CHEMISTRY 31 FINAL EXAM - SOLUTIONS Dec. 10, 2012 150 points total A. Multiple Choice and Short answer Section. Multiple Choice Answers in bold 1. A student calculates their unknown soda ash % NaCO 3 as 38.331 + 0.091%. When expressed with the proper number of significant figures, this should be: a) 40 + 0.09% b) 38.3 + 0.1 c) 38.33 + 0.09 d) 38.331 + 0.091% 2. It is difficult to prepare sodium hydroxide solutions of known concentration by weighing solids (because solid NaOH absorbs water so quickly). What can be done to determine the concentration of NaOH for use in accurate titrations? _Perform a standardization titration 3. A sample is known to have a matrix effect (a different slope results if using the sample matrix instead of water to make a calibration curve). One should use a calibration method. a) standard addition b) external standard c) internal standard d) one point calibration curve 4. Write one complex ion forming reaction used in a quantitative analysis lab. Reaction: 1) AgCl(s) + 2NH 3 [Ag(NH 3 ) 2 ] + + Cl -, 2) Ca 2+ + Y 4- CaY 2- (Y = EDTA) 3) Mg 2+ + HIn 2- MgIn - + H + 4) Co 2+ + 6H 2 O [Co(H 2 O) 6 ] 2+ 5. When ultra-violet light interacts with matter, the predominant type of transition is: a) nuclear level transitions b) electron level transitions c) molecular vibrational transitions d) molecular rotation transitions 6. In gas chromatography, the two factors affecting a compound's retention factor are: a) the compound's volatility and polarity b) the compound's polarity and the mobile phase's polarity c) the compound's volatility and the flow rate d) the carrier gas used and its flow rate 7. Which of the following mixtures will make a traditional buffer solution: a) 0.010 M HCl + 0.010 M NaOH b) 0.010 M HCl + 0.0050 M NH 4 Cl c) 0.030 M HCl + 0.008 M NaCH 3 CO 2 d) 0.0040 M HCl + 0.0090 M NH 3 (NH 4 + and CH 3 CO 2 H are weak acids) 8. A chemist is measuring the concentration of benzoic acid (a weak acid) in sodas using UV absorption. Both benzoic acid standards and soda sample are diluted in a buffer. The main purpose of the buffer is to: a) keep the fraction of benzoic acid in the ionized form constant b) provide additional ions to make ion-pair species that absorb light c) to increase the ionic strength of the solution d) to keep the acid from eating through the cuvette walls 1
9. The compound 2-aminophenol, which exists as NH 3 C 6 H 4 OH + in its most acidic form, has pk a1 and pk a2 values of 4.70 and 9.97 for the functional groups -NH 3 + and -OH, respectively. What would be the best ph to use as a buffer so that 2-aminophenol can be retained using anion chromatography (where it is retained as an anion)? a) 2.0 b) 5.0 c) 7.0 d) 11.0 10. When a weak base is titrated with a strong acid, the ph at the equivalence point will be a) below 7 b) exactly 7 c) greater than 7 11. Based on the titration curve shown to the right (with grey box showing where indicator changes color), the titration can be considered to be: a) precise and accurate b) precise but not accurate c) accurate but not precise d) neither accurate nor precise ph V(strong base) Problem Section. Show all needed calculations to receive full credit. The number of points are shown in parentheses. Use the back side of the page if needed. 1. Calculate the concentration of a 6.0 M formaldehyde (H 2 CO) solution in mass percent. Atomic weights (H, C, and O are 1.01, 12.01, and 16.00 g/mol, respectively). The density of the 6.0 M solution is 1.06 g/ml. (8 pts) mass % = (g H 2 CO/g sol'n) 100 mass % = (6.0 mol /L sol'n)(30.03 g H 2 CO/mol H 2 CO)(1 L/1000 ml)(1 ml/1.06 g) 100 mass % = 17% 2
2. Calcium fluoride is a sparingly soluble salt that dissociates in water as: CaF 2 (s) Ca 2+ + 2F -. The K sp for this reaction is 3.9 x 10-11. For both parts of this question assume F - and Ca 2+ do NOT significantly react with water or form an ion pair. a) Calculate the solubility of calcium fluoride in water without considering the effects of ionic strength of the dissolved ions. (8 pts) This can be set up using an ICE approach: CaF 2 (s) Ca 2+ + 2F - initial 0 0 change +x +2x equilibrium x 2x K sp = 3.9 x 10-11 = [Ca 2+ ][F - ] 2 = (x)(2x) 2 x = solubility x = (K sp /4) 1/3 = 2.1 x 10-4 M b) Determine the solubility of calcium fluoride in 0.050 M KNO 3 if the activity coefficients, γ(ca 2+ ) and γ(f - ), are 0.485 and 0.81, respectively, in the KNO 3 solution. (6 pts) The set up will be the same as in a) except that the equilibrium equation is modified to: 3.9 x 10-11 = γ(ca 2+ )[Ca 2+ ]{γ(f - )[F - ]} 2 = 0.485(x)(0.81) 2 (2x) 2 x = {3.9 x 10-11 /[4 0.485(0.81) 2 ]} = 3.1 x 10-4 M 3. A compound is known to have a molar absorbtivity of 731 + 1 M -1 cm -1 at a wavelength of 382 nm in water (solvent). A cell with path length of 0.200 + 0.005 cm is filled with the compound and the absorbance is measured to be 0.103 + 0.004. Determine the concentration of the compound and the absolute uncertainty in the concentration and report the numbers with the correct number of significant figures. (16 pts) A = ε b C or C = A/(ε b) = (0.103 + 0.004)/[(0.200 + 0.005 cm)(731 + 1 M -1 cm -1 )] C = 7.045 x 10-4 M S C /C = [(0.004/0.103) 2 + (0.005/0.200) 2 + (1/731) 2 ] 0.5 = 0.0462 S C = C(0.0462) = 3.3 x 10-5 M To report this with the correct number of significant figures, we want to have 1 sig fig in the uncertainty, to go to the same place in the value and to have the same exponent: This works out as 7.0+0.3 x 10-4 M 3
4. Using the HPLC chromatograms for separating components A and B below and the fact that the separation occurring on column 1 (lower chromatogram) was a reversed phase separation, answer the questions below. Also, the flow rate for both columns was 1.5 ml/min. and the small first peak seen is for unretained compounds. (12 pts) 3.5 3 2.5 A B response 2 1.5 1 B A column 1 column 2 0.5 0 0.0 0.5 1.0 1.5 2.0 2.5 time (min.) Determine: a) Which compound is more polar: A or B? Explain your answer. B it elutes earlier on the reversed phase column. Reversed phase (non-polar) will retain less polar compounds more b) The retention factor (k) for A for the separation on column 1. k = (2.1 0.5)/0.5 = 3.2 c) Which column has better resolution? Explain your answer. Column 1. The response returns to the baseline between the two peaks. Bonus 1)Is the separation occurring on column 2 most likely reversed phase or normal phase? You can assume that the main factor affecting separation for both columns is polarity. Explain your answer. (3 pts) Normal phase because the elution order is reversed from the reversed phase. 4
5. A solution of 0.050 M NaBrO is made by dissolving NaBrO into 1.00 L of solution. The K a of HBrO (the conjugate acid of BrO - ) = 2.3 x 10-9 and K w = 1.0 x 10-14. Do not consider activity or any metal complexes to solve this problem to 4 significant figures. a) Determine the ph of the solution. (10 pts) NaBrO dissolves and dissociates to give [BrO - ] = 0.050 M (theoretical concentration, before BrO - reacts further) Since HBrO is a weak acid, BrO - is a weak base and will react as one, allowing use of the ICE method to determine the ph: BrO - + H 2 O(l) HBrO(aq) + OH - initial 0.050 0 0 change - x +x +x equilibrium 0.050 x x x K = K b = K w /K a = 4.348 x 10-6 = [HBrO][OH - ]/[BrO - ] = x 2 /(0.050 x) can ignore x in 0.050 x so x = [(4.348 x 10-6 )(0.050)] 0.5 = 4.66 x 10-4 M (which is << 0.050) [OH - ] = 4.66 x 10-4 M or poh = 3.33 ph = 14 poh = 10.67 b) If 25 ml of 0.820 M HNO 3 is added to the above solution, what will be the ph? (12 pts) Once HNO 3 is added, we convert some of the BrO - to HBrO, which we can determine by making a mole table: The moles of BrO - = (0.050 mol/l)(1.00 L) = 0.050 mol. The moles of HNO 3 = (0.025 L)(0.820 mol/l) = 0.0205 mol BrO - + H + HBrO(aq) initial 0.050 0.0205 0 change -0.0205-0.0205 +0.0205 full right 0.0295 0 0.0205 Now, the Henderson-Hasselbalch equation can be used: ph = pk a + log[n(bro - )/n(hbro)] ph = 8.638 + log(0.0295/0.0205) = 8.80 6. Glycine, NH 3 + CH 2 CO 2 -, is an amino acid with an acid (-CO 2 H) and base ( NH 2 ) functional group. If the pk a s for the CO 2 H functional group and NH 3 + (conjugate acid of the base functional group) are 2.35 and 9.78, respectively, calculate the ph and concentration of NH 3 + CH 2 CO 2 H (acid form) when 0.010 moles of glycine are used to make a 1.00 L solution. (12 pts) The source of glycine is the intermediate form (HL), meaning we can use the equation for [H + ]: [H + ] = [(K a1 K a2 [HL] o + K w [HL] o )/(K a1 + [HL] o )] 0.5 [H + ] = [(10-2.35 10-9.78 0.010 + 10-14 0.010)/(10-2.35 + 0.010)] 0.5 [H + ] = (7.513 x 10-15 /0.01447) 0.5 = 7.21 x 10-7 M or ph = 6.14 K a1 = 0.00447 = [H + ][HL]/[H 2 L + ] or [H 2 L + ] = [H + ][HL]/0.00447 = (7.21 x 10-7 )(0.010)/0.00447 [H 2 L + ] = 1.6 x 10-6 M 5
Bonus #2) 10 ml of the above solution were added to 90 ml of a buffer. The concentration of the base form of glycine, NH 2 CH 2 CO 2 -, was measured and found to be 1.2 x 10-4 M. What is the ph of the buffer assuming the addition of glycine did not affect the ph of the buffer? (3 pts) [HL] o = 0.010*10/100 = 0.0010 M Can solve this by 1) using the H-H equation or 2) using α(l - ) to determine ph Method 1) ph = pk a2 + log{[l - ]/[HL]} we also need to assume [H 2 L + ] is insignificant [L - ] = 1.2 x 10-4 M and [HL] = 0.0010 1.2 x 10-4 M = 0.00088 M ph = 9.78 + log(1.2 x 10-4 /0.00088) = 8.91 Method 2) α(l - ) = 1.2 x 10-4 /0.0010 = 0.12 = K a1 K a2 /{[H + ] 2 + K a1 [H + ] + K a1 K a2 } 0.12 = 7.413 x 10-13 /([H + ] 2 + 0.00447[H + ] + 7.413 x 10-13 ) 0.12[H + ] 2 + 5.36 x 10-4 [H + ] + 8.90 x 10-14 = 7.413 x 10-13 [H + ] 2 + 0.00447[H + ] 5.44 x 10-12 = 0 or [H + ] = {-0.00447 + [(0.00447) 2 4(1)( 5.44 x 10-12 )] 0.5 }/2 [H + ] = 1.4 x 10-10 M; ph = 9.84 (method 2 is not as accurate because we need to subtract two similar values in the quadratic equation). We can get a better answer by removing the [H + ] 2 term, which results in ph = 8.91 7. A 50.0 ml aliquot of an unknown solution containing the weak acid, formic acid (HCO 2 H, with K a =1.80 x 10-4 ) is titrated with 0.100 M NaOH and found to require 33.3 ml to reach the equivalence point. Calculate: (22 pts) a) The concentration of formic acid in the unknown solution. At equivalence point, n(hco 2 H) = n(naoh) or [HCO 2 H]V HCO2H = [NaOH]V NaOH [HCO 2 H] = (0.100 M)(33.3 ml)/(50.0 ml) = 0.0666 M b) The ph at the equivalence point. At the equivalence point, the forward reaction, HCO 2 H + OH - HCO 2 - + H 2 O, has gone to completion. We need to look at the backwards reaction to determine [OH - ] and the ph. The backwards reaction, HCO 2 - + H 2 O HCO 2 H + OH -, tells us we have a weak base problem. To solve the weak base problem, we need to 1) determine [HCO 2 - ] o (conc. after mixing and reacting full to the right but before we consider the backwards reaction starting and 2) use the ICE method. 1) [HCO 2 - ] o = n(hco 2 H)/V Total = (0.100 M)(33.3 ml)/(33.3 ml + 50.0 ml) = 0.0400 M HCO - 2 + H 2 O HCO 2 H + OH - Intitial 0.0400 0 0 Change -x +x +x Equil. 0.0400 x x x K = K b = K w /K a = 5.556 x 10-11 = [HCO 2 H][OH - ]/[HCO - 2 ] = x 2 /(0.040 x) 5.556 x 10-11 = x 2 /0.040 (assume x << 0.040) or x = (2.222 x 10-12 ) 0.5 x = [OH - ] =1.49 x 10-6 M or ph = 14 + log[oh - ] = 8.17 6