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Qz2KeyR - CHM 2046 QUIZ 2 50 MINUTE TIME LIMIT T DIRECTIONS...

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Unformatted text preview: CHM 2046 QUIZ 2 50 MINUTE TIME LIMIT T: 10/7/08 DIRECTIONS: Same rules as for uiz 1. Point value for each is in the ieft ma in. Name (pledged): 9514 (if? a DISC Instr. Name: Hgdfi Meflméyeufl 1. The olubility of CaF2(s) at 25 °C in 1.0 M Cam2 is about the same as it is in aqueous NaF soin at (12) @4 x10'3 M (2) 0.50M (3) 5.8 x 10"!" (4) 2.0M (5) 0.06001. 2. For 0.0010 M Nazco3 the correct, “charge balance” relationship for principal species is (12) (1)[Na*1= 2030321 (2) [Nail + mm = [6032'] @Naq = mag-1+ [H.203] + [HO'] (4) 2(Na‘] = [0032'] + [HCO3'] + [H0'] (5) 2(Na”) + [H3 1 = 2[co,2‘] + [HCOa‘] + [H0‘]. 3. Indicator HIN is yellow (Y) at pH 4.7 or less, green (G) at pH 5.3, and blue (B) at pH 5.9 or above. So, if HIN as a BIL acid is ten times weaker than HX, and treating 100 mL of a colorless HX soln with one drop HIN indicator gives a distinct YG color, M, for the HX soln ii closest to (12) (1)3x10'5M (2)1x10'3M (3)3x10‘5M _ x10'5M (5)5x10'4M. 4. if, for aqueous (H0)2802 soln, [H3041 = 0.8 5042'], then, for this soln (12) (1) pH is ~3.0 (2) pH is ~4.0 @ is ~2.0 (4) pH is ~2.5 (5) pH is ~3.5. 5. If a buffer soln is prepared by mixing 20.0 mL 0.50 M HAc with 40.0 mL 0.45 M NaAc, the amount of 4.0 M HAc which must -_-. ._dded to this buffer to dOuble [H30+] is ciosest to (12) (1) 7.0 mL : mL (3) 5.0 mL (4)1.0 mL (5) 3.5 mL. 6. Consider: Ca003(s) + HZO(I) ‘-—- CaZ+(aq) + HCO3'(aq) + H0"(aq). if, in this system, [Ca2+] = 4[HCOs'], and the system is basic, [HO'] for the system is closest to (12) (1) 6.8 x10"5 (2) 3.2 x10"8 @3 x10"1 (4) 3.5 x10’3 (5) 2.7 x 10's. 7. The foliowing mixture which results in a large extent BIL acid-base rxn is (12) (1) HF(I) &1 M HCN (2) 603(3) & 0.01 M “H2603” (3) FeCl3(s) & 1 M Nazso, (4) 1 M NaHS . M Nazo (5) 0.8 M HOCI 8: KF(s). _---—--------—-------—-—--- _--- —_-—-_-_ _——.-.-----—b.dau.wan—-u.u.-—-ni.a--—-m.hu-..anuhum-------—--u—---—---------u---—--n-- ---------------- ------ 8. Consider: “H2003”(aq) + Ca003(s) + SO42'(aq) + 2 H200) = CaSO4°2H20(s) + 2 HCOflaq). This EQ can be established by bubbling C02(g) into 1 M Na2804 slurried with Ca003(s). i. Write and calculate K for this EQ. For K calculation be sure to indicate ail K's (such as any Ka, Kb, Ksp, or Kw) of which your overall K is composed. (Note: The correct value for K is within 1.0 - 9.0) “K “’fl i {2.343 7, i’ 1'] (FY1501? ,2 W __ (15) Y9“? mg)? {504.1%} [Gnu] {Cd-32"] E13353 («spasoy 2W)H<RHC033 _ (5,?21'115-61 .27fil5'10 : GLO (“dam I _. 1 191993;) UL) it w“ 1)“), 9 x it“) Wm§ 23K“; _ ii. Consider mlxmg 1.0 mole CaCO3(s) with a 1.0 L soln containing [304 ] at 1.0 and [H003 ] at 0.10. If the system is closed and 1.0 mols 602(9) is bubbled in, caiculate mols CaSO4-2H20(s) that can be obtained. Show pertinent work. (If you couldn’t solve part i., use K = 5.0.) . a m. __ ’3— ' ‘2— «”6“, we .... (cans-213 25(ng E-F" (15) [ng’l {362'} (1.0vx)tl.o~qé§ R ”l? ‘7' w :2, ass—2.9.5 x = 0 .10 MM 79 4.05;): 23.35“ 1.0-— 74 x: 0.52 2': Wu Canvas—ems) awnabb iii. Quaiitatively, what wiii happen to this system if, after the change applied in ii., excess Ca(0H)2(s) 'ddd?Wh?-.-.. , " ' i " his; scam): H0 Mn Ww. song at “when? WWW-w) -e ed) 47> C03 NWJ), 0M Wyatt Need w. my (4300),, 0d omits/15' do 010% --——— (15) . g". 3mm) 5354mm win be C3603 .3650“,sz L: my 50., C; (an), (at) [W cm Mai/wax) iv. Write the correct equation for each rxn of consequ‘ence which occurs if he, treated with excess 1 M HF . (Hint: You should write three equations.) 6) “9:06) «i 000;“? évflgwgafli "1' 12:35.5) 2HFufl+ (‘9‘ ld?fi%)d Ckwgtk) 9t¢?x14—'l+“)11033w%9 60(3):“) 1' “flea-1H- 030%; an “'20 m +00...“ "1' 0.0 U!) .2 wola- wag can...“ «i Turf] 2. 2 tea“? a! system is CHM 2046 QUIZ 2 50 MINUTE TIME LIMlT T: 10I7I03 DIRECTIONS: Same rules as for uiz 1. Point value for each is in the left mar in. Name (pledged): Mm «gel: DISC Instr. Name: flay RM 1. If, for aqueous (H0)2SO2 soln, [H8041 = 0.8[8042'], then, for this soln ([email protected]‘| is ~2.0 (2) pH is ~3.0 (3) pH is 14.0 (4) pH is ~3.5 (5) pH is ~2.5. 2. If a buffer soln is prepared by mixing 20.0 mL 0.50 M HAc with 40.0 mL 0.45 M NaAc, the amount of 4.0 M HAc which must be added to this buffer to double [H301 is closest to (12) (1) 1.0 mL (2) 5.0 ml. (3) 3.5 mL (4) 7.0 mL @5 mL. 3. Consider: CaCO3(s) + H200) v3 Ca2+(aq) + HCO3‘(aq) + H0'(aq). if, in this system, [Ca2+] = 4EHCOs'], and the system is basic, [HO'] for the system is closest to (12) (1) 2.7 1110'5 (2) 3.5 x10"3 (3) 3.2 x 10'8 .3 x 10'4 (5) 6.8 x10'5. 4 The solubility of CaF2(s) at 25 °C in 1.0 M CaCl2 is about the same as it is in aqueous NaF soln at (12) @4 x 10'3 M (2) 0.060 M (3) 5.8 x 10‘4 M (4) 0.50 M (5) 2.0 M. 5. The following mixture which results in a large extent BlL acid-base rxn is (12) (1) 1 M NaHS .1 1.5 M NaZO (2) HF(l) 81 1 M HCN (3) 0.8 M HOCI 81 KF(s) 1.003(5) 81 0.01 M “H2003” (5) FeCl3(s) 8: 1 M Na2804. 6. For 0.0010 M Nazco3 the correct, “charge balance” relationship for principal species is (12)@la*1 = 2(6032'] + [HCOa'] + [H0'] (2) [Na+] = 2(0032‘] (3) [Na+] + [H301 = [0032‘] (4) 2[Na*] + [H301 = 2[c032*] + [H603] + [HO‘] (5) 2[Na*] = [c032'1+ [HCO3'] + [H0']. 7. Indicator HlN is yellow (Y) at pH 4.7 or less, green (G) at pH 5.3, and blue (B) at pH 5.9 or above. So, if HlN as a BlL acid is ten times weaker than HX, and treating 100 mL of a colorless HX soln with one drop HiN indicator gives a distinct YG color, Miter the HX soln is closest to (12) (1)1x10'3M (2)5x10'4M @xw'w (4)8x10‘5M (5)3x10'6M. _-_-—-___-___-__--_---——-----—----—----—--___--_--_----_.----_-_.---.---_--.__.1.1an..m-auua-n—u-n---—----—---———___u___-___--_-_—_---—_-__-— ...... 8. Consider: “H2C03”(aq) + CaCO3(s) + SO42'(aq) + 2H20(l) # Casey-2H20(s) + 2 HCOfiaq). This EQ can be established by bubbling 002(9) into 1 M NaZSOfi slurried with CaCOs(s). i. Write and calculate K for this EQ. For K calculation be sure to indicate all K's (such as any Ka, Kb, Ksp, or Kw) of which your overall K is composed. (Note: The correct value for K is within 1.0 - 9.0) (15) Sm Dm V1265 ii. Consider mixing 1.0 mois CaCOs(s) with a 1.0 L soln containing [8042‘] at 1.0 and [H0031 at 0.10. If the system is closed and 1.0 mols C02(g) is bubbled in, calculate mols CaSO4-2H20(s) that can be obtained. Show pertinent work. (If you couldn't solve part i., use K = 5.0.) (15) iii. Qualitatively, what will happen to this system if, after the change applied in ii., excess Ca(0H)2(s) is added? Why? (15) iv. Write the correct equation for each rxn of consequence which occurs if the oirginal system is treated with excess 1 M HF . (Hint: You should write three equations.) (15) ...
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