Qz2KeyHA - CHM 2046 QUIZ 2 50 MINUTE TIME LIMIT T: 10/7/08...

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Unformatted text preview: CHM 2046 QUIZ 2 50 MINUTE TIME LIMIT T: 10/7/08 I DIRECTIONS: Same rules as for uiz 1. Point value for each is in the left mar in. Name (pledged): flaw 9a’tr'e3e. DISC Instr. Name: Cal C” s UWW: 1. The solubility of CaF2(s) at 25 °C in 0.10 M CaCl2 is about the same as it is in aqueous NaF soln at (12) (1)20 M (2) 0.50 M (3) 2.5 x 104 M (4) 3.4 x10'3 M (5_ 6.0x 10“ M. 2. For 0.00001 M (HO)2302, (H2804), the correct “charge balance” relationship for principal species is (12) (1) [H301 = [HSOd'] + [5042'] (2) 2[H30+] = [HSO4'] @H30+] = 2(3042'] (4) [H301 = [H8041 + [5042'] + 010'] (5)0130? = [H8041 + 2180.21. 3. Indicator HIN is yellow (Y) at pH 4.5 or less, green (6) at pH 5.3, and blue (B) at pH 6.1 or above. So. if HIN as a BIL acid is five times stronger than HX, and treating 100 mL of a colorless HX soln with one drop HIN indicator gives a distinct 36 color, Mi for the HX soln is closest to (12) (1)3x10‘6M x10'6M (3)8x10‘5M (4)4x10'5M (5)5x10"M. 4. If, for aqueous (HO)ZSO2 soln, [HSO4'] = 0.4[3042‘], then, for this soln (12) (1) pH is ~3.0 (2) is ~2.3 (3) pH 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 2.0 M HAc which must be added to this buffer to double [H30+] is closest to (12) (1) 7.0 mL (2) 2.5 mL .0 mL (4) 1.0 mL (5) 3.5 mL. 6. Consider: CaCOs(s) + H20“) = Ca2+(aq) + HCOs'laq) + H0'(aq). If, in this system, [Ca2+] = 5[HCO3'], and the system is basic, [HO'] for the system is closest to (1. (1)“83 x10"4 (2) 7.2 x1El'a (3) 6.3 x 10" (4) 3.9 x10'5 (5) 9.7 X102. 7. The following mixture which does not result in a large extent BIL acid-base rxn is (12) (1) HF(|) & (2) K2C03(s) & 0.01 M “H2C03” (3) FeCl3(s) & 1 M NaZS - NaHS & 0.5 M Map (5) 0.8 M HOCI & KCN(s). 8. Consider: CaSO4-2H20(s) + 2 HCOfiaq) us “H2C03”(aq) + Ca603(s) + SO42'(aq) + 2H20(|). This EQ can be established by treating gypsum, Ca804°2H20(s), with an aqueous soln containing HC03'. 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 value for K is within 0.010- 2.0) 1<-== 44:1 .. Lag .. {Mi} .4611 a E (15) st'T W 3 W9? 7 WW] thaws”) (1<.._ 11.693] (asixmfili 41.3 x 1.13"“ 0 m C i ... Visited”) . u ""3 v ‘m'i’e Hazel: +——--- A ___~_._3~ ($.2X10"II (4.2 26107) (5.153 «2:32.503 "K02 ' ii. Consider mixing 1.0 mols CaSO4~2 20(5) with a 1.0 L soln containing [HCOg'] at 0.10 and [5042"] at 0.50. If the system is closed so that 002(9) cannot escape calculate mols CaCO3(s) that can be obtained. Show pertinent work. (If you couldn’t solve part i., use K = 0.50) JR: 0.1‘] ‘11 (“mwgfij H (x) (050 (He) ‘1':- ‘kuId Cam-5 oIfiaimeLIEJ H d v-v 3 (15) mug-31 Lupe-qt)“ T9wa ’9’? 0.00m WM— [Mrsg 35.4 14 2. 3.003) £11m:- 0.0.1:.) 190%"; 64-403"— r‘émw- 0.1111) iii. Qualitatively, what will happen to this system if, after the change applied in ii., excess Ca(OH)2(s) Is added? Why? £131.) * Mb} 0‘ 11% M ‘ ‘6’ Cfififlfl r QIOQ'RimesI ) law-l (gig-0W MMW w, me cal/I WW3 (twill—LU]; A (15} 00?. A "" 1343 M632 (3450?, Wé‘fi'vvw‘eé AHIKIJIE Kasai-Hue HM biz/MC ijwi'li was) pla‘rchh NOJO "YMIA (“4.00.3 1. (nab? 'II’lmJe. 0.10 who am; "is a 101mm C83 * Walt mean c-yfljwh (4504124410) iv. Write the correct equation for each rxn of consequence which occurs If the original EQ system IS treated with excess 1 M HF . (Hint: You should write three equati ns.) Q.) 003%) «— WQW) —-.. “ismfigps ‘Flsp U A 9‘ “I: I m9? “‘0 W T Calais) 4 1150., (up 4: 03031.1() (15) ., ,. _. . . 2 00.5.5134 £4403“) -—> (atlas i “vsws’by 4 H10 02) (Mere: M61 HAM u “Lt—Oglimtj , 111.0 I.“ «t 6,0115) UK) CHM 2046 QUIZ 2 50 MINUTE TIME LIMIT T: 10/7703 DIRECTIONS: Same rules as for uiz 1. Point value for each is in the left mar in. Name (pledged): Q DISC Instr. Name: - bb‘h5“m 1. If, for aqueous (HO)ZSCJO2 soln, [HSO4'] = 0.4[8042'], then, for this soln (12) (1) pH is ~2.0 (2) pH is ~3.5 (3) pH is ~3.0 (4) p, is ~2.3 (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 2.0 M HAc which must be added to this buffer to double [H30+] is closest to (12) mL (2) 3.5 mL (3) 7.0 mL (4)1.0 mL (5) 2.5 mL. 3. Consider: CaCO3(s) + H200) -—‘ Ca2+(aq) + HCOflaq) + HO'(aq). If, in this system, [Ca2+] = 5[HCOS'], and the system is basic, [HO‘] for the system is closest to (12) (1) 6.3 )(10'4 (2) 7.2 x10'3 X104 (4) 3.9 x10”5 (5) 9.7 X102. 4. The following mixture which does not result in a large extent BIL acid-base rxn is (12) (1) 0.8 M HOCI & KCN(s) (2) ((20045) a 0.01 M “H2003” @ M NaHS & 0.5 M flap (4) FeCI3(s) &1 M NaZS (5) HF(|) &1 M H3N. 5. The solubility of CaF2(s) at 25 °C in 0.10 M CaCl2 is about the same as it is in aqueous NaF soln at (12) (1) 3.4 x 10'3 M (2) 0.50 M x10"1 M (4) 2.0 M (5)2.5x10'4 M. 6. For 0.00001 M (H0)2302 (H2304) the correct, “charge balance" relationship for principal species is (12) (1) [0130*] = 2503'] (2) 2[H30*] = [HSO4'] (3) [1430*] = [1130;] + [3042‘] (4) [1130*] = [H504] + [3042'] + [H0] (5) [1130*] = [HSO4‘] + 2(303']. ?. Indicator HIN is yellow (Y) at pH 4.5 or less, green (G) at pH 5.3, and blue (B) at pH 6.1 or above. So, if HIN as a BIL acid is five times stronger than HX, and treating 100 mL of a colorless HX soln with one drop HIN indicator gives a distinct BG color, ME for the HX soln is closest to (12)@~«x10‘5M (2) 3x10'6M (3)5x10'4M (4)3x10'5M (5)4x10"5M. 8. Consider: CaSO4-2H2O(s) + 2 HCO3‘(aq) w-—- “H2C03”(aq) + CaCO3(s) + SO42'(aq) + 2 H200). This EQ can be established by treating gypsum, CaSO4-2H20(s), with an aqueous soln containing HC03'. 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 value for K is within 0.010 - 2.0) . 3" f“\ i (15} SM "951160136 i ii. Consider mixing 1.0 mols 03804-2H20(s) with a 1.0 L soln containing [HCOa'] at 0.10 and [8042'] at 0.50. If the system is closed so that C02(g) cannot escape calculate mols CaCO3(s) that can be obtained. Show pertinent work. (If you couldn’t solve part i., use K = 0.50) (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 original EQ system is treated with excess 1 M HF . (Hint: You should write three equations.) {15) ...
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Qz2KeyHA - CHM 2046 QUIZ 2 50 MINUTE TIME LIMIT T: 10/7/08...

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