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Chemfinalcheat - Arrhenius concept says that an acid...

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Unformatted text preview: Arrhenius concept says that an acid supplies H to an aqueous solution. A base supplies OH to an aq solution. Bronsted-Lowry concept = acid is a H+ donor, base = proton acceptor. HA + H2P A + H30 : HA = acid 1, H2O = base 2, A = conjugate base 1, H3O = conj. Acid 2. Stronger acid = weaker conjugate base. Stronger base, weaker its conj. Acid. Kw = [H][OH] pH = -log [H] pOH = -log[OH] pX = -log[X] [X] = 10-pX At 25*, Kw = 1.0x10-14, pKw = 14, pH + pOH = 14. percent dissociated = (amount dis. / init. Conc) x100 B + H2O -> BH + OH : B = base1, H2O = acid2, BH = acid 1, OH = base2. A polyprotic acid can furnish more than one proton to a solution. Fraction = concen. Of specie / total conc. Of all. + -6 Kw = Ka + Kb When [H ] >= 10 use: Ka = [H] / [HA]o [H] ([HA]o = conc. Of ori.) else, if [HA]o > [H] - Kw / [H] use: Ka = [H] - Kw / [HA]o. A Buffer solution resist change in its pH. If [HA] = [A], then pH = pKa and pOH = pKb. in order for a titration to be feasible, it must be cmplt and fast, in order to be complete, value : k >= 10-7 K = 1/Kw = 1 x 1014. if Q >= Ksp precipitation will occur. Q = [conc.]o[conc.2]o. complex ion is a charged species consisting of a metal ion surrounded by a ligand. A ligand is a moluclure or an ion having al one pair of electron. pH scale = 0 6= acidic. 7 is pure water. Vinegar have pH of 3, stomach acid have pH of 2. HCl ph = 0. 8 14 = basic. NaOH = 14. Example : Calculate the pH of ..500M aqueous solution of formic acid, HCOOH (Ka = 1.77 x 10-4)) Acid HF HCl HBr HI HCN H2S HNO3 HNO2 H2SO4 H2SO3 H3PO4 HC2H3O2 Name Hydrofluoric acid Hydrochloric acid Hydrobromic acid Hydroiodic acid Hydrocyanic acid Hydrosulfuric acid Nitric acid Nitrous acid Sulfuric acid Sulfurous acid Phosphoric acid Acetic acid Caculating pH of a Buffer: Buffer of weak acid and conjugate base: Start with both weak acid, [HA]init, and salt, [A]ini. The equilibrium Concetraion are goverened by Ka = [H]eq [A]eq / [HA]eq HA <==> H + A Init [HA]init 0 [A]init Equil [HA]init a a [A]init + a ! Method: Substitute the "equil" expression into Ka and solve equation for a. Ka = a * ([A]init + a) / ([HA]init a) Balancing redox using half-react. Method: split the reaction into oxidation and reduction half-reactions. You may need to add H2O and H for acidic, or H2O and OH for basic reaction conditions. Balance these separately, then combine to balance the exchange of electrons. Example: Balance the following for acidic solution: MnO4- + Fe2+ => Mn2+ + Fe3+ 1. in acidic solution: add H to the left and H2O to the right side: MnO4- + Fe2+ + H+ => H2O + Mn2+ + Fe3+ 2. identify the half-reactions: Fe2+ => Fe3+ (oxidation) MnO4- + H+ => Mn2+ + H2O(reduction) 3. Add electrons to account for valence changes: Fe2+ => Fe3+ + 1eFe(II) to Fe(III) Cation H+ Li+ Na+ K+ Cs+ Be2+ Mg2+ Ca2+ Ba2+ Al3+ Ag+ Zn2+ Name Hydrogen Lithium Sodium Potassium Cesium Beryllium Magnesium Calcium Barium Aluminum Silver Zinc Anion HFClBrIO2S2N3P3- Name Hydride Fluoride Chloride Bromide Iondide Oxide Sulfide Nitride phosphide Ion NH4+ NO2No3SO32SO42HSO4OHCNPO43HPO42H2P04CO32- 5e- + MnO4- + H+ => Mn2+ + H2O Mn(VII) to Mn(II) 4. Balance each half- reaction: a. oxidation: multiply by a factor of 5 to match electrons in reduction step: 5 Fe2+ => 5 Fe3+ 5echarge: +10 on each side, balanced! b. Reduction: Balance O, then H+, check charge: 5e- + MnO4- + 8H+ => Mn2+ + 4H2O charge: +2 on each side, balanced! 5. Combine half-reactions to eliminate the 5e-: PV = nRT n=moles R=0.08206 Latm/Kmol 5Fe2+ + MnO4- + 8H+ => Mn2+ + 4H2O + 5Fe3+ Boyle's Law (constant temp) Partial Pressure @ STP N = PV/RT Charles Law (constant pressure) PHe = nRT/V1+V(constant temp and pressure) Avogadros Law 2 xi = ni / ntot = pi/ptot P1V1 = P2V2 V1/T1 = V2/T2 V1/n1 = V2/n2 Name Ammonium Nitrite Nitrate Sulfite Sulfate Hydrogen sulfate Hydroxide Cyanide Phosphate Hydrogen phosphate Dihydrogen phosphate Carbonate Ion HCO3ClOClO2ClO3ClO4C2H3O2MnO4Cr2O72CrO42O22SeO4 BrO3 C2O4 Name Hydrogen carbonate Hypochlorite Chlorite Chlorate Perchlorate Acetate Permanganate Dichromate Chromate Peroxide Selenate Bromate Oxalate 1 atm=760 torr 1 atm=101, 325Pa=101KPa 1 m3 = 1000 L Molar volume of gas @ STP = 22.4L Molar mass (M.M.) = dPT/P KEavg = (3/2)RT Distance between collisions is called mean free path, about 10-7 Effusion passage of gas Diffusion mixing (rate of effusion)2/(rate of effusion)1 Distance = M1/M2 = rms2 / rms1 Distance traveled1/ The higher the molar mass, the slower the rate of effusion Distance traveled2 = M1/M2 Root mean square velocity average velocity of gas particles p rms = (3RT/M) R = 8.3145 J/Kmol = kgm2 /s2 Most probably velocity p mp = (2RT/M) Ratio of mp : avg : rms 1.00:1.128:1.225 Effusion Rate Up Effusion Rate Down Molar Mass Down Up Effusion time Down Up Temperature Up Down Collisions per second = ZA = A (N/V)(RT)/(2piM) M=kg/mol n/V = P/RT = 1.00atm / (0.08206)(330) = 0.0309 mol/L Real Gases P = nRT/(V-nb) Pobs = Pideal a[n/V]2 a = constant, depends on gas n = # of moles of gas Van Der Waals [P+a[n/V]2](V-nb) = nRT Interactions is greatest @ low temp, temp, high pressure, low volumes, the deviation goes up N/V = 0.0369 mol/L x 6.02 x 1023 molecules/mol x 1000L/m3 = 2.22 x 1025 molecules/m3 Mean Free Path = wavelength = 1/(2) (N/V)(pi d2) Acidication 2NO2 (g) + H2O (l) HNO2 (aq) + HNO3 (aq) 2SO2 (g) + O2 (g) 2SO3 (g) SO3 (g) + H2O (l) H2SO4 (aq) K = [product]/[reactant] Collision rate (per second)= large K = most products @ equil Z = 4(N/V)d(piRT)/M small K = most reactants @ equil (N/V) = number of particles Q = initial concentrations A = area of wall section Q=K, no shift Q>K, shift left Q<K, shift right x = -bb2 4ac / 2a exo gives off heat cooling Kp = K(RT)n n = sum of coefficients of the gaseous products minus endo absorbs heat cooling the sum of the coefficients of the gaseous reactants R=0.08206 molarity based equilibrium constant Kc = Kp/(RT) n K = Kp only when the sum of the powers in the numerator and denominator equal K>1, products, lies to the right K<1, reactants, lies to the left A sample of methane gas (CH4) @ 0.848 atm [N2]0 = moles/total Volume and 4.0C occupies a volume of 7.0L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11.0C. 0.500 L of H2 (g) are reacted with 0.600 L of P1 = 0.848 atm P2 = 1.52 atm n 1 = n2 O2 (g) at STP according to the equation : 2H2 V1 = 7.0L V2 = ? R=R (g )+ O2 (g) 2H2O (g) T1 = 277K T2 = 284 K What volume will the H2O occupy at 1.00 atm P1V1 = n1RT1 and P2 V2 =n2 RT2 and 350C. P1V1/T1 = n1R and P2 V2 /T2 = n2R Sol: limiting reactant problem because two P1V1/T1 = P2 V2 /T2 reactants are present. H2 = 1 I 1 2HI V2(g)P+V2 T 2 / P2T1 (g) K=7.1 x 10 @ 25C. Calculate the equilibrium concentrations if a 5.00L 0.5L = Moles H2 at STP = 1mol/22.4L x vessel initially contains 15.7 g of 4.0L = (0.848)(7.0)(284)/(1.52)(277) =H2 and 294 g of I2. 0.0223 Sol: We first determine the direction of the reaction, compare Q to K. mol H2 Moles O2 = 1mol/22.4L x 0.6L = 0.026 [H2]0 = 15.7g/5.00L x 1mol/2.016g = 1.56M Then determine limiting reactant: [I2]0 = 294g / 5.00L x 1mol/256.8g = 0.232M 0.0223molH2 x 2molH2O/2molH2 = [HI]0 = 0.00M 0.0223molH2O Q = [HI]0 / [H2]0[I2]0 = 0 / 1.56(0.232) 0.0268 mol O2 x 2molH2O/1molO2 Q<K, so the reaction will go to the right. We then look at the K value. We assume it will complete= 0.0536molH2O and I2 is the limiting reactant. Limiting reactant is H2, 0.0223 molH2O form H2 I2 2HI Initial 1.56M 0.232M VolumeH2O=nRT/P=0.0223(0.08206)(623)/1 0M = 1.14L Change -0.232 + x -0.232 + x + 0.464 2x Equilibrium 1.328 + x +x 0.464 - 2x We assume that x will be <5% of 1.328 and 0.464. K = [HI] / [H2][I2] = 710 = (0.464)/1.328x x = [I2] = 2.3 x 10-4M [H2] = 1.33M [I2] = 2.3 x 10-4M [HI] = 0.464M A volume of 2.0L of He at 46C and 1.2 atm pressure was added to a vessel that contained 4.5L of N2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added? Sol: First determine # of moles n=PV/RT nHe= (1.2)(2.0L)/0.08206)(319) = 0.091 mol He When the gases are combined under STP conditions, the partial pressure of He wil change. That of N2 (already at STP) will remain the same. PHe = nRT/V = (0.091mol)(0.08206)(273)/(4.5) = 0.457 atm PN2 = 1.00atm Ptot = 1.00atm + 0.46atm = 1.5 atm Calculate the number of moles of N2 present Sol: 1molN2/22.4L = xmolN2/4.5L x=0.20molN2 Calculate the mole fractions of N2 and He given the following data i. mole data ii. pressure data i. the total number of moles = mol N2 + mol He = 0.20 + 0.091 = 0.293mol xN2 = 0.20mol/0.293mol = 0.69 xHe = 0.091mol/0.293mol = 0.31 ii. Using partial pressure data, the total pressure was 1.46 atm x N2 = 1.00atm/1.45atm = 0.69 xHe = 0.45atm/1.45atm = 0.31 "if a change in conditions is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in condition." "if a gaseous reactant or product is added to a system at equilibrium, the system will shift away from the added component. If a gaseous reactant or product is removed, the system will shift toward the removed component." "the addition of an inert gas increases the total pressure but has no effecto on the concentrations or partial pressures of the reactants or products." "when the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume." ...
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