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
|
|
- Rosamund Wilson
- 6 years ago
- Views:
Transcription
1 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 M ~0 0 x x x equilibrium x x x 8 [OH ][HONH ] (x)(x) x b = 1.1 x 10 = = = [OHNH ] (0.100 x) x If you ssume tht x << 0.100, then x x x x x 10 8 x = 1.1 x 10 9 x =. x 10 5 M = [OH ] (ssumption good) poh = log[oh ] = 4.48 ph = poh = 9.5 b. HONH Cl(q) HONH (q) Cl (q) (Cl conj. bse SA) HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil M 0 ~0 x x x equilibrium x x x x b = w for conjugte cid/bse b for HONH is 1.1 x x 10 [H O ][HONH ] (x)(x) x 9.1 x 10 (0.100 x) x 14 7 = = = = = x 10 [HONH ]
2 If you ssume tht x << 0.100, then x x x x x 10 7 x = 9.1 x 10 8 x =.0 x 10 4 M = [H O ] (ssumption good) ph = log[h O ] =.5 c. Pure H O ph = 7.00 d. Using either equilibrium: HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil M ~0 x x x equilibrium x x x [H O ][HONH ] x 10 7 (x)(0.100 x) = = = = x 10 [HONH ] (0.100 x) 9.1 x 10 If you ssume tht x << 0.100, then x x (0.100 x) x (0.100) x x 10 7 x = 9.1 x 10 7 M (ssumption good) ph = log[h O ] = The dded HCl (s H O ) rects with HONH, forming HONH (s HONH Cl): HONH (q) H O (q) HONH (q) H O(l) HONH H O HONH H O I mole/l x = mole 0.00 mole F = mole mole
3 0.080 mole HONH [ HONH ] = = M HONH [HONH ] 0.00 mole HONH = = 0.00 M HONH HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil 0.00 M ~0 x x x equilibrium 0.00 x x x [H O ][HONH ] x 10 7 (x)(0.080 x) = = = = x 10 [HONH ] (0.00 x) 9.1 x 10 If you ssume tht x is smll: x (0.080 x) x (0.080) 0.00 x x 10 7 x =. x 10 7 M (ssumption good) ph = log[h O ] = 6.64 b. This is solution of SA nd WA. The H O from the WA is going to be negligible: [H O ] = 0.00 mole/ = 0.00 M ph = log[h O ] = 1.70 c. Sme s b. (even more so becuse H O is n even weker cid then HB ). ph = log[h O ] = 1.70 d. The dded HCl (s H O ) rects with just the HONH, forming HONH (s HONH Cl). This dds to the conjugte cid nd tkes wy from the bse:
4 HONH (q) H O (q) HONH (q) H O(l) HONH H O HONH I mole/l x = mole/l x mole 0.00 mole = mole F = = mole 0.10 mole H O mole HONH [ HONH ] = = M HONH [HONH ] 0.10 mole HONH = = 0.10 M HONH HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil 0.10 M ~0 x x x equilibrium 0.10 x x x [H O ][HONH ] x 10 7 (x)(0.080 x) = = = = x 10 [HONH ] (0.10 x) 9.1 x 10 If you ssume tht x is smll: x (0.080 x) x (0.080) 0.10 x x 10 7 x = 1.4 x 10 6 M (ssumption good) ph = log[h O ] = This is solution of SB nd WB. The OH from the WB is going to be negligible: [OH ] = 0.00 mole/ = 0.00 M poh = log[oh ] = 1.70; ph = poh = 1.0 b. The dded NOH (s OH ) rects with HONH (s HONH Cl), forming HONH :
5 HONH (q) OH (q) HONH (q) H O(l) HONH OH HONH H O I mole/l x = mole 0.00 mole F = mole mole 0.00 mole HONH [HONH ] = = 0.00 M HONH [HONH ] mole NH 4 = = M HONH HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil M 0.00 ~0 x x x equilibrium x 0.00 x x [H O ][HONH ] x 10 7 (x)(0.00 x) = = = = x 10 [HONH ] (0.080 x) 9.1 x 10 If you ssume tht x is smll: x (0.00 x) x (0.00) x x 10 7 x =.6 x 10 6 M (ssumption good) ph = log[h O ] = 5.44 c. Sme s. (even more so becuse H O is n even weker bse then B). poh = log[oh ] = 1.70; ph = poh = 1.0 d. The dded NOH (s OH ) rects with HONH (s HONH Cl), forming HONH. This dds to the conjugte bse nd tkes wy from the cid:
6 HONH (q) OH (q) HONH (q) H O(l) HONH OH HONH H O I mole/l x = mole/l x mole 0.00 mole = mole F = mole = 0.10 mole 0.10 mole HONH [HONH ] = = 0.10 M HONH [HONH ] mole NH 4 = = M HONH HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil M 0.10 ~0 x x x equilibrium x 0.10 x x [H O ][HONH ] x 10 7 (x)(0.10 x) = = = = x 10 [HONH ] (0.080 x) 9.1 x 10 If you ssume tht x is smll: x (0.10 x) x (0.10) x x [NH ] = 0.75 M x = 6.1 x 10 7 M (ssumption good) ph = log[h O ] = 6. NH 4 Cl(q) [NH 4 ] = NH 4 (q) Cl (q) 1 mole NH4Cl 1 mole NH g NH4Cl x x 5.49 g NH Cl 1 mole NH Cl 4 4 = 0.95 M
7 NH (q) H O(l) OH (q) NH 4 (q) NH H O OH NH 4 initil 0.75 M ~ M x x x equilibrium 0.75 x x 0.95 x 5 [OH ][NH 4 ] (x)(0.95 x) b = 1.8 x 10 = = [NH ] (0.75 x) If you ssume tht x smll, then: x (0.95) 0.75 = 1.8 x 10 5 x = 1.4 x 10 5 M = [OH ] (ssumption good) poh = log[oh ] = 4.84 ph = poh = ph [HO ] = 10 = 10 =.98 x 10 M [HCO ] = [HO ] = 4. x 10 [H CO ] 7 [HCO ] 4. x x 10 = 10.8 [H CO ] [H O ].98 x = = 8 [H CO ] 1 = [HCO ] 10.8 = 0.09 ph b. [ H O ] = 10 = 10 = 7.08 x 10 M [HPO ] = [H O ] = 6. x [HPO 4 ] = = 8 4 [HPO ] 6. x x 10 = [H PO ] [H O ] 7.08 x 10 [H PO ] 1 = [HPO ] = 1.1
8 c. The best buffer is one with 1:1 rtio of the cid (or bse) to the conjugte bse (or conjugte cid). This will give ph close to p. The p of H PO 4 is.1, which is fr removed from There would be very little H PO 4 t tht bsic ph SA CB of SA not buffer n cidic solution b. SA WA not buffer no conjugte bse produced. Acidic. c. SA CB of WA yes, buffer. Hlf of the F ion would rect with the H O from HNO to give HF. This results in mixture of HF ( WA) nd F (its conjugte bse) buffer. d. SA SB not buffer bsic solution (NOH in excess). 50. Henderson Hsselblch is convenient for this type of clcultion. [A ] ph = p log [HA] The rection is: C H O (q) HCl(g) HC H O (q) Cl (q). For ph to equl p, the rtio must be 1:1. There re 1.0 mole CHO x 1.0 L = 1.0 mole C H O L To convert hlf (0.50 mole) into HCHO would require 0.50 mole HCl. b. [A ] 4.00 = 4.74 l og [HA] [A ] log = 0.54 ; [HA] 10 [A ] log [HA] [A ] 0.54 = = 10 = 0.88 [HA] Since x moles of moles CHO HCHO re produced from x moles of HCl(g) (by removing x ), the eqution bove becomes: 1.0 x x = 0.88 ; x = 0.78 = mole HCl(g)
9 c. [A ] 5.00 = 4.74 log [HA] [A ] l og = 0.6; [HA] 10 [A ] log [HA] [A ] 0.6 = = 10 = 1.8 [HA] 1.0 x x = 1.8 ; x = 0.5 = mole HCl(g) HC H O (q) H O(l) H O (q) C H O (q) HC H O H O H O C H O initil 0.00 M ~0 0 x x x equilibrium 0.00 x x x 5 [HO ][CHO ] (x)(x) x = 1.8 x 10 = = = [HC H O ] (0.00 x) 0.00 x If you ssume tht x << 0.00, then 0.00 x 0.00 x x 0.00 x x 10 5 x =.6 x 10 6 x = 1.9 x 10 M = [H O ] (ssumption good) ph = log[h O ] =.7
10 b. The dded OH (s OH ) rects with HC H O, forming C H O : HC H O (q) OH (q) C H O (q) H O(l) HC H O OH C H O I 0.00 mole/l x L = mole/l x L = mole mole F = mole mole mole HC H O [HC H O ] = = M HC H O L mole CHO [CHO ] = = 0.0 M CHO L HC H O (q) H O(l) C H O (q) H O (q) HC H O H O C H O H O initil M 0.0 ~0 x x x equilibrium x 0.0 x x 5 [HO ][CHO ] (x)(0.0 x) = 1.8 x 10 = = [HC H O ] (0.100 x) ssuming tht x is smll; x (0.0 x) x (0.0) x x 10 5 x = 5.4 x 10 5 M (ssumption good) ph = log[h O ] = 4.7 c. The dded OH (s OH ) rects with HC H O, forming C H O : HC H O (q) OH (q) C H O (q) H O(l) HC H O OH C H O I 0.00 mole/l x L = mole/l x L = mole mole F = mole mole
11 mole HC H O [HC H O ] = = M HC H O 0.00 L mole CHO [CHO ] = = M CHO 0.00 L HC H O (q) H O(l) C H O (q) H O (q) HC H O H O C H O H O initil M ~0 x x x equilibrium x x x 5 [HO ][CHO ] (x)( x) = 1.8 x 10 = = [HC H O ] ( x) ssuming tht x is smll; x ( x) x (0.0500) x x 10 5 x = 1.8 x 10 5 M (ssumption good) ph = log[h O ] = 4.74 d. The dded OH (s OH ) rects with HC H O, forming C H O : HC H O (q) OH (q) C H O (q) H O(l) HC H O OH C H O I 0.00 mole/l x L = mole/l x L = mole mole F = mole mole [HC H O ] mole HC H O = = M HC H O 0.50 L mole C H O [C H O ] M C H O 0.50 L = = HC H O (q) H O(l) C H O (q) H O (q)
12 HC H O H O C H O H O initil M ~0 x x x equilibrium x x x 5 [HO ][CHO ] (x)( x) = 1.8 x 10 = = [HC H O ] (0.000 x) ssuming tht x is smll; x ( x) x (0.0600) x x 10 5 x = 6.0 x 10 6 M (ssumption good) ph = log[h O ] = 5. d. The dded OH (s OH ) rects with HC H O, forming C H O : HC H O (q) OH (q) C H O (q) H O(l) HC H O OH C H O I 0.00 mole/l x L = mole/l x 0.00 L = mole mole F = 0.0 mole mole mole C H O [C H O ] M C H O 0.00 L = = C H O (q) H O(l) HC H O (q) OH (q) C H O H O HC H O OH initil M 0 ~0 x x x equilibrium x x x [OH ][HC H O ] x b = = 5.6 x 10 = = ( x) w 10 [CHO ] x x ssuming tht x is smll; 5.6 x x x =.7 x ; x = 6.1 x 10 6 M (ssumption good) poh = log[oh ] = 5.1; ph = poh =
13 f. This is mole/l x 0.50 L = mole OH excess mole OH mole OH rected = mole OH [ OH ] mole OH = = 0.50 L M OH poh = log[oh ] = 1.85; ph = poh = Ag CO (s) Ag (q) CO (q) b. Ce(IO ) (s) Ce (q) IO (q) c. BF (s) B (q) F (q) 8.. PbI (s) Pb (q) I (q) PbI Pb I initil 0 0 x x equilibrium x x sp = [Pb ][I ] = 1.4 x 10 8 = x(x) = 4x x = 1.5 x 10 M = [Pb ] = [PbI ] = molr solubility b. CdCO (s) Cd (q) CO (q) CdCO Cd CO initil 0 0 x x equilibrium x x sp = [Cd ][CO ] = 5. x 10 1 = x x =. x 10 6 M = [Cd ] = [CO ] = [CdCO ] = molr solubility c. Sr (PO 4 ) (s) Sr (q) PO 4 (q) Sr (PO 4 ) Sr PO 4 initil 0 0 x x equilibrium x x sp = [Sr ] [PO 4 ] = 1 x 10 1 = (x) (x) = 108x 5 x = x 10 7 M = [Sr (PO 4 ) ] = molr solubility
14 90.. Ag SO 4 (s) Ag (q) SO 4 (q) Ag SO 4 Ag SO 4 initil 0.10 M 0 x x equilibrium 0.10 x x sp = [Ag ] [SO 4 ] = 1. x 10 5 = (x) x = 4x x = 1.4 x 10 M = [SO 4 ] = molr solubility b. AgNO (q) Ag (q) NO (q) (ionizes completely) [Ag ] = [AgNO ] = 0.10 M Ag SO 4 (s) Ag (q) SO 4 (q) Ag SO 4 Ag SO 4 initil 0.10 M 0 x x equilibrium 0.10 x x sp = [Ag ] [SO 4 ] = 1. x 10 5 = (0.10 x) x Assume x smll: 1. x 10 5 (0.10) x x = 1. x 10 M (ssumption good) = [SO 4 ] = molr solubility c. SO 4 (q) (q) SO 4 (q) (ionizes completely) [SO 4 ] = [ SO 4 ] = 0.0 M Ag SO 4 (s) Ag (q) SO 4 (q) Ag SO 4 Ag SO 4 initil M x x equilibrium x 0.0 x sp = [Ag ] [SO 4 ] = 1. x 10 5 = (x) (0.0 x) Assume x smll: 1. x x (0.0) x =.9 x 10 M (ssumption good) = [SO 4 ] = molr solubility
15 90.. AgF, becuse F is the CB of WA b. Pb(OH), becuse OH is SB c. Sr(NO ), becuse NO is the CB of WA d. Ni(CN), becuse CN is the CB of WA HC H O OH C H O H O 1 =? The reverse is b of the CB of WA: C H O H O HC H O OH 1.0 x 10 = = = 5.6 x w b x = = = 1.8 x b 5.6 x 10 b. C H O H O HC H O H O 1 =? The reverse is of WA: HC H O H O C H O H O 1 1 = = = 5.6 x x 10 c. This rection is just: H O OH H O 1 =? This is the reverse of the w rection: H O H O OH w = 1.0 x = = = 1.0 x w 1.0 x 10
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
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)
More informationChapter 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
More informationCu 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. ()
More informationHydronium 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,
More informationWhich 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.
More informationInitial 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
More information[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
More informationChapter 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.
More informationFundamentals 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
More informationCHEMISTRY 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
More informationDr. 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.
More information1. 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,
More informationFind 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
More informationCHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA
Advanced Chemistry Name Hour Advanced Chemistry Approximate Timeline Students are expected to keep up with class work when absent. CHAPTER 15 APPLICATIONS OF AQUEOUS EQUILIBRIA Day Plans for the day Assignment(s)
More informationReview: Acid-Base Chemistry. Title
Review: Acid-Base Chemistry Title Basics General properties of acids & bases Balance neutralization equations SA + SB water + salt Arrhenius vs. Bronsted-Lowry BL plays doubles tennis match with H+) Identify
More informationHomework #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
More informationAcids 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
More informationChapter 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)
More informationChapter 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
More information9-1 (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
More informationCHAPTER 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
More informationMixtures 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
More informationHomework 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
More informationCHAPTER 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)
More informationChapter 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
More informationCHEM 1412 Zumdahl & Zumdahl Practice Exam II (Ch. 14, 15 & 16) Multiple Choices: Please select one best answer. Answer shown in bold.
CHEM 1412 Zumdahl & Zumdahl Practice Exam II (Ch. 14, 15 & 16) Multiple Choices: Please select one best answer. Answer shown in bold. 1. Consider the equilibrium: PO -3 4 (aq) + H 2 O (l) HPO 2-4 (aq)
More information7/19/2011. 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
More informationChapter 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
More informationBCIT Winter Chem Exam #2
BCIT Winter 2017 Chem 0012 Exam #2 Name: Attempt all questions in this exam. Read each question carefully and give a complete answer in the space provided. Part marks given for wrong answers with partially
More informationFind 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
More informationCHEMGURU.ORG YOUTUBE ; CHEMGURU. Syllabus. Acids and Bases, ph, Common ion effect, Buffer solutions, Hydrolysis of salts and Solubility Product.
Syllbus Acids nd Bses, ph, Common ion effect, Buffer solutions, Hydrolysis of slts nd Solubility Product. Acids nd Bses Here we discuss some importnt definitions of cids nd bses. Arrhenius Definition Arrhenius
More informationConjugate 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
More informationFormation of a salt (ionic compound): Neutralization reaction. molecular. Full ionic. Eliminate spect ions to yield net ionic
Formation of a salt (ionic compound): Neutralization reaction molecular Full ionic Eliminate spect ions to yield net ionic Hydrolysis/ reaction with water Anions of Weak Acids Consider the weak acid HF
More informationDougherty Valley High School AP Chemistry Chapters 14 and 15 Test - Acid-Base Equilibria
Dougherty Valley High School AP Chemistry Chapters 14 and 15 Test - Acid-Base Equilibria This is a PRACTICE TEST. Complete ALL questions. Answers will be provided so that you may check your work. I strongly
More informationCHEMISTRY 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
More informationExam 2 Practice (Chapter 15-17)
Exam 2 Practice (Chapter 15-17) 28. The equilibrium constant Kp for reaction (1) has a value of 0.112. What is the value of the equilibrium constant for reaction (2)? (1) SO2 (g) + 1/2 O2(g) SO3 (g) Kp
More informationCH 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
More informationACID-BASE REACTIONS. Titrations Acid-Base Titrations
Page III-b-1 / Chapter Fourteen Part II Lecture Notes ACID-BASE REACTIONS Chapter (Part II A Weak Acid + Strong Base Titration Titrations In this technique a known concentration of base (or acid is slowly
More informationAPEF 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
More informationChapter 16 Aqueous Ionic Equilibrium
Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Chapter 16 Aqueous Ionic Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall The Danger of Antifreeze
More informationAnswer 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),
More informationSecondary Topics in Equilibrium
Secondary Topics in Equilibrium Outline 1. Common Ions 2. Buffers 3. Titrations Review 1. Common Ions Include the common ion into the equilibrium expression Calculate the molar solubility in mol L -1 when
More informationPrimary Topics in Equilibrium
Primary Topics in Equilibrium Outline 1. Equilibrium Expression 2. Calculating Concentration Given K 3. Calculating K Given Concentration Review 1. Equilibrium Expression (only gas and aqueous do not include
More information[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 )]
More informationChem1120pretest2Summeri2015
Name: Class: Date: Chem1120pretest2Summeri2015 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. When the system A + B C + D is at equilibrium, a. the forward
More informationAcids, 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
More informationBuffers. How can a solution neutralize both acids and bases? Beaker B: 100 ml of 1.00 M HCl. HCl (aq) + H 2 O H 3 O 1+ (aq) + Cl 1 (aq)
Buffers How can a solution neutralize both acids and bases? Why? Buffer solutions are a mixture of substances that have a fairly constant ph regardless of addition of acid or base. They are used in medicine,
More informationAcids 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
More informationStrong 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
More informationHomework #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)
More informationCHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA. Questions
CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA Questions 8. MX(s) M n+ (aq) + X n (aq) K sp = [M n+ [X n ; the K sp reaction always refers to a solid breaking up into its ions. The representations all
More informationAcid 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
More informationAcid-Base Equilibria. 1.NH 4 Cl 2.NaCl 3.KC 2 H 3 O 2 4.NaNO 2. Acid-Ionization Equilibria. Acid-Ionization Equilibria
Acid-Ionization Equilibria Acid-Base Equilibria Acid ionization (or acid dissociation) is the reaction of an acid with water to produce hydronium ion (hydrogen ion) and the conjugate base anion. (See Animation:
More informationCalorimetry, 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
More informationChapter 17 Answers. Practice Examples [H3O ] 0.018M, 1a. HF = M. 1b. 30 drops. 2a.
Chapter 17 Answers Practice Examples 1a. + [HO ] 0.018M, 1b. 0 drops [HF] = 0.8 M. [H O + ] = 0.10 M, HF = 0.97 M. a. + HO 1.10 M, CHO = 0.150 M. b. 15g NaCHO a. The hydronium ion and the acetate ion react
More informationReally 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
More informationProblems -- Chapter Write balanced chemical equations for the important equilibrium that is occurring in an aqueous solution of the following.
Problems -- Chapter 1 1. Write balanced chemical equations for the important equilibrium that is occurring in an aqueous solution of the following. (a) NaNO and HNO answers: see end of problem set (b)
More informationFIRST 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
More informationChapter 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
More information= ) = )
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
More information( x) [ ] ( ) ( ) ( ) ( )( ) ACID-BASE EQUILIBRIUM PROBLEMS. Because HCO / K = 2.8 < 100 the xin the denominator is not going to be negligible.
ACID-BASE EQUILIBRIUM PROBLEMS 1. Clculte the ph of solutio tht cotis 0.165 M olic cid. Clculte the cocetrtio of the olte io i this solutio. 1 H C O 4 + H O Ý H 3O + + HC O 1 5.9 HC O + H O Ý H 3O + +
More informationtemperature 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]
More information#89 Notes Unit 11: Acids & Bases and Radiochemistry Ch. Acids, Bases, and Radioactivity
#89 Notes Unit 11: Acids & Bases and Radiochemistry Ch. Acids, Bases, and Radioactivity Common Strong Acids Common Strong Bases HCl hydrochloric acid Group #1 + OH HNO 3 nitric acid NaOH, KOH etc. H 2
More informationIn the Brønsted-Lowry system, a Brønsted-Lowry acid is a species that donates H + and a Brønsted-Lowry base is a species that accepts H +.
16.1 Acids and Bases: A Brief Review Arrhenius concept of acids and bases: an acid increases [H + ] and a base increases [OH ]. 16.2 BrønstedLowry Acids and Bases In the BrønstedLowry system, a BrønstedLowry
More informationCHM 112 Dr. Kevin Moore
CHM 112 Dr. Kevin Moore Reaction of an acid with a known concentration of base to determine the exact amount of the acid Requires that the equilibrium of the reaction be significantly to the right Determination
More informationNorthern Arizona University Exam #3. Section 2, Spring 2006 April 21, 2006
Northern Arizona University Exam #3 CHM 152, General Chemistry II Dr. Brandon Cruickshank Section 2, Spring 2006 April 21, 2006 Name ID # INSTRUCTIONS: Code the answers to the True-False and Multiple-Choice
More informationChemistry 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.
More information1. (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.
More informationNorthern Arizona University Exam #3. Section 2, Spring 2006 April 21, 2006
Northern Arizona University Exam #3 CHM 152, General Chemistry II Dr. Brandon Cruickshank Section 2, Spring 2006 April 21, 2006 Name ID # INSTRUCTIONS: Code the answers to the True-False and Multiple-Choice
More informationCHE 107 Summer 2017 Exam 3
CHE 107 Summer 2017 Exam 3 Question #: 1 What is the ph of a 0.10 M hydrocyanic acid (HCN) solution. Ka = 4.9 10-10. A. 2.56 C. 4.04 B. 3.17 D. 5.15 Question #: 2 Original Windex has a ph = 11.60 and [H
More informationChapter 15. Properties of Acids. Structure of Acids 7/3/08. Acid and Bases
Chapter 15 Acid and Bases Properties of Acids! Sour taste! React with active metals! React with carbonates, producing CO 2! Change color of vegetable dyes!blue litmus turns red! React with bases to form
More informationAcid-Base Equilibria. 1.NH 4 Cl 2.NaCl 3.KC 2 H 3 O 2 4.NaNO 2. Solutions of a Weak Acid or Base
Acid-Base Equilibria 1 Will the following salts be acidic, basic or neutral in aqueous solution? 1.NH 4 Cl.NaCl.KC H O 4.NaNO A = acidic B = basic C = neutral Solutions of a Weak Acid or Base The simplest
More informationChemistry 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
More informationb. H 2 SO 4!! d. KOH!! f. HCl!
Chemistry L3 Name Page 1 Worksheet: Acids and Bases (Chapter 20, p. 576) 1. Classify these as an Arrhenius acid ( starts with H + ) or Arrhenius base (ends in OH ). Then name each one. (p 577-579, 594-596)
More informationDepartment of Chemistry University of Texas at Austin
Polyprotic and Special Cases Calculations Supplemental Worksheet KEY For the following polyprotic acid questions: Citric acid (H3C6H5O6) Ka1 = 8.4 x 10 4 Ka2 = 1.8 x 10 5 Ka3 = 4.0 x 10 6 Oxalic acid (H2C2O4)
More informationCHM 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
More informationLecture Presentation. Chapter 16. Aqueous Ionic Equilibrium. Sherril Soman Grand Valley State University Pearson Education, Inc.
Lecture Presentation Chapter 16 Aqueous Ionic Equilibrium Sherril Soman Grand Valley State University The Danger of Antifreeze Each year, thousands of pets and wildlife species die from consuming antifreeze.
More informationArrhenius, Bronstead-Lowry, Intro to ph scale
Unit 9 Acid/Base Equilibrium In Class Problems and Notes Arrhenius, Bronstead-Lowry, Intro to ph scale Arrhenius acid and Arrhenius base. An acid is a substance which, when dissolved in water, increases
More informationExperiment 9: DETERMINATION OF WEAK ACID IONIZATION CONSTANT & PROPERTIES OF A BUFFERED SOLUTION
Experiment 9: DETERMINATION OF WEAK ACID IONIZATION CONSTANT & PROPERTIES OF A BUFFERED SOLUTION Purpose: Prt I: The cid ioniztion constnt of wek cid is to be determined, nd the cid is identified ccordingly.
More informationPart 01 - Assignment: Introduction to Acids &Bases
Part 01 - Assignment: Introduction to Acids &Bases Classify the following acids are monoprotic, diprotic, or triprotic by writing M, D, or T, respectively. 1. HCl 2. HClO4 3. H3As 4. H2SO4 5. H2S 6. H3PO4
More information[H + ] OH - Base contains more OH - than H + [OH - ] Neutral solutions contain equal amounts of OH - and H + Self-ionization of Water
19.1 Acids & Bases 1. Compare and contrast the properties of acids & bases. 2. Describe the self-ionization of water & the concept of K w. 3. Differentiate between the Arhennius & Bronsted-Lowry models
More informationCHAPTER FIFTEEN APPLICATIONS OF AQUEOUS EQUILIBRIA. For Review
CHAPTER FIFTEEN APPLICATIONS OF AQUEOUS EQUILIBRIA For Review 1. A common ion is an ion that appears in an equilibrium reaction but came from a source other than that reaction. Addition of a common ion
More informationOperational Skills. Operational Skills. The Common Ion Effect. A Problem To Consider. A Problem To Consider APPLICATIONS OF AQUEOUS EQUILIBRIA
APPLICATIONS OF AQUEOUS EQUILIBRIA Operational Skills Calculating the common-ion effect on acid ionization Calculating the ph of a buffer from given volumes of solution Calculating the ph of a solution
More informationGuide 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
More informationFind 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 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. Still having trouble understanding the material? Check
More informationAcid/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
More informationExample 15.1 Identifying Brønsted Lowry Acids and Bases and Their Conjugates
Example 15.1 Identifying Brønsted Lowry Acids and Bases and Their Conjugates For Practice 15.1 In each reaction, identify the Brønsted Lowry acid, the Brønsted Lowry base, the conjugate acid, and the conjugate
More informationContents 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
More informationChem1120pretest2Summeri2015
Chem1120pretest2Summeri2015 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. When the system A + B C + D is at equilibrium, a. the forward reaction has
More informationWksht 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 -?
More informationANALYTICAL 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
More informationName AP CHEM / / Chapter 15 Outline Applications of Aqueous Equilibria
Name AP CHEM / / Chapter 15 Outline Applications of Aqueous Equilibria Solutions of Acids or Bases Containing a Common Ion A common ion often refers to an ion that is added by two or more species. For
More informationChem 112, Fall 05 Exam 3A
Before you begin, make sure that your exam has all 10 pages. There are 32 required problems (3 points each, unless noted otherwise) and two extra credit problems (3 points each). Stay focused on your exam.
More informationChemistry 12 Provincial Exam Workbook Unit 04: Acid Base Equilibria. Multiple Choice Questions
R. Janssen, MSEC Chemistry 1 Provincial Workbook (Unit 0), P. 1 / 69 Chemistry 1 Provincial Exam Workbook Unit 0: Acid Base Equilibria Multiple Choice Questions 1. Calculate the volume of 0.00 M HNO needed
More informationChapter 17. Additional Aspects of Aqueous Equilibria 蘇正寬 Pearson Education, Inc.
Chapter 17 Additional Aspects of Aqueous Equilibria 蘇正寬 chengkuan@mail.ntou.edu.tw Additional Aspects of Aqueous Equilibria 17.1 The Common-Ion Effect 17.2 Buffers 17.3 Acid Base Titrations 17.4 Solubility
More informationChapter 15 - Applications of Aqueous Equilibria
Neutralization: Strong Acid-Strong Base Chapter 15 - Applications of Aqueous Equilibria Molecular: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) SA-SB rxn goes to completion (one-way ) Write ionic and net ionic
More informationCopyright 2018 Dan Dill 1
TP The expression for the equilibrium constant for the solubility equilibrium M 2 X 2 M X 2 is 1. sp 2 M X 2 / M 2 X 2. sp 2 M 2 X 2 / M 2 X 3. sp 2 M 2 X 2 4. sp M 2 X 2 Lecture 21 CH102 A1 (MWF 9:05
More informationACID-BASE EQUILIBRIA. Chapter 14 Big Idea Six
ACID-BASE EQUILIBRIA Chapter 14 Big Idea Six Acid-Base Equilibria Common Ion Effect in Acids and Bases Buffer SoluDons for Controlling ph Buffer Capacity ph-titradon Curves Acid-Base TitraDon Indicators
More informationUnit 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.
More informationPlease print: + log [A- ] [HA]
Please print: Last name: First name: Chem 1062 Exam 3 Spring 2005 Andy Aspaas, Instructor Thursday, April 7, 2005 Equations: K c = [C]c [D] d [A] a [B] b ph =! log[h 3 O + ] poh =! log[oh! ] ph + poh =
More information