Formular weight of K 2 Cr 2 O 7 = 2( ) + 2( ) + 7( ) = (7 S.F.) (1 pt/each factor)

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1 Chem 222. Midterm Exam Answer Key Q1. a) Using the proper number of significant figures what is the formula weight of K 2 Cr 2 O 7 and the uncertainly? Use atomic weights as: K ± , Cr ± , and O ± (7 pts) Formular weight of K 2 Cr 2 O 7 = 2( ) + 2( ) + 7( ) = (7 S.F.) (1 pt/each factor) Uncertainly = [2(0.0003) 2 + 2(0.0006) 2 + 7(0.0003) 2 ] 1/2 = (3 pts for uncertainly) Formular weight of K 2 Cr 2 O 7 = ± Wrong significant figure (S.F.) 1 pt/ each answer of formular weight and uncertainly b) How many grams of K 2 Cr 2 O 7 are required to prepare ml of M Cr 2 O 7 2-? (8 pts) L * mol/l * 1 mol K 2 Cr 2 O 7 * g/mol = = 3.68 g 1 pt/each factor and 4 pts/final answer Wrong significant figure (S.F.) 1 pt c) Sometimes we will dry chemicals such as KHP, EDTA before we use them, Is this process (drying) necessary to do? Why or why not? (5pts) Yes, if those chemicals are the primary standard and hydroscopic compound then you need to dry them first before you use them. Because their masses will be interfered by air moisture (water), your results will not be nearly 100% accurate. No, if you correct mass of those chemicals with buoyancy equation and use true mass for calculation such as you did with Tris in exp.5 If your answer is either of these answers, you get 5 pts If your answer mentions to other purities or other substances, you have indicate them: What are they?, Why? And How? Up to 4pts take off, it depends your answer Totally wrong (No pts)

2 Q2. 2. A certain procedure required the preparation of ~0.1 M HCl and ~0.01 M NaOH. The molarities of these solutions were determined by standardization with Na 2 CO 3 and KHP, respectively. Four titrations of Na 2 CO 3 with ~0.1 M HCl and KHP with ~0.01 M NaOH were performed, and the calculated molarities for the HCl and NaOH solutions were as follows: , , , and mol/l for HCl solution; , , , mol/l for the NaOH solution. a. Perform the Q test to determine if it is necessary to discard any of the calculated molarities for HCl or NaOH. (6 points) For HCl: Q calc =gap/range=( )/( ) =0.43 keep(<q table (0.76)) Q calc =( )/( ) =0.14 keep(<q table (0.76)) +2 for performing Q test for HCl (+3 if only Q test for HCl) For NaOH: Q calc =( )/( ) =0.95 discard(>q table (0.76)) Q calc =( )/( ) =0.02 keep(<q table (0.76)) +2 for performing Q test for NaOH (+3 if only Q test for NaOH) Discard M +2 for Q test result (+3 if only result for HCl or NaOH) Q (90% confidence) Number of Observations b. Calculate the mean molarities and standard deviations of the HCl and NaOH solutions. (8 points) x avg (HCl)=Σx i /N=( )/4=0.1011M *Answer same if Q test was or was not performed on HCl. s(hcl)= [Σ(x avg -x i )/(N-1)]= [( ) 2 + )/(4-1)]= *Answer same if Q test was or was not performed on HCl. x avg (NaOH)=( )/3= M *x avg = M if Q test was not performed on NaOH.

3 s(naoh)= [( ) 2 + )/(3-1)]= *s= if Q test was not performed on NaOH. c. Using the mean molarities of the HCl and NaOH solutions, calculate the H + concentration and the ph in each solution. K w =1.01*10-14 (6 points) M HCl H + +Cl - [H + ]=0.1011M since HCl completely dissociates ph=-log(0.1011)= for correct answer and setup M NaOH Na + +OH - [OH - ]= M since NaOH completely dissociates [H + ]=1.01*10-14 / M=9.21*10-13 *[H + ]=9.09*10-13 M if M was used for [NaOH]. ph=-log(9.21*10-13 )= for correct answer and setup *ph= if 9.09*10-13 was used for [H + ]. only 1 point off for SF for entire question

4 Q3 1) The formation constant (K f ) of CaY 2- is Calculate the concentration of free Ca 2+ in a solution of 0.15 M CaY 2- at ph 10. At ph 10, α 4- Y is You may use that [Ca 2+ ] is much smaller than 0.15 M and so 0.15 M- x ~ 0.15 M, where [Ca 2+ ] = x. 4- K f = K f * α Y (2pts) K f = [CaY -2 ] / ([Ca 2+ ] [EDTA]) (2pts) K f * α Y 4- = [CaY -2 ] / ([Ca 2+ ] [EDTA]) (2pts) 4.9x10 10 * 0.36 = {[0.15 x] / ([x] [x])} (2pts) [0.15 x] = [0.15] 1.764x10 10 = 0.15/x 2 (2pts) (No more than 6pts for eqs.; 3 2 pts) x = 2.9x10-6 M (2pts) SF for square roots was not covered. So no points off for SF on this answer. 2) A solution containing 25.00mL of ~5 M HCl is titrated by adding 50.00mL of 3.267M NaOH, and then the excess NaOH is titrated with 23.24mL of a KHP solution (MW= ). The KHP solution was prepared by dissolving g of KHP in 40.00mL of H 2 O. a) Calculate the Molarity of the KHP solution. 1pt for Eq. 3pts for Answer g KHP / (g/mole)(1/ l) = M b) Calculate the number of moles of NaOH added. 1pt for Eq. 3pts for Answer L * 3.267M NaOH = moles NaOH c) Calculate the number of moles of HCl in the initial solution. Moles of HCl = moles of initial NaOH - moles of excess NaOH (2pts) moles inaoh moles enaoh = moles HCl moles HCl / L = M (2pts)

5 Q4 (1) Total 20 points (a) 5 points (b) 10 points (c) 5 points A compound has a molecular mass of was dissolved in a 5-ml volumetric flask. A 1.00 ml aliquot was withdrawn, placed in a 50.0-ml volumetric flask, and diluted to the mark. The absorbance at 340 nm was in a 1.00 cm cuvet. The molar absorptivity for this compound at 340 nm is 6454 M -1 cm -1. A blank solution was also measured at 340 nm and the absorbance of the blank was found to be a) Calculate the concentration of the compound in the cuvet. A=( ) = (1point) A=εbC (1point) C=0.414/(6454M -1 cm -1 )(1.00cm) =6.41*10-5 M (1 point for math, 1 point for answer, 1 point for unit) (-1 point if you don t have the right Sig.Fig) b) How many milligram of compound were used to make the 5-ml solution? Sample had been diluted *50.0ml/1.0ml C=(50/1)(6.41*10-5 M = M (3 points for dilution, 1 point for answer) X g/(450.83gmol -1 )(5.0*10-3L)= mol L -1 ( 4points for calculations) X=0.0072g = 7.22mg (1 point for answer, 1 point for units) c) If one prepare a M standard solution of the same compound. What is the absorbance of the standard solution measured at 340 nm? A=εbC A-0.013=(6454 M -1 cm -1 )(1.00cm)(0.005M) A=32.3 (1 point) (2points for math, 1 point for A=A-A b ) (1 point)

6 Q5. 1) (a) 8 points (b) 4 points 2) 8 points 1) Consider the titration of ml of M Ce 4+ in 1 M HClO 4 by M Fe 2+ to give Ce 3+ and Fe 3+, using Pt and calomel reference Hg Hg 2 Cl 2 electrodes to find the end point. E = Log([Fe 2+ ]/[Fe 3+ ]) (V) E = Log([Ce 3+ ]/[Ce 4+ ]) (V) a) Calculate E at the following volume (V Fe ) of Fe 2+, V Fe = 2 ml and 30.0 ml. b) Calculate the concentration of Ce 3+ when V Fe = 10.0 ml. For a mistake in the significant figure, 1 point is subtracted. 1. a) First, obtain V Fe at the equivalence point (V e ). Ve = C Ce4+ V Ce4+ /C Fe2+ = ml*0.0100m/ M =20.0 ml (2 points) When V = 2 ml, Use [Ce3+]/[Ce4+] and obtain the ratio. E = Log([Ce 3+ ]/[Ce 4+ ]) = Log{(V/(V-Ve)) = Log{(2/18)} (2 point) (If [Ce 3+ ] and [Ce 4+ ] are calculated without V e (3 points) ) Note: No points if the equation for Fe is used or the concentration ratio is incorrect. E = V (Correct calculation 1 points) When V = 30 ml, Use [Fe2+]/[Fe3+] and obtain the ratio E = Log([Fe 2+ ]/[Fe 3+ ]) = Log((V-Ve)/Ve) = Log((V-Ve)/Ve) = Log((10.0 ml)/(20.0 ml) (2 points; No points if the equation for Ce is used) = V (1 point)

7 1. b) Calculate the concentration of Ce 3+ when V = 10.0 ml. [Ce 3+ ] = [Ce 4+ ] (V- V e = V) [Ce 3+ ] = ( M ml M 10.0 ml) /( ) ml (3 points) = moles /0.11 L = mm (Calculation 1 point) Q5. 2) Find the ph and concentration of (CH 3 )N and (CH 3 )NH + in a M solution of trimethylaminium chloride ((CH 3 )NH + Cl - ). The pka of (CH 3 )NH + is You can neglect the effect of Cl -. Ka = 10 -pka = = is incorrect) (2 points; No points if the definition of pka (CH 3 )NH + Ka (CH 3 )N + H + F x x x F = M Ka = x 2 /(F-x) ~ x 2 /F (assume F >> x) x = (KaF) 1/2 = ( ) 1/2 ( 3 points; No points if Ka is not obtained with a proper equation; 2 points if the calculation of K a is incorrect) x = M << F (1 point) Or you can use x 2 + K a x K a F = 0. x = {-K a ± (K a 2 + 4K a F) 1/2 }/2 = (3 points) ph = -Log(x) = -Log( ) (1 point) (no points if x is not properly obtained) ph = (1 point) Alternative answer ph = pka + Log[CH 3 N]/[CH 3 NH + ] (2 points) (10-16) = pka + Log[CH3N] Log[CH3NH+] ph = pka - ph log(f-x) (ph = -log[x]) ph = {pka -log(f-x)}/2 (4 points) ~ (pka - logf)/2 = {9.800 log(0.080)}/2 = X= 10 -ph = (2 point) If you just answer [(CH 3 )N ] = F without answering as indicated above (2 points).

8 Q6. Extra question Total 5 points g sample contained only NaCl and KBr. It was dissolved in water and required 2.00 L of M AgNO 3 to precipitate all Cl - and Br -. The molecular weights of NaCl and KBr are 58.4 and 119.0, respectively. Calculate how many moles of KBr are contained in the solid sample. The question is from Home work Assume the contained KBr is x mmol. Ag - +Cl - AgCl Ag - +Br - AgBr Total Ag moles is 2.00 L M = 3.00 mmol (2 points) Total weight of NaCl and KBr is x mg (3.00 x) mg = mg (2 points) ( )x = x = 90.9/( ) = mmol (1 point) No points will be given to 3.00 mmol/ 2 =1.5 mmol. The assumption that the AgCl and AgBr are equal in the number of moles is incorrect.

Chem 222 #20 Ch 11 Nov 1, 2004

Chem 222 #20 Ch 11 Nov 1, 2004 Announcement The answers for Midterm will be uploaded sometime tomorrow. Please check your grade and talk to your TA about general questions If you have specific questions