25 Solved Problems on Conjugate Acid Base Pairs - Exam 3 | CHEM 1036, Exams of Chemistry

Download 25 Solved Problems on Conjugate Acid Base Pairs - Exam 3 | CHEM 1036 and more Exams Chemistry in PDF only on Docsity! Chemistry 1036 AMfl ii,t, Test 3 THIS IS FORM A April 3, 2006 Name K*: l.o x lo-ra R:8.314 x lo-3 kJ/mol Choose the single best answer to each question. I . Calculate the pH of a buffer that is composed of 0.21 M HCzH:Oz and 0.17 M NaCzH:Oz. (Ka = 1.8 x l0-') t. 4.54 2. 4.83 3. 4.7r 4. 4.65 Answer: 4. 4.65 Buffer: HCzHrOz (acid) 0.21 M pKa: -log[.8 x t0-s] :4.74 CzHtOz- @ase) 0.17 M pH = pKa + tog !f'92- pH : 4.74 + tog 9Ll4' HC"H"O" ', - 0.21 M pH = 4.65 2. Calculate the pH of the buffer in the previous question after 0.035 mol of HNO: has been added to 1.0 L of the buffer. l. 4.55 2. 5.00 3. 4.48 4. 4.60 Answer: 3. 4.48 The added HNO: (acid, H*) will react with CzH:O2-, the base component of the buffer. The base component of the buffer decreases in concentration and the acid component of the buffer increases in concentration: H* + CzH:Oz- -+ HCzH:Oz 0.035 mol 0.17 mol 0.21 mol -0.035 mol -0.035 mol + 0.035 mol 0 mol 0.135 mol 0.245 mol pH = pKa * to* T*f pH = 4 7+ + ros 911:y" HC.HrO, - 0.245 M pH = 4.48 3. Which of the following mixtures would make the best buffer? l. HCzH:Oz and NH+CI 2. HCI and NaOH 3. NaCzH:Oz and NH: 4. HNO: and NaNOr 5. NH3 and NH+CI Answer: 5. NH3 and NHICI A buffer is a weak acid-base conjugate pair. HCzHrOz and NH4CI are both weak acids, NaCzHrOz and NH: are both weak bases, HCI and NaOH are a strong acid and a strong base and HNO3 and NaNO3 is a strong acid/base conjugate pair. Only NH3 and NH+CI is a weak acid/base conjugate pair. 4. Which of the following solutions would be best to buffer a solution near a pH of 4? l. 1.0 x l0-a M HCl. 2. 1.0 x l0-a M NaoH. 3. Approximately equal concentrations of HCOOH (Ka = L8 x l0-a; and NaOOCH. 4. Approximately equal concentrations of HCIO (Ka = 2.9 x l0-E) and NaClO. 5. Approximately equalconcentrations ofNHg (Kb: 1.8 x l0-s and NHaCl. Answer: 3. Approximately equal concentrations of HCOOH (Ka: 1.8 x l0-4) and NaOOCH. The pKa of the acid component of a buffer tells you the pH range at which buffer will work. Buffer pH range : pKa * l. If you want to buffer at a pH of 4, then you need a buffer with a weak acid whose pKa value is about 4. 1. This is a strong acid and cannot function as a buffer. 2. This is a strong base and cannot function as a buffer. 3. pKa: -log[Ka] : -log[1.8 x l0-4] :3.7 *'**This will buffer at a pH of 4**s'r' 4. pKa : -log[Ka] : -logl2.9 x 10-61 : 7.5 Will buffer around a pH of 7-8 5. pKb: -log[Kb]: -log[.8 x l0-5] :4.74; pKa + pKb: 14 14 - 4.74:9.26: pKa Will buffer at a pH of about 9 For NH: you are given the Kb value; you cannot use the pKb to find the buffer range,. You must find the pKa value of the acid component of the buffer. 5. Calculate the mass of NaCzH:Oz$2.0 g/mol) that must be added to 1.0 L of 0.15 M HCzH:Oz (Ka: 1.8 x 10-') to prepare a buffer solution with a pH of 5.12. l. 29.s g 2. 5.13 g 3. 17.2 g 4. 36.0 g Answer: l. 29,5g Find the pH of 0.025 M CN-, a weak base: you will need to find the Kb of this base using the Ka of the conjugate acid: Ka x Kb : Kw Kw: l.0x 10-ra (6.2x lO-roxKb) = 1.0 x lg-ra Kb: 1.6 x l0-5 CN- + HzO +> HCN + OH- weak base equilibrium 0.025 M 0 0 -X *x +X 0.025-x x x Ku : 1.6 x l0-5 : *' Assume 0.025 -x = 0.025" 0.025 - x l.6x lO-s : *' 0.025 M x:6.3 x l0-a M: [OH-] pOH: -lo9[6.3 x l0-5] pOH:3.20 14_3.20:pH:10.9 9. The indicator methyl red has a Ku value of I x l0-s. For which of the following titrations would methyl red be a good indicator? 1. 0.10 M HCzHTO: (Ku = t.8 x-I0-s) with 0.100 M NaOH 2. 0.100 M NHr (Ku: 1.8 x 10-') with 0.100 M HCI 3. 0.100 M HF (K^:7.2 x l0-" with 0.100 M NaOH 4. 0.100 M CoHsNHz (Ku:3.8 x l0-'0) with 0.100 M HCI Answer: 2. 0.100 M NHr (K6: 1.8 x lO-s) with 0.100 M HCI An indicator's color change happens-around pH : pKa + I of the indicator. pKa: -log[Ka] : -log[1 x l0-'] : 5. This indicator changes color around a pH of 4-6. Therefore methyl red will be a good indicator for a titration whose equivalence point is around 4-6. 1. and 3. are titrations of weak acids with strong bases. At the equivalence point. all of the weak acid has reacted and the conjugate base is present. Therefore, the pH is greater thanT since the solution is a weak base. Methyl red will not work as an indicator for these titrations since it will change color around a pH of 4 which is too early. 2. and 4. are titrations of weak bases with strong acids. At the equivalence point, all of the weak base has reacted and the conjugate acid is present. So the pH will be less than 7 since an acid is present. The difference between the two titrations is the Ka of the conj ugate acid present. 2. The Kb of NH:: 1.8 x l0-5. Ka x Kb : Kw Kw: l.0x l0-ra (Ka)(I.8 x l0-s): 1.0 x l0-ra Ka of NFIa* = 5'6 x l0-lo 4. The Kb of CoHsNHz :3.8 x l0-ro. Ka x Kb : Kw Kw: l.0x lO-la (Ka)(3.Sx l0-ro)= l.0x 1g-ra Ka of CoHsNH3+ : 2.6 x l1-s In both of these titrations, 0.100 M base is titrated with 0.100 M base. Since the molarities of the acid and base is equal in each titration, equalvolumes of acid and base will be required to reach the equivalence point. So the final volume will be twice the initial volume of the base to be titrated. Therefore, the conjugate acid produced will have a molarity that is half the molarity of the base: 0.050 M. Calculate the pH of 0.050 M NFIa* and then of 0.050 M CoHsNHr* NFIn*: -.2 5.6 x l0-ro 0.050 M x: 5.3 x 10-6 M: [H*] pH: 5.3 Methyl red turns color at a pH:4-6 so this indicator will be turning color at the equivalence point of this titration. CoHsNH:*: -.2 2.6 x 10-' 0.050 M x: 1.1 x 10-3 M: [H*] pH:3 Methyl red turns color at a pH:4-6 so this indicator will be turning color after the equivalence point of this titration. 10. Calculate the solubility of Ag2CrOa in water. (K,p: 2.6 x l0-t2) 1. 8.1 x l0-7 M 3. 1.6 x 10-6 M Answer: 4. 8,7 x 10-s M 2. 1.1x l0-a M 4. 8.7 x lO-s M The equilibrium reaction is Ag2CrOa (s) e 2Ag*(aq) + CrO+-(aq) I- 00 C -x + 2x + x l:2:l mole ratio E 2xx When x amount of Ag2CrOa dissolves, 2x amount of Ag* is produced and x amount of CrO+- is produced due to the mole ratio. Ksp = [eg-]'[ CrOa-] 2.6xlo-r2:12x121x1 2.6xlA-tz:4x3 6.5 x 10-13: x3 8.7 x 10-s: M I l. Calculate the solubility of CaF2 (K,p = 4.0 x 10-1r; in a 0.050 M NaF solution. 1 l.0 x 10-6 M 3. 3.4 x 10-6 M 2. 8.0 x 10-ro M 4. 1.6 x l0-8 M Answer: 4. 1.6 x 10-8 M CaFz is only slightly soluble but NaF is completely soluble: NaF(aq) ---) Na+(ad + F-(aq) 0.050 M 0 0 -0.050M +0.050M +0.050M 0 0.050 M 0.050 M So you are dissolving CaF2 in a solution of 0.050 M F-. The equilibrium reaction is CaF2 (s) ++ Car*1aq.; + 2F'(aq) I- C-x 0 0.050 M +x *2x 1:l 2moleratio E - x 2x+0.050M (Note the two sources of F-: x amount from CaFz and 0.050 M from NaF) Ksp -- [ca2.][ r'12 4.0 x 10-r' : [*][ 2x + 0.050]2 AssumeZx+ 0.050: 0.050 M 4.0 x 1o-'1 :1x11o.oso12 1.6 x 10-8 M: x 12. AgI (K,o:8.3 x l0-r7; is least soluble in 1. water 2.0.50MNaCl 3. 0.10 M AgNO3 4. 0.050 M KI Answer: 3. 0.f 0 M AgNO3 2. Ag*(aq) + Cl-(aq) ---+ AgCl(s) 2 moles of aqueous solution I mole of solid The solid on the product side has lower entropy than the aqueous solutions on the reactant side, so entropy has decreased. 3.2H2G) + Oz(g) --' zHzO(E) 3 moles of gas 2 moles of gas Since there are more moles of gas on the reactant side, the reactants have more entropy than the products and entropy has decreased 4. NHr(g) + HzO($ ---' NHa*(aq) + OH-(aq) I mole gas + I mole liquid 2 moles of aqueous solution I mole of high entropy gas has been converted to lower entropy aqueous solution. Entropy has decreased. 5. Na2CO3(s) + 2HCl(aq) ---+ 2NaCl(aq) + HzO(O + COz(S) I mole solid + 2 moles solution 2 moles solution, I mole liquid,l mole gas Low entropy solid is converted to I mole of very high entropy gas. Entropy has increased. 17. Which of the following statements is always true? 1. lf K> 1, AGo:0 3. IfK<1, AGo<0 5. IfK<I,AGO:Q Answer: 2. If K> l, AGo < 0 2. lfK>l,aco<0 4. IfK>1,AGo>0 A negative AGo value (AG" < 0) means that the reaction is spontaneous. Spontaneous reactions have large values of K (K > l) since reactants are consumed and products are formed. Therefore the [product]/[reactant] ratio is large (large amount of products, small amount of reactants). A positive AGo value (AGo > 0) means that the reaction is NOT spontaneous. Nonspontaneous reactions have smallvalues of K (K < 1) since reactants are not consumed and products are not formed. Therefore the [product]/[reactant] ratio is small (small amount of products. large amount of reactants). 18. Consider the following reaction: 2C(s) + Hz(g) --) CzHz(g) AGo : 209.2 kJ at 25.}uC. Calculate AG when the pressure of H2 : I 00 atm and the pressure of CzHz:0.10 atm. 1. l92kl 2. 208k1 3. 232k1 4. r77 kJ Answer: l. 192 kJ You are being asked to calculate AG at nonstandard conditions - the pressures for the reactants and products are not I atm. Use the relationship AG=AG" +RT ln [products] AGu = 20g.2kJ: T: 25uC +273 :2ggK Ireactants] LG =209.2kJ + (8.3 14 x l0'3kJ/KmolX29SK;tn!9-Hr PH: LG = 209.2 kJ + (8.3 l4 x I 0-3 kJ / KmolX2gsrl m 9J 9 AG = 192 kJ 19. Which of the following reactions has the largest equilibrium constant, K, at25"C? l. 2HzO(g) - 2Hz(e) + OzG) 2. Hz(g) + coz(g) -.-- Hzo(g) + co(g) 3. 1\.111*; + Oz(s) -:-- 2 NO(g) 4. CO(g) + Clz(g) -\ COCI2@) AGo : +458 kJ AG":'24.6kJ AGo : +173 kJ AGo: -341 kJ Answer: a. Co(g) + Clz(g) -L COCI2(g) AGo: -341 kJ AGo : -RT ln K A large equilibrium constant means that the [product]/[reactant] ratio is large (large amount of products, srnallamount of reactants). This occurs when reactants are consumed and products are formed. ln other words, when a reaction is spontaneous (AG" = negative), K is large. The more negative the value of AG" :, the more spontaneous the reaction and the larger the equilibrium constant. Reaction 4 has the largest K value since AGo is the most negative for this reaction. , 20. Consider the following reaction for which AHo : +170.44kJ: i ua(OH)z'8HzO(s) + 2NF{+NO:(s) ---+ Ba(NOs)z(aq) + 2NH3(g) + lOH2O(l) Which of the following statements must be true for this reaction? l. Enthalpy and entropy are both favorable for spontaneity. 2. Neither enthalpy nor entropy is favorable for spontaneity. 3. Enthalpy is favorable and entropy is not favorable for spontaneity. 4. Enthalpy is not favorable and entropy is favorable for spontaneity. Answer: 4. Enthalpy is not favorable and entropy is favorable for spontaneity. AHo is positive. The reaction is endothermic. A positive value does NOT favor spontaneity, ASo is also positive since three moles of low entropy solid are converted to higher entropy liquids. Entropy increases. A positive entropy change does favor spontaneity. 21. Which of the following lists the compounds in order of increasing molar entropy at25oC? l. ZnS(s) < HzO(l) < C:Hs(g) < CzH+(g) 2. CzH+(g) < H2O(l) < CsH8(g) < ZnS(s) 3. ZnS(s) < C:Hs(g) < CzHl(g) < HzO(g) 4. ZnS(s) < HzO(l) < CzH+(g) < CrHs(g) Answer: 4. ZnS(s) < HzO(4 < CzHr(g) < CrHs(g) First note that So sold ( So I'qu,d ( Sosu, So ZnS(s) should have the lowest entropy value of the choices; the liquid HzO would have more entropy but not nearly as much entropy as the two gases, CzFI+(g) and C:Hs(g). Of the two gases, CrHr has more entropy than CzH+ because there are more bonds in C3Hs. 22. A particular reaction has a AGo value of -100. kJ. Which of the following MUST be true? l. AHo is negative. 2. The reaction is fast. 3. The reaction is spontaneous. 4. ASo is positive. 5. l, 3 and 4 must be true. 6. Allof these must be true. Answer: 3. The reaction is spontaneous. The values of AGo is negative. A negative AGo value means that the reaction is spontaneous For a spontaneous reaction such as this one, ASo and AHo can be either positive or negative. A spontaneous reaction can have a negative value of ASo if the AHo is negative; but the value of ASo may be positive if AHo is positive. The sign of AGo does not tell us anything about the rate of the reaction.