H+][A-] [HA] How Does a Buffer Work? How Does a Buffer Work? Add a strong acid to a weak acid / salt buffer and see what happens: H+ + A- HA Original buffer pH Final pH of buffer close to original- Added H+ ions Replaced by HA Ka = [H+] [A-] [HA] [H +] = Ka [HA] [A-] pH and Capacity of Buffered Solutions The pH of a buffered solution is determined: mainly by the pKa of the acid, and much more weakly by the ratio [A-]/[HA]: pH = pKa + log ([A-]/[HA]) The capacity of a buffered solution is determined: by the lower magnitude of [HA] or [A-]. Buffers well when: [added base or acid] << [HA] and [A-] and [HA] ≈ [A-] BUFFERS and The Henderson-Hasselbalch equation: pH = pKa + log ([base]/[acid]) For basic buffers this becomes: pH = (14.00-pKb) + log ([base]/[acid]) This named equation can be used for calculating the pH of buffered solutions of acids or bases, and even after adding strong acid or strong base, provided the conc. is small compared to buffer ingredients. Assume they react 100%. Titration of a strong acid (HCl) solution by a strong base (NaOH). Equivalence point: mol acid = mol base mol H+ = mol OH- Quantitative Analysis How concentrated is the acid sample? Quantify! pH=7.0 for strong acid / strong base titration Eq. Pt. TABLE 8.2 Summary of Results for the Titration of 100.0 mL 0.050 M NH; with 0.10 M HCl Volume of 0.10 M HCl Added (mL) [NH3]o INH," Jo ([H*] pH 0 0.050 M 0 11x10°''M 10.96 4.0 mmol 1.0 mmol TO i 4X ‘ ane (Won Comim i" “~ 7 : 2.5 mmol 2.5 mmol m0 .Q* 6% : 2s) (100 + 25) mL (100 + 25) mL St i a= 5.0 mmol = i SEE EESEREe x 6 50.0 0 (100 + 50) mL 43x10 °M 5.36 5.0 mmol 1.0 mmol + nae 60,0 ° (100 + 60) mL 160 mL 2. =62x103°M *Halfway point. tEquivalence point. '[H*] determined by the 1.0 mmol of excess H*. weak Base: ‘a satretbeniot 20,00 mil ot 2-120 ENE rong Aci Titration 12} — INHy]=INH,*) Curve Ph = 5.27 at equivalence point 0 T T T T T T 1 10 20 30 40 50 #60 70 £80 Volume of HCI added (mL) Figure 8.11: A summary of the important equilibria at various points in the titration of a triprotic acid Table 8.5: Ksp Values at 25 C for Common Ionic Solids Like Example 8.12 (P 320) The Ksp value for the mineral fluorite, CaF2 is 3.4 x 10-11 . Calculate the solubility of fluorite in units of grams per liter. Concentration (M) CaF2 (s) Ca2+(aq) + 2 F-(aq) Starting 0 0 Change +x +2x Equilibrium x 2x Substituting into Ksp: [Ca2+][F-]2 = Ksp (x) (2x)2 = 3.4 x 10-11 4x3 = 3.4 x 10-11 x = x = 2.0 x 10-4 3.4 x 10-11 4 3 The solubility is 2.0 x 10-4 moles CaF2 per liter of water. To get mass we must multiply by the molar mass of CaF2 (78.1 g/mol). 2.0 x 10-4 mol CaF2 x = 1.6 x 10-2 g CaF2 per L 78.1 g CaF2 1 mol CaF2 The Effect of a Common Ion on Solubility PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO42-(aq; added)