30. a. Al(H 2 O) 6 + H 2 O Co(H 2 O) 5 (OH) 2+ + H 3 O + acid base conjugate base conjugate acid
|
|
|
- Horatio Morgan
- 6 years ago
- Views:
Transcription
1 Chpter 14 : xx,3,40,4,xx,xx,47,50,54,55,xx,64,xx,83,85,91,93,95,10,11,134,138 xx.. HC H 3 O (q) H O(l) C H 3 O (q) H 3 O (q) = [H O ][C H O ] 3 3 [HC H O ] 3 b. Co(H O) 6 3 (q) H O(l) Co(H O) 5 (OH) (q) H 3 O (q) [H3O ][Co(HO) 5(OH) ] = 3 [Co(HO) 6 ] c. CH 3 NH 3 (q) H O(l) CH 3 NH (q) H 3 O (q) [H O ][CH NH ] 3 3 = [CH3NH 3 ] Al(H O) 6 H O Co(H O) 5 (OH) H 3 O cid bse conjugte bse conjugte cid b. HONH 3 H O HONH H 3 O cid bse conjugte bse conjugte cid c. HOCl C 6 H 5 NH OCl C 6 H 5 NH 3 cid bse conjugte bse conjugte cid 38. [H 3 O ][OH ] = 1.0 x ;. b. c. d. [H O ] 3 [H O ] 3 = 1.0 x x 10 [OH ] x 10 [OH ] = = =.8 x 10 M bsic x x 10 [H3O ] = 9 [OH ] = 9.7 x 10 = x 10 M cidic x x 10 [H3O ] = = = 3 [OH ]. x x x 10 [OH ] 1.0 x x 10 M bsic = = = 1.0 x 10 M neutrl 7 [H O ]
2 40.. t 40 o C: H O(l) H 3 O (q) OH (q) H O H 3 O OH initil 0 0 equilibrium =.9 x 10 = [H O ][OH ] = x x = 1.71 x 10 7 M = [H 3 O ] = [OH ] 14 3 b. ph = log[h 3 O ] = log(1.71 x 10 7 ) = c. [H 3 O ][OH ] =.9 x ; [H O ] 3 =.9 x 10 [OH ] x 10.9 x [H3O ] = = =.9 x 10 M [OH ] 0.10 ph = log[h 3 O ] = log(.9 x ) = xx x x 10 [OH ] [H3O ] = = =.8 x 10 M ph = log[h 3 O ] = log(.8 x ) = poh = log[oh ] = log(3.6) = 0.56 b. 1.0 x x 10 [OH ] 9.7 x [H3O ] = = = 1.0 x 10 M 9 ph = log[h 3 O ] = log(1.0 x 10 6 ) = 6.00 (5.99 if use ll digits bove) poh = log[oh ] = log(9.7 x 10 9 ) = 8.01 c. 1.0 x x 10 [OH ]. x [H3O ] = = = 4.5 x 10 M 3 ph = log[h 3 O ] = log(4.5 x 10 1 ) = poh = log[oh ] = log(. x 10 3 ) =.66 d. 1.0 x x 10 [OH ] 1.0 x [H3O ] = = = 1.0 x 10 M 7 ph = log[h 3 O ] = log(1.0 x 10 7 ) = 7.00 poh = log[oh ] = log(1.0 x 10 7 ) = 7.00
3 xx. ph = 6.77 ph poh = p = 14.00; poh = ph poh = = 7.3 [H 3 O ] = 10 ph = = 1.7 x 10 7 M [OH ] = 10 poh = = 5.9 x 10 8 M or x x 10 [OH ] = = = 5.9 x 10 8 M 7 [H O ] 1.7 x HClO 4 H O H 3 O ClO 4 (strong cid, goes to completion) Species in solution: H 3 O nd ClO 4 [H 3 O ] = [HClO 4 ] = 0.50 M ph = log[h 3 O ] = log(0.50) = 0.60 b. HNO 3 H O H 3 O NO 3 (strong cid, goes to completion) Species in solution: H 3 O nd NO 3 [H 3 O ] = [HNO 3 ] = 0.50 M ph = log[h 3 O ] = log(0.50) = ml of 5.00 M HCl SA [H 3 O ] = [HCl] = 5.00 M 30.0 ml of 8.00 M HNO 3 SA [H 3 O ] = [HNO 3 ] = 8.00 M Finl volume = 1.00 L 5.00 mole H O 8.00 mole H O x L L L 1.00 L 3 3 ph = log[h 3 O ] = log(0.690) = x L = M H 3 O ph poh = p = 14.00; poh = = [OH ] = 10 poh = = 1.45 x M
4 54.. Wek cid, HOC 6 H 5 in solution HOC 6 H 5 (q) H O(l) H 3 O (q) OC 6 H 5 (q) HOC 6 H 5 H O H 3 O OC 6 H 5 initil 0.50 M ~0 0 x equilibrium 0.50 x 10 [H3O ][OC6H 5 ] (x)(x) x = 1.6 x 10 = = = [HOC H ] (0.50 x) 0.50 x 6 5 If you ssume tht x << 0.50, then 0.50 x x x x = 4.0 x x = 6.3 x 10 6 M = [H 3 O ] (ssumption good) ph = log[h 3 O ] = 5.0 b. Wek cid, HCN in solution HCN(q) H O(l) H 3 O (q) CN (q) HCN H O H 3 O CN initil 0.50 M ~0 0 x equilibrium 0.50 x [H O ][CN ] (x)(x) x [HCN] (0.50 x) 0.50 x 10 3 = = = = 6. x 10 If you ssume tht x << 0.50, then 0.50 x x x 0.50 x = 1.6 x x = 1. x 10 5 M = [H 3 O ] (ssumption good) ph = log[h 3 O ] = 4.90
5 55. [HC H 3 O ] = 1.05 g HCH3O 1 mole HCH3O 0.0 ml HCH3O ml 6.06 g HC H O 0.50 L 3 = 1.40 M HC H 3 O (q) H O(l) H 3 O (q) C H 3 O (q) HC H 3 O H O H 3 O C H 3 O initil 1.40 M ~0 0 x equilibrium 1.40 x 5 [H3O ][CH3O ] (x)(x) x = 1.8 x 10 = = = [HC H O ] (1.40 x) 1.40 x 3 If you ssume tht x << 1.40, then 1.40 x x x 10 5 x =.5 x 10 5 x = 5.0 x 10 3 M = [H 3 O ] (ssumption good) ph = log[h 3 O ] =.30 xx. HIO 3 (q) H O(l) H 3 O (q) IO 3 (q) HIO 3 H O H 3 O IO 3 initil 0.0 M ~0 0 x equilibrium 0.0 x [H3O ][IO 3] (x)(x) x = 0.17 = = = [HIO ] (0.0 x) 0.0 x 3 If you ssume tht x << 0.0, then 0.0 x x
6 x = x = 0.18 M = [H 3 O ] (ssumption NOT good!) If you do not mke the ssumption, solving the qudrtic: x x C HA = x 0.17x = 0 x = 0.1 M nd 0.30 M ph = log[h 3 O ] = HNO 3 100% dissocition (by definition strong cid). b. Wek cid, HNO in solution HNO (q) H O(l) H 3 O (q) NO (q) HNO H O H 3 O NO initil 0.0 M ~0 0 x equilibrium 0.0 x 4 [H3O ][NO ] (x)(x) x = 4.0 x 10 = = = [HNO ] (0.0 x) 0.0 x If you ssume tht x << 0.0, then 0.0 x x x 10 4 x = 8.0 x 10 3 x = 8.9 x 10 5 M = [H 3 O ] (ssumption good) % dissocition = x 10 M x 100 = 0.0 M 4.4 %
7 c. Wek cid, HOC 6 H 5 in solution HOC 6 H 5 (q) H O(l) H 3 O (q) OC 6 H 5 (q) HOC 6 H 5 H O H 3 O OC 6 H 5 initil 0.0 M ~0 0 x equilibrium 0.0 x 10 [H3O ][OC6H 5 ] (x)(x) x = 1.6 x 10 = = = [HOC H ] (0.0 x) 0.0 x 6 5 If you ssume tht x << 0.0, then 0.0 x x x x = 3. x x = 5.7 x 10 6 M = [H 3 O ] (ssumption good) % dissocition = x 10 M x 100 = 0.0 M.8 x 10 3 % d. % dissocition increses ith for sme initil concentrtion. xx.. Sr(OH) Sr OH strong bse stoichiometric initil M 0 ~0 finl M (0.006 M) [OH ] = M poh = log[oh ] = 1.91; ph = poh = 1.09 b. [OH ] = (0.75 M) poh = log[oh ] = 0.18; ph = poh = c. [OH ] = (5.0 x M) poh = log[oh ] = 9.00; ph = poh = 5.00!! This cnnot be correct becuse this suggests tht dding bse to ter ill give n cidic ph. The problem is tht the 1.0 x 10 7 M OH from ter is not insignificnt in this cse. The ph ould be pproximtely 7.00.
8 83. NH NH (q) H O(l) OH (q) NH NH 3 (q) NH NH H O OH NH NH 3 initil.0 M ~0 0 x equilibrium.0 x 6 [OH ][NH NH 3 ] (x)(x) x b = 3.0 x 10 = = = [NH NH ] (.0 x).0 x If you ssume tht x <<.0, then.0 x.0.0 x x 10 6 x = 6.0 x 10 6 x =.4 x 10 3 M = [OH ] (ssumption good) poh = log[oh ] =.61 ph = poh = [H 3 O ] = = 4.1 x 10 1 M [OH ] =.4 x 10 3 M [NH NH ] =.0 M.4 x 10 3 M =.0 M [NH NH 3 ] =.4 x 10 3 M 85.. C 6 H 5 NH (q) H O(l) OH (q) C 6 H 5 NH 3 (q) C 6 H 5 NH H O OH C 6 H 5 NH 3 initil 0.0 M ~0 0 x equilibrium 0.0 x 10 [OH ][C6H5NH 3 ] (x)(x) x b = 3.8 x 10 = = = [C H NH ] (0.0 x) 0.0 x 6 5 If you ssume tht x << 0.0, then 0.0 x x x 10 10
9 x = 7.6 x x = 8.7 x 10 6 M = [OH ] (ssumption good) poh = log[oh ] = 5.06 ph = poh = 8.94 [H 3 O ] = = 1.1 x 10 9 M b. C 5 H 5 N(q) H O(l) OH (q) C 5 H 5 NH (q) C 5 H 5 N H O OH C 5 H 5 NH initil 0.0 M ~0 0 x equilibrium 0.0 x 9 [OH ][C5H5NH ] (x)(x) x b = 1.7 x 10 = = = [C H N] (0.0 x) 0.0 x 5 5 If you ssume tht x << 0.0, then 0.0 x x x 10 9 x = 8.5 x x = 1.8 x 10 5 M = [OH ] (ssumption good) poh = log[oh ] = 4.73 ph = poh = 9.7 [H 3 O ] = = 5.4 x M 91. C 4 H 8 NH(q) H O(l) OH (q) C 4 H 8 NH (q) ph = 10.8; poh = ph = 3.18 [OH ] = = 6.6 x 10 4 M C 4 H 8 NH H O OH C 4 H 8 NH initil M ~0 0 x equilibrium x = M x = 6.6 x 10 4 M x = 6.6 x 10 4 M [OH ][C H NH ] (6.6 x 10 ) = = = 1.3 x b 4 [C4H8NH] (3.4 x 10 )
10 93. 1 H 3 C 6 H 5 O 7 H O H C 6 H 5 O 7 H 3 O H C 6 H 5 O 7 H O HC 6 H 5 O 7 H 3 O 3 HC 6 H 5 O 7 H O C 6 H 5 O 7 3 H 3 O 95. Use 1 much lrger thn the other. H 3 AsO 4 (q) H O(l) H 3 O (q) H AsO 4 (q) H 3 AsO 4 H O H 3 O H AsO 4 initil 0.0 M ~0 0 x equilibrium 0.0 x 3 [H3O ][HAsO 4] (x)(x) x 1 = 5 x 10 = = = [H AsO ] (0.0 x) 0.0 x 3 4 Solving the qudrtic: x x C HA = x 5 x 10 3 x 0.01 = 0 x = [H 3 O ] = 0.03 M [OH ] x x 10 = = = [H O ] x 10 M [H 3 AsO 4 ] = 0.0 M 0.09 M = 0. M [H AsO 4 ] = 0.03 M [H O ][HAsO ] (0.03)(x) = 8 x 10 = = (0.03) [HAsO 4 ] x = [HAsO 4 ] = 8 x 10 8 M [H O ][AsO ] (0.03)(x) [HAsO ] (8 x 10 ) = = = x 10 x = [AsO 4 3 ] = x M
11 10. NH 3 H O NH 4 OH b = 1.8 x 10 5 NH 4 H O NH 3 H 3 O 1.0 x 10 = = = 5.6 x b 1.8 x 10 CH 3 NH H O CH 3 NH 3 OH b = 4.4 x 10 4 CH 3 NH 3 H O CH 3 NH H 3 O 1.0 x 10 = = =.8 x b 4.38 x 10 The stronger cid hs the lrgest NH 4. xxx.. NO (q) (q) NO (q) ( not n cid or bse) NO (q) H O(l) HNO (q) OH (q) NO H O HNO OH initil 0.1 M 0 ~0 x equilibrium 0.1 x x b = for conjugte cid/bse for nitrous cid (HNO ) is 4.0 x x 10 [OH ][HNO ] (x)(x) x.5 x 10 (0.1 x) 0.1 x 14 = 11 b x 10 = = [NO ] = = If you ssume tht x << 0.1, then 0.1 x x x x = 3.0 x 10 1 x = 1.7 x 10 6 M = [OH ] (ssumption good) poh = log[oh ] = 5.76 ph = poh = 8.3
12 b. NOCl(q) N (q) OCl (q) (N not n cid or bse) OCl (q) H O(l) HOCl(q) OH (q) OCl H O HOCl OH initil 0.45 M 0 ~0 x equilibrium 0.45 x x b = for conjugte cid/bse for HOCl is 3.5 x x 10 7 [OH ][HOCl] (x)(x) x = b.8 x x 10 = = [OCl ] = (0.45 x) = 0.45 x If you ssume tht x << 0.45, then 0.45 x x x 10 7 x = 1.3 x 10 7 x = 3.6 x 10 4 M = [OH ] (ssumption good) poh = log[oh ] = 3.44 ph = poh = c. NH 4 ClO 4 (q) NH 4 (q) ClO 4 (q) (ClO 4 conj. bse SA) NH 4 (q) H O(l) NH 3 (q) H 3 O (q) NH 4 H O NH 3 H 3 O initil 0.40 M 0 ~0 x equilibrium 0.40 x x b = for conjugte cid/bse b for NH 3 is 1.8 x x 10 [H O ][NH ] (x)(x) x 5.6 x 10 (0.40 x) 0.40 x = = = = = x 10 [NH 4 ] If you ssume tht x << 0.40, then 0.40 x 0.40
13 0.40 x x x =. x x = 1.5 x 10 5 M = [H 3 O ] (ssumption good) ph = log[h 3 O ] = Cl neutrl not n cid or bse, Cl is the conjugte bse of strong cid. b. NH 4 C H 3 O NH 4 C H 3 O NH 4 C H 3 O conj. cid of NH 3 conj. bse of cetic cid NH 4 : C H 3 O : 1.0 x 10 = = = 5.6 x b 1.8 x x 10 = = = 5.6 x b x 10 As strong n cid s bse neutrl. c. CH 3 NH 3 Cl CH 3 NH 3 Cl CH 3 NH 3 conj. cid of CH 3 NH Cl conj. bse of SA CH 3 NH 3 : 1.0 x 10 = = =.8 x cidic 14 4 b 4.38 x 10 d. F F not n cid or bse conj. bse of ek cid HF F : 1.0 x 10 = = = 1.4 x bsic 14 b 4 7. x 10 e. NH 4 F NH 4 F conj. cid of NH 3 conj. bse of HF
14 NH 4 : F : 1.0 x 10 = = = 5.6 x b 1.8 x x 10 = = = 1.4 x b 4 7. x 10 cidic cid stronger thn the bse f. CH 3 NH 3 CN CH 3 NH 3 conj. cid of CH 3 NH CN conj. bse of HCN CH 3 NH 3 : CN : 1.0 x 10 = = =.8 x b 4.38 x x 10 = = = 1.6 x b x 10 bsic bse stronger thn the cid HI < HF < NF < NI SA WA CB of WA CB of SA b. HBr < NH 4 Br < Br < NH 3 SA CA of WB CB of SA WB. HNO 3 < C 6 H 5 NH 3 < C 6 H 5 OH < NNO 3 < C 6 H 5 NH < SA CA of WB WA CB of SA WB C 6 H 5 O < CB of WA NOH SB 14.. NH 3 H 3 O NH 4 H O 1 =? The reverse is of the CA of WB: NH 4 H O NH 3 H 3 O 1.0 x 10 = = = 5.6 x b 1.8 x = = = 1.8 x x 10
15 b. NO H 3 O HNO H O 1 =? The reverse is of the WA: HNO H O NO H 3 O = 4.0 x = = =.5 x x 10 c. NH 4 OH NH 3 H O 1 =? The reverse is b of the WB: NH 3 H O NH 4 OH b = 1.8 x = = = 5.6 x b 1.8 x 10 d. HNO OH NO H O 1 =? The reverse is b of the CB of WA: NO H O HNO OH 1.0 x 10 = = =.5 x b x = = = 4.0 x b.5 x 10
HONH 3 initial M - ~0 0 -x - +x +x equilibrium x - x x. ] (x)(x) x b. If you assume that x << 0.100, then x
![HONH 3 initial M - ~0 0 -x - +x +x equilibrium x - x x. ] (x)(x) x b. If you assume that x << 0.100, then x](/thumbs/72/67573254.jpg "HONH 3 initial M - ~0 0 -x - +x +x equilibrium x - x x. ] (x)(x) x b. If you assume that x << 0.100, then x")
Chpter 15 : xx, 4, 8, 0, 8, 44, 48, 50, 51, 55, 76, 8, 90, 96(90), 1(111) xx. NH 4 OH NH H O NH H O NH 4 H O 4.. HONH (q) H O(l) OH (q) HONH (q) HONH H O OH HONH initil 0.100 M ~0 0 x x x equilibrium 0.100
Hydronium or hydroxide ions can also be produced by a reaction of certain substances with water:
Chpter 14 1 ACIDS/BASES Acids hve tste, rect with most metls to produce, rect with most crbontes to produce, turn litmus nd phenolphthlein. Bses hve tste rect very well well with most metls or crbontes,
Dr. Steward s CHM152 Exam #2 Review Spring 2014 (Ch ) KEY
Dr. Stewrd s CHM152 Em #2 Review Spring 2014 (Ch. 16. 17) KEY 1. Eplin the commonion effect. Wek cid or wek bse s % dissocition will decrese if they re plced in solutions with one of the products of dissocition.
Chapter 17: Additional Aspects of Aqueous Equilibria
1 Chpter 17: Additionl Aspects of Aqueous Equilibri Khoot! 1. Adding Br to sturted queous solution of decreses its solubility in wter. BSO 4, Li CO 3, PbS, AgBr. Which of the following mitures could be
Which of the following describes the net ionic reaction for the hydrolysis. Which of the following salts will produce a solution with the highest ph?
95. Which of the following descries the net ionic rection for the hydrolysis of NH4Cl( s)? A. NH4 ( q) Cl & ( q) NH4Cl( s) B. NH Cl & 4 ( s) NH4 ( q) Cl ( q) C. Cl ( q) H O & 2 ( l) HCl( q) OH ( q) D.
1. Weak acids. For a weak acid HA, there is less than 100% dissociation to ions. The B-L equilibrium is:
th 9 Homework: Reding, M&F, ch. 15, pp. 584-598, 602-605 (clcultions of ph, etc., for wek cids, wek bses, polyprotic cids, nd slts; fctors ffecting cid strength). Problems: Nkon, ch. 18, #1-10, 16-18,
Homework #6 Chapter 7 Homework Acids and Bases
Homework #6 Chapter 7 Homework Acids and Bases 20. a) 2H 2O(l) H 3O (aq) OH (aq) K [H 3 O ][OH ] Or H 2O(l) H (aq) OH (aq) K [H ][OH ] b) HCN(aq) H 2O(l) H 3O (aq) CN (aq) K [H 3O ][CN ] [HCN] Or HCN(aq)
Cu 3 (PO 4 ) 2 (s) 3 Cu 2+ (aq) + 2 PO 4 3- (aq) circle answer: pure water or Na 3 PO 4 solution This is the common-ion effect.
CHEM 1122011 NAME: ANSWER KEY Vining, Exm # SHORT ANSWER QUESTIONS ============= 4 points ech ============ All work must be shown. 1. Wht re [H O ] nd [OH ] for solution tht hs ph of 9.0? Choose one. ()
Chapter 16 Acid Base Equilibria
Chpter 16 Acid Bse Equilibri 16.1 Acids & Bses: A Brief Review Arrhenius cids nd bses: cid: n H + donor HA(q) H(q) A(q) bse: n OH donor OH(q) (q) OH(q) Brønsted Lowry cids nd bses: cid: n H + donor HA(q)
Acids and Bases. H + (aq) + Cl - (aq) 100 molecules HCl 100 H+ ions Cl- ions 100% HCl molecules dissociate in water.
Acids nd Bses Acids: in generl cid is the substnce tht produces H ions when it is dissolved in wter. Acids cn be divided into two different clsses: Strong cid: is one tht completely dissocites into its
Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this mteril useful? You cn help our tem to keep this site up nd bring you even more content consider donting vi the link on our site. Still hving trouble understnding the mteril? Check out our Tutoring
[HCO ] log = M. + log = log = log [CH
![[HCO ] log = M. + log = log = log [CH](/thumbs/75/72569575.jpg "[HCO ] log = M. + log = log = log [CH")
Chpter 17: 7, 9, 11, 1, 15, 1, 7,, 9, 4, 49, 50, 51, 79, 81, 8, 87, 9, 99, 105, 11, 117 [HCO ] 7. ph = pk + log [H CO ] [HCO ] [H CO ] log = 1.6 [HCO ] = 0.04 [HCO ] [NH ] 0.15 M 9. ph = pk + log = 9.5
temperature is known as ionic product of water. It is designated as K w. Value of K w
Ionic product of ter The product of concentrtions of H nd OH ions in ter t prticulr temperture is knon s ionic product of ter. It is designted s K. H O H 1 OH ; H 57.3 kjm The vlue of [H ][OH ] K ; K[HO]
Chapter 17 Homework Problem Solutions
Chapter 17 Homework Problem Solutions 17.40 D 2 O D + + OD, K w = [D + ] [OD ] = 8.9 10 16 Since [D + ] = [OD ], we can rewrite the above expression to give: 8.9 10 16 = ([D + ]) 2, [D + ] = 3.0 10 8 M
Fundamentals of Analytical Chemistry
Homework Fundmentls of nlyticl hemistry hpter 9 0, 1, 5, 7, 9 cids, Bses, nd hpter 9(b) Definitions cid Releses H ions in wter (rrhenius) Proton donor (Bronsted( Lowry) Electron-pir cceptor (Lewis) hrcteristic
Homework 04. Acids, Bases, and Salts
HW04 - Acids, Bses, nd Slts! This is preview of the published version of the quiz Strted: Feb 21 t 8:59m Quiz Instruc!ons Homework 04 Acids, Bses, nd Slts Question 1 In the reversible rection HCN + H O
CHAPTER 08: MONOPROTIC ACID-BASE EQUILIBRIA
Hrris: Quntittive Chemicl Anlysis, Eight Edition CHAPTER 08: MONOPROTIC ACIDBASE EQUILIBRIA CHAPTER 08: Opener A CHAPTER 08: Opener B CHAPTER 08: Opener C CHAPTER 08: Opener D CHAPTER 08: Opener E Chpter
Initial Change x +x +x Equilibrium x x x
Chpter 8 Homework Solutions 8.1 () n(h ) 0.144 mol/l x 0.05 L 3.60x10-3 mol n(oh - ) 0.15 mol/l x 0.05 L 3.1x10-3 mol After neutrliztion, excess n(h ) 3.6x10-3 mol - 3.1x10-3 mol 4.8x10-4 mol [ H n( H
CHEMISTRY 16 HOUR EXAM III March 26, 1998 Dr. Finklea - Answer key
CHEMISTRY 16 HOUR EXAM III Mrch 26, 1998 Dr. Finkle - Answer key 1. Phenol, lso known s crbolic cid, is wek monoprotic cid (HA). A 1.0 M solution of phenol hs ph of 4.95. Wht is the K of phenol? 1. 2.0
Strong acids and bases. Strong acids and bases. Systematic Treatment of Equilibrium & Monoprotic Acid-base Equilibrium.
Strong cids nd bses Systemtic Tretment of Equilibrium & Monoprotic cid-bse Equilibrium onc. (M) 0.0.00 -.00-5.00-8 p Strong cids nd bses onc. (M) p 0.0.0.00 -.0.00-5 5.0.00-8 8.0? We hve to consider utoprotolysis
CHAPTER 14 ACIDS AND BASES. Questions
CHAPTER 1 ACIDS AND BASES Questions 19. Acids are proton (H donors, and bases are proton acceptors. HCO as an acid: HCO (aq H O(l CO (aq H O (aq HCO as a base: HCO (aq H O(l H CO (aq OH (aq H PO as an
[base (aq)][oh (aq)] [base(aq)] [OH ( aq )]
![[base (aq)][oh (aq)] [base(aq)] [OH ( aq )]](/thumbs/80/81026565.jpg "[base (aq)][oh (aq)] [base(aq)] [OH ( aq )]")
Section 8.5: Clcultions Involving Bsic Solutions Tutoril 1 Prctice, pge 527 1. Given: [OH (q)] 0.000 mol/l Required: [H + (q)], [OH (q)] [OH (q)] 0.000 mol/l ecuse OH is strong se. + 1.0 [H ] [OH ( q )]
CHEMISTRY - BROWN 13E CH.16 - ACID-BASE EQUILIBRIA - PART 2.
!! www.clutchprep.com CONCEPT: ph and poh To deal with incredibly small concentration values of [H + ] and [OH - ] we can use the ph scale. Under normal conditions, the ph scale operates within the range
The Lewis definition is most general. The Lewis definition can apply to all Arrhenius and +
CHAPTER FOURTEEN ACIDS AND BASES Questions 16. The Arrhenius definitions are: acids produce H in water and bases produce OH in water. The difference between strong and weak acids and bases is the amount
Completion of acid/base/buffer chemistry. Hanson Activity Clicker quiz 3/11/2013. Chs 7 8 of Zumdahl
Completion of acid/base/buffer chemistry Chs 7 8 of Zumdahl Hanson Activity 16 3 Discuss Key Questions 1 of Activity 16 3, page 301, with your partner for three minutes. The clicker quiz will commence
Homework #7 Chapter 8 Applications of Aqueous Equilibrium
Homework #7 Chapter 8 Applications of Aqueous Equilibrium 15. solution: A solution that resists change in ph when a small amount of acid or base is added. solutions contain a weak acid and its conjugate
Ch 7 Practice Problems
Ch 7 Practice Problems 1. For the equilibrium that exists in an aqueous solution of nitrous acid (HNO 2, a eak acid), the equilibrium constant expression is [H ] [NO 2 ] = [HNO ] 2 [H ][N][O] [HNO 2] =
7/19/2011. Models of Solution Chemistry- III Acids and Bases
Models of Solution Chemistry- III Acids nd Bses Ionic Atmosphere Model : Revisiting Ionic Strength Ionic strength - mesure of totl concentrtion of ions in the solution Chpter 8 1 2 i μ ( ) 2 c i z c concentrtion
APEF Acids and Bases - Answers
APEF Acids nd Bses - Answers 1. It requires 3.0 ml of 0.0500 mol/l NOH(q) to neutrlize 100.0 ml of gstric juice. We cn ssume tht HCl(q) is the only cid present in gstric juice. ) Clculte the concentrtion
Advanced Chemistry. Approximate Timeline. Students are expected to keep up with class work when absent. CHAPTER 14 ACIDS & BASES
Advanced Chemistry Approximate Timeline Students are expected to keep up with class work when absent. CHAPTER 14 ACIDS & BASES Day Plans for the day Assignment(s) for the day 14.1 The Nature of Acids &
Acids and Bases Written Response
Acids and Bases Written Response January 1999 4. Consider the salt sodium oxalate, Na2C2O4. a) Write the dissociation equation for sodium oxalate. (1 mark) b) A 1.0M solution of sodium oxalate turns pink
FIRST EXAM. Answer any 4 of the following 5 questions. Please state any additional assumptions you made, and show all work.
CEE 680 8 Mrch 018 FIRST EXAM Closed ook, one pge of notes llowed. Answer ny 4 of the following 5 questions. Plese stte ny dditionl ssumptions you mde, nd show ll work. Miscellneous Informtion: R = 1.987
Acids, Bases, and ph. ACIDS, BASES, & ph
I. Arrhenius Acids and Bases ACIDS, BASES, & ph Acid any substance which delivers hydrogen ion (H + ) _ to the solution. Base any substance which delivers hydroxide ion (OH ) to the solution. II ph ph
Acid/Base Definitions
Acids and Bases Acid/Base Definitions Arrhenius Model Acids produce hydrogen ions in aqueous solutions Bases produce hydroxide ions in aqueous solutions Bronsted-Lowry Model Acids are proton donors Bases
Advanced Chemistry Practice Problems
Finding ph 1. Question: Determine the ph for each of the given solutions. a. 0.150 M HNO3 b. 0.150 M CH3COOH, a = 1.8 10-5 c. 0.150 M CHOOH, a = 3.5 10-4 Answer: The method to determine the ph of a solution
CHEMISTRY - BROWN 14E CH.16 - ACID-BASE EQUILIBRIA.
!! www.clutchprep.com CONCEPT: ACID IDENTIFICATION The most common feature of an acid is that many possess an H + ion called the. When it comes to acids there are 2 MAJOR TYPES that exist: are acids where
Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.
Find this mteil useful? You cn help ou tem to keep this site up nd bing you even moe content conside donting vi the link on ou site. Still hving touble undestnding the mteil? Check out ou Tutoing pge to
1. (3) The pressure on an equilibrium mixture of the three gases N 2, H 2 and NH 3
1. (3) The pressure on an equilibrium mixture of the three gases N 2, H 2 and NH 3 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) is suddenly decreased by doubling the volume of the container at constant temperature.
CHAPTER FOURTEEN ACIDS AND BASES. For Review
CHAPTER FOURTEEN ACIDS AND BASES For Review 1. a. Arrhenius acid: produce H + in water b. Brnsted-Lowry acid: proton (H + ) donor c. Lewis acid: electron pair acceptor The Lewis definition is most general.
A) Arrhenius Acids produce H+ and bases produce OH not always used because it only IDs X OH as basic species
3 ACID AND BASE THEORIES: A) Arrhenius Acids produce H+ and bases produce OH not always used because it only IDs X OH as basic species B) Bronsted and Lowry Acid = H + donor > CB = formed after H + dissociates
Chem. 1B Midterm 1 Version B February 1, 2019
First initial of last name Chem. 1B Midterm 1 Version B February 1, 2019 Name: Print Neatly. You will lose 1 point if I cannot read your name or perm number. Perm Number: All work must be shown on the
Unit 4-1 Provincial Practice Questions Page 1
Page 1 Page 2 Page 3 Page 4 Page 5 1.00 Page 6 Page 7 Page 8 55. The conjugate base of HAsO 4 2 is A. H 3 O + B. 3 AsO 4 C. H 3 AsO 4 D. H 2 AsO 4 56. Consider the following acidbase equilibrium: HC 6
Conjugate Pairs Practice #1
Name: Key Skill: Learning to Draw Tie Lines Conjugate Pairs Practice #1 Look at each example drawn below. Sets of partners (called s) are matched with tie lines. HNO3 + OH - NO3 - + H2O CH3NH2 + H2O CH3NH3
CHM 152/154 HOUR EXAM II Diebolt Summer multiple choice 52 Parts II and III 73 Total Pts 125
CHM 152/154 HOUR EXAM II Diebolt Summer 2010 pts earned name pts possible multiple choice 52 Parts II and III 73 Total Pts 125 Part One: Multiple choice. Mark the correct answers on the provided scantron
CH 15 Summary. Equilibrium is a balance between products and reactants
CH 15 Summary Equilibrium is a balance between products and reactants Use stoichiometry to determine reactant or product ratios, but NOT reactant to product ratios. Capital K is used to represent the equilibrium
CHE 107 Spring 2017 Exam 3
CHE 107 Spring 2017 Exam 3 Your Name: Your ID: Question #: 1 What is the ph of a 0.20 M solution of hydrocyanic acid at 25ºC? The Ka of HCN at 25ºC is 4.9 10 10. A. 2.08 B. 5.00 C. 3.89 D. 8.76 Question
CHEMISTRY - CLUTCH CH.15 - ACID AND BASE EQUILIBRIUM.
!! www.clutchprep.com CONCEPT: ACID IDENTIFICATION The most common feature of an acid is that many possess an H + ion called the. When it comes to acids there are 2 MAJOR TYPES that exist: are acids where
Judith Herzfeld 1996,1998. These exercises are provided here for classroom and study use only. All other uses are copyright protected.
Judith Herzfeld 1996,1998 These exercises are provided here for classroom and study use only. All other uses are copyright protected. 3.3-010 According to Bronsted-Lowry Theory, which of the following
Really useful information = H + = K w. K b. 1. Calculate the ph of a solution made by pouring 5.0 ml of 0.20 M HCl into 100. ml of water.
Acid Base Equilibrium Putting it all together HA H + + A H + A incomingsa HA +incomingsa Strong Acids HCl HNO3 HBr H2SO4 HI HClO4 HClO3 Really useful information K w H + OH K w M V M V B + H2O OH + HB
Unit 9: Acid and Base Multiple Choice Practice
Unit 9: Acid and Base Multiple Choice Practice Name June 14, 2017 1. Consider the following acidbase equilibrium: HCO3 H2O H2CO3 OH In the reaction above, the BrönstedLowry acids are: A. H2O and OH B.
Chemistry 1A Fall 2013 MWF 9:30 Final Test Form A
Chemistry 1A Fall 2013 MWF 9:30 Final Test Form A 1. How many moles of P 4 molecules are in 141.4 g of phosphorus? A) 4.566 mol B) 1.752 x10 4 mol C) 1.141 mol D) 2.348 x 10 1 mol E) 1.414 x 10 1 mol 2.
Chapter 15. Applications of Aqueous Equlibria. A bag of mostly water - Star Trek -
Chpter 15. Applictions of Aqueous Equlibri A bg of mostly wter - Str Trek - Solutions of Acids or Bses Contining Common Ion Until now, we were concerned with the equilibrium of solution with n cid or bse.
Guide to Chapter 15. Aqueous Equilibria: Acids and Bases. Review Chapter 4, Section 2 on how ionic substances dissociate in water.
Guide to Chapter 15. Aqueous Equilibria: Acids and Bases We will spend five lecture days on this chapter. During the first two class meetings we will introduce acids and bases and some of the theories
Chapter 17 Additional Aspects of Aqueous Equilibria (Part A)
Chapter 17 Additional Aspects of Aqueous Equilibria (Part A) Often, there are many equilibria going on in an aqueous solution. So, we must determine the dominant equilibrium (i.e. the equilibrium reaction
AP Chemistry Chapter 14 Answers Zumdahl ph = -log[h + ]; poh = -log[oh - ]; At 25 o C, ph + poh = 14.00; For Exercise 13.
![AP Chemistry Chapter 14 Answers Zumdahl ph = -log[h + ]; poh = -log[oh - ]; At 25 o C, ph + poh = 14.00; For Exercise 13.](/thumbs/77/75028194.jpg "AP Chemistry Chapter 14 Answers Zumdahl ph = -log[h + ]; poh = -log[oh - ]; At 25 o C, ph + poh = 14.00; For Exercise 13.")
AP Chemistry Chapter 14 Answers Zumdahl 14.41 ph = log[h + ]; poh = log[oh ]; At 25 o C, ph + poh = 14.00; For Exercise 13.37: ph = log[h + ] = log(1.0 x 10 7 ) = 7.00; poh = 14.00 ph = 14.00 7.00 = 7.00
Chapter 16 Homework Solutions
//05 Chapter 16 Homework Solutions 6. a) H AsO b) CH 3 NH 3 + c) HSO d) H 3 PO 8. acid base conj. base conj. acid a) H O CHO OH CH O a) HSO HCO 3 SO H CO 3 b) H 3 O + HSO 3 H O H SO 3 10. a) H C 6 H 7
NATURE OF ACIDS & BASES
General Properties: NATURE OF ACIDS & BASES ACIDS BASES Taste sour Bitter Change color of indicators Blue Litmus turns red no change Red Litmus no change turns blue Phenolphtalein Colorless turns pink
Chemistry 12. Bronsted Acids and Equilibria
Worksheet 42 Bronsted Acids and Equilibria Name Date Due Hand In With Corrections by Chemistry 12 Worksheet 42 Bronsted Acids and Equilibria 82 1. Write the formula for a proton 2. Write the formula for
Chapter 16 exercise. For the following reactions, use figure 16.4 to predict whether the equilibrium lies predominantly. - (aq) + OH - (aq)
1 Chapter 16 exercise Q1. Practice exercise page 671 Write the formula for the conjugate acid of the following, HSO 3, F, PO 4 3 and CO. HSO 3 H H 2 SO 4 F H HF PO 4 3 H HPO 4 2 CO H HCO Q2. Practice exercise
9-1 (a) A weak electrolyte only partially ionizes when dissolved in water. NaHCO 3 is an
Chpter 9 9- ( A ek electrolyte only prtilly ionizes hen dissolved in ter. NC is n exmple of ek electrolyte. (b A Brønsted-ory cid is cule tht dontes proton hen it encounters bse (proton cceptor. By this
Acid Base Equilibrium Review
Acid Bse Equilirium Review Proof of true understnding of cid se equilirium culmintes in the ility to find ph of ny solution or comintion of solutions. The ility to determine ph of multitude of solutions
CHAPTER 17 ADDITIONAL ASPECTS OF ACID BASE EQUILIBRIA
CHAPTER 17 ADDITIONAL ASPECTS OF ACID BASE EQUILIBRIA PRACTICE EXAMPLES 1A (D) Orgnize the solution round the lnced chemicl eqution, s we hve done efore. Initil: 0.500M 0M 0M Eqution: HF(q) HO(l) HO (q)
1. Strengths of Acids and Bases 2. K a, K b 3. Ionization of Water 4. Relative Strengths of Brønsted-Lowry Acids and Bases
Chemistry 12 Acid-Base Equilibrium II Name: Date: Block: 1. Strengths of Acids and Bases 2. K a, K b 3. Ionization of Water 4. Relative Strengths of Brønsted-Lowry Acids and Bases Strengths of Acids and
CHAPTER 13: ACIDS & BASES. Section Arrhenius Acid & Bases Svante Arrhenius, Swedish chemist ( ).
CHAPTER 13: ACIDS & BASES Section 13.1 Arrhenius Acid & Bases Svante Arrhenius, Swedish chemist (1839-1927). He understood that aqueous solutions of acids and bases conduct electricity (they are electrolytes).
Mixtures of Acids and Bases
Mixtures of Acids and Bases CH202, lab 6 Goals : To calculate and measure the ph of pure acid and base solutions. To calculate and measure the ph of mixtures of acid and base solutions. Safety : Hydrochloric
ph scale [H 3 O + ] varies over a very wide range. The ph scale simplifies expressing acidity/basicity.
![ph scale [H 3 O + ] varies over a very wide range. The ph scale simplifies expressing acidity/basicity.](/thumbs/96/126526725.jpg "ph scale [H 3 O + ] varies over a very wide range. The ph scale simplifies expressing acidity/basicity.")
ph scale [H 3 O + ] varies over a very wide range. The ph scale simplifies expressing acidity/basicity. ph = log[h 3 O + ] = log 1 + [H3O ] (really an approximation; just as we did for K expressions we
CHEMISTRY 1AA3 Tutorial 2 Answers - WEEK E WEEK OF JANUARY 22, (i) What is the conjugate base of each of the following species?
CHEMISTRY 1AA3 Tutorial 2 Answers - WEEK E WEEK OF JANUARY 22, 2001 M.A. Brook B.E. McCarry A. Perrott 1. (i) What is the conjugate base of each of the following species? (a) H 3 O + (b) NH 4 + (c) HCl
[HIO 4 ] (mol/l) [IO 4 ] (mol/l) [H 3 O + ] (mol/l) I C x +x +x E x x x
![[HIO 4 ] (mol/l) [IO 4 ] (mol/l) [H 3 O + ] (mol/l) I C x +x +x E x x x](/thumbs/74/70636472.jpg "[HIO 4 ] (mol/l) [IO 4 ] (mol/l) [H 3 O + ] (mol/l) I C x +x +x E x x x")
54. Determine the ph of 0.084 mol/l periodic cid solution, HIO 4 (q). The vlue for the rection below is.9 10. HIO 4 (q) + H O(l) IO 4 (q) + H 3 O + (q) Wht Is Required? You must determine the ph of the
Acids and Bases Review Worksheet II Date / / Period. Molarity. moles L. Normality [H 3 O +1 ] [OH -1 ] ph poh
![Acids and Bases Review Worksheet II Date / / Period. Molarity. moles L. Normality [H 3 O +1 ] [OH -1 ] ph poh](/thumbs/75/72206853.jpg "Acids and Bases Review Worksheet II Date / / Period. Molarity. moles L. Normality [H 3 O +1 ] [OH -1 ] ph poh")
Honors Chemistry Name Acids and Bases Review Worksheet II Date / / Period Solute Name of Solute Molar Mass grams mole Molarity moles L Normality [H 3 O +1 ] [OH ] ph poh Acidic or Basic 1. HCl Hydrochloric
Calorimetry, Heat and ΔH Problems
Calorimetry, Heat and ΔH Problems 1. Calculate the quantity of heat involved when a 70.0g sample of calcium is heated from 22.98 C to 86.72 C. c Ca= 0.653 J/g C q = 2.91 kj 2. Determine the temperature
1. What do a chemical indicator and a buffer solution typically both contain?
Acids, Bases & Redox 2 - Practice Problems for Assignment 9 1. What do a chemical indicator and a buffer solution typically both contain? (a) A strong acid and its conjugate acid (b) A strong acid and
Acids, Bases, and Salts Review for Sections
1. Consider the following: Review for Sections 4.1 4.9 I H 2 CO 3 + F HCO 3 + HF 2 II HCO 3 + HC 2 O 4 H 2 CO 3 + C 2 O 4 2 III HCO 3 + H 2 C 6 H 6 O 7 H 2 CO 3 + HC 6 H 5 O 7 The HCO 3 is a base in A.
CHM Third Hour Exam Spring 2003
CHM 1143 Third Hour Exam Spring 2003 Each question is worth 10 points. You get six free misses. Write the letter of your choice to the right of the answer choices. MULTIPLE CHOICE. Choose the one alternative
Unit 2 Acids and Bases
Unit 2 Acids and Bases 1 Topics Properties / Operational Definitions Acid-Base Theories ph & poh calculations Equilibria (Kw, K a, K b ) Indicators Titrations STSE: Acids Around Us 2 Operational Definitions
Wksht 3.3 Acid-Base Equilibria II Supplemental Instruction Iowa State University
Wksht 3.3 Acid-Base Equilibria II Supplemental Instruction Iowa State University Leader: Deborah Course: CHEM 178 Instructor: Bonaccorsi/Vela Date: 2/13/18 1. What are the conjugate bases of a. HCO3 -?
Chapter 17 Additional Aspects of Aqueous Equilibria (Part A)
Chapter 17 Additional Aspects of Aqueous Equilibria (Part A) What is a dominant equilibrium? How do we define major species? Reactions between acids and bases 1. Strong Acids + Strong Base The reaction
Organic Acids - Carboxylic Acids
Orgnic Acids - rboxylic Acids Orgnic cids - crboxylic cid functionl group rboxylic cids re redily deprotonted by bses such s NO eg 3 O O - + O - + O 3 O O Acid Bse onjugte Bse onjugte Acid This rection
To determine if our protein unfolds or not is it sufficient to know the K a and K b Values of the functional groups? - O H
A students work (without solutions mnul) ~ problems/night To determine if our protein unfolds or not is it sufficient to know the nd b Vlues of the functionl groups? Dr Alnh Fitch Flnner Hll 40 508-119
Problem 22: Buffer solutions 1. The equilibrium, which governs the concentration of H + within the solution is HCOOH! HCOO + H + + Hence K
Problem : Buffer solutions. The equilibrium, hich governs the concentrtion of H ithin the solution is HCOOH! HCOO H [HCOO ] 4 Hence. [HCOOH] nd since [HCOOH] 0.00 M nd [HCOO ] 0.50 M -4 0.00 4..8 M 0.50
First Exam December 19---Christmas Break begins December 21. Silberberg Chapter 17-18, Skoog 2-7, 11-13
Announcements First Exam December 19---Christmas Break begins December 21. Silberberg Chapter 17-18, Skoog 2-7, 11-13 Please keep up with the work (lots of problems in this Chapter) and see me if you have
Non Sibi High School
Non Sibi High School Andover s Chem 300: Accelerated/Honors Chemistry Chapter 17, Review Quiz 1 Answers 1 A 65 ml sample of HBr gas, measured at 35 C and 722 mmhg, was dissolved in water to yield 275 ml
= ) = )
Basics of calculating ph 1. Find the ph of 0.07 M HCl. 2. Find the ph of 0.2 M propanoic acid (K a = 10-4.87 ) 3. Find the ph of 0.4 M (CH 3 ) 3 N (K b = 10-4.20 ) 4. Find the ph of 0.3 M CH 3 COO - Na
Chemistry 400 Homework #3, Chapter 16: Acid-Base Equilibria
Chemistry 400 Homework #3, Chapter 16: Acid-Base Equilibria I. Multiple Choice (for those with an asterisk, you must show work) These multiple choice (MC) are not "Google-proof", but they were so good
Practice test Chapters 15 and 16: Acids and Bases
Name: Class: Date: Practice test Chapters 15 and 16: Acids and Bases 1. Which of the following pairs of species is not a conjugate acid base pair? A) HOCl, OCl B) HNO 2, NO + 2 C) O 2, OH D) HSO 4, SO
Progressive Science Initiative. Click to go to website:
Slide 1 / 33 New Jersey Center for Teaching and Learning Progressive Science Initiative This material is made freely available at www.njctl.org and is intended for the non-commercial use of students and
2 NO, has reached a state of dynamic equilibrium, which statement below is true?
Chemistry 11-014, Vining Exam #3 NAME: Answer Key Take Home Version 1. When the reversible reaction, N + O NO, has reached a state of dynamic equilibrium, which statement below is true? (a) Both the forward
Worksheet 4.1 Conjugate Acid-Base Pairs
Worksheet 4.1 Conjugate AcidBase Pairs 1. List five properties of acids that are in your textbook. Acids conduct electricity, taste sour, neutralize bases, change the color of indicators, and react with
AP Chemistry: Acids & Bases Notes
AP Chemistry: Acids & Bases Notes Objectives Definition of Acids-Bases Acid Strength Base Strength ph-poh Scale Calculating ph of Strong Acids-Bases Calculating ph of Weak Acids-Bases Calculating Ka from
ANALYTICAL CHEMISTRY - CLUTCH 1E CH.8 - MONOPROTIC ACID-BASE EQUILIBRIA.
!! www.clutchprep.com CONCEPT: ARRHENIUS ACIDS AND BASES The most general definition for acids and bases was developed by Svante Arrhenius near the end of the 19 th century. According to him, the cation
Chapter 16. Acids and Bases. Copyright Cengage Learning. All rights reserved 1
Chapter 16 Acids and Bases Copyright Cengage Learning. All rights reserved 1 Section 16.1 Acids and Bases Models of Acids and Bases Arrhenius: Acids produce H + ions in solution, bases produce OH ions.
REVIEW QUESTIONS Chapter Identify the Lewis acid and base in each of the following reactions:
Chemistry 102 REVIEW QUESTIONS Chapter 16 1. For each reaction below, identify the Brønsted-Lowry acid and base and their conjugates: A) NH 4 + (aq) + CN (aq) HCN (aq) + NH 3 (aq) B) (CH 3 ) 3 N (aq) +
Unit 6: ACIDS AND BASES
Unit 6: Acids and Bases Honour Chemistry Unit 6: ACIDS AND BASES Chapter 16: Acids and Bases 16.1: Brønsted Acids and Bases Physical and Chemical Properties of Acid and Base Acids Bases Taste Sour (Citric
Contents and Concepts
Chapter 16 1 Learning Objectives Acid Base Concepts Arrhenius Concept of Acids and Base a. Define acid and base according to the Arrhenius concept. Brønsted Lowry Concept of Acids and Bases a. Define acid
Solutions are aqueous and the temperature is 25 C unless stated otherwise.
Solutions are aqueous and the temperature is 25 C unless stated otherwise. 1. According to the Arrhenius definition, an acid is a substance that produces ions in aqueous solution. A. H C. OH B. H + D.
Answer Key, Problem Set 6 (With explanations)
Chemistry 1 Mines, Spring 18 Answer ey, Problem Set 6 (With explanations) 1. 16.58(a,d)*;. 16.6(a); 3. 16.66(a,c) Assume 5 C. Also find % ionization; 4. 16.7; 5. 16.8(a,b) For (a), assume 5 C, for (b),
Chp 13, 14, 15 SHOW ALL WORK AND CIRCLE FINAL ANSWERS. a) 1 only b) 2 only c) 3 only d) 1 and 2 only e) 1, 2, and H 2
Chp 13, 14, 15 Name: SHOW ALL WORK AND CIRCLE FINAL ANSWERS 1. Which of the following factors affect the initial rate of a reaction? 1) The nature of the reactants. 2) The concentration of the reactants.
CHM 2046 Test #3 Review: Chapters , 15, & 16
Chapter 14 1. For the following reaction Kc = 0.513 at 500 K. N 2 O 4 (g) 2 NO 2 (g) If a reaction vessel initially contains an N 2 O 4 concentration of 0.0500 M at 500 K, what are the equilibrium concentrations
[base] [acid] ph = pka + log
![[base] [acid] ph = pka + log](/thumbs/82/86175854.jpg "[base] [acid] ph = pka + log")
Fall 2013 CCBC-Catonsville (Wed 10/30/13) Use your time wisely. Do not get stuck on one question. NO CREDIT WILL BE GIVEN UNLESS WORK IS SHOWN CLEARLY. UNITS MUST BE INCLUDED IN YOUR SETUPS. Answers must
ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA
ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA Acids- taste sour Bases(alkali)- taste bitter and feel slippery Arrhenius concept- acids produce hydrogen ions in aqueous solution while
CHEM Dr. Babb s Sections Exam #3 Review Sheet
CHEM 116 Dr. Babb s Sections Exam #3 Review Sheet Acid/Base Theories and Conjugate AcidBase Pairs 111. Define the following terms: Arrhenius acid, Arrhenius base, Lewis acid, Lewis base, BronstedLowry