Unit The mw of Na 2 CO 3 is : Na=23, O=16, C=12 A) 140 B) 106 C) 96 D) 100 E) 60
|
|
|
- Rosalyn Carroll
- 6 years ago
- Views:
Transcription
1 Unit 2 1- The mw of Na 2 CO 3 is : Na=23, O=16, C=12 A) 140 B) 106 C) 96 D) 100 E) 60
2 2- How many grams of Na 2 CO 3 (mw = 106 ) A) 318 B) C) 134 D) 201 E) 67 in 3 moles,
3 3- Calculate the normal concentration (N ) of 0.1 M solution of HCl A) 0.2 N B) 0.3 N C) 0.05 N D) 0.1 N
4 4- The eq.wt of H 3 PO 4 (mw=98) is: A) 98 B) 49 C) 32.7 D) 196 E) 294
5 5-2.5 g of Na 2 SO 4 ( mw = 106 ) has been dissolved in water and the volume was completed to 500 ml, calculate the followings : The molar concentration of Na 2 SO 4 A) 5 B) 500 C) 0.5 D) 0.05
6 Unit3: STOICHIOMETRY AND EQUILIBRIUM The limiting reactant is: A) present in lower of moles B) present in higherof moles C) present in lower of mass D) present in lower of mass
7 Unit3: STOICHIOMETRY AND EQUILIBRIUM The limiting reactant is: A) It is comletly used up in the reaction B) It is not comletly used up in the reaction
8 Unit3: STOICHIOMETRY AND EQUILIBRIUM In chemical equilibrium: A) rate (forward= reverse) B) rate (forward< reverse) C) rate (forward> reverse)
9 Unit3: STOICHIOMETRY AND EQUILIBRIUM If Keq >> 1 (M=0.1 of product) A) Product 0.1-x B) product 0.1+x
10 Unit3: STOICHIOMETRY AND EQUILIBRIUM If Keq << 1 (M=0.1 of product) A) Product 0.1-x B) Product x
11 Unit3: STOICHIOMETRY AND EQUILIBRIUM Calculate the molar concentration of [ H + ] in 0.16 M solution of dichloroacetic acid Cl 2 CHCOOH ( Ka = 5 X 10-3 ). The concentration of the acid is initial ( before dissociation) Cl 2 CHCOOH Cl 2 CHCOO + H + The concentration of Cl 2 CHCOO and H + A) 0.16-x B) 0.16+x C) X
12 Unit3: STOICHIOMETRY AND EQUILIBRIUM Calculate the molar concentration of [ H + ] in 0.16 M solution of dichloroacetic acid Cl 2 CHCOOH ( Ka = 5 X 10-3 ). The concentration of the acid is initial ( before dissociation) Cl 2 CHCOOH Cl 2 CHCOO- + H + The concentration of Cl 2 CHCOOH and H + A) 0.16-x B) 0.16+x C) X
13 Unit3: STOICHIOMETRY AND EQUILIBRIUM Calculate the molar concentration of [ H + ] in 0.16 M solution of dichloroacetic acid Cl 2 CHCOOH ( Ka = 5 X 10-3 ). The concentration of the acid is initial ( before dissociation) Cl 2 CHCOOH Cl 2 CHCOO- + H + The value of x A) 5x10-3 B) 0.16 C) 5.9x10 3
14 Unite 4: ACID BASE EQUILIBRIUM Acid and base strong A) high conc. B) high vol. C) Completely dissociate
15 Unite 4: ACID BASE EQUILIBRIUM ph A) 1-7 B) 1-14 C) 7-14 D) 0-14
16 Unite 4: ACID BASE EQUILIBRIUM POH A) 1-7 B) 1-14 C) 7-14 D) 0-14
17 Unite 4: ACID BASE EQUILIBRIUM ph of HCl(0.01M) A)pH= 1 B) ph=2 ph= 3
18 Unite 4: ACID BASE EQUILIBRIUM poh of Ba(OH)2 (0.01M) A)pOH= 0.7 B) poh=1.7 C) poh= 3
19 Unite 4: ACID BASE EQUILIBRIUM poh of NaOH(0.03M) A)pOH= 1 B) poh=1.5 poh= 3 poh of HCl(0.01M) A)pOH= 2 B) poh=5 poh= 12
20 Unite 4: ACID BASE EQUILIBRIUM ph of strong acid : A) B) C)
21 Unite 4: ACID BASE EQUILIBRIUM ph of strong base : A) B) C)
22 Unite 4: ACID BASE EQUILIBRIUM ph of weak acide : A) B) C)
23 Unite 4: ACID BASE EQUILIBRIUM ph of weak base: A) B) C)
24 Unite 4: ACID BASE EQUILIBRIUM K w : A) B) C) D)
25 Unite 4: ACID BASE EQUILIBRIUM A) B) C) D)
26 Unite 4: ACID BASE EQUILIBRIUM Calculate the ph of. 0.5 M solution of NH 4 Cl? Kb ( NH 3 ) = 1.75 X 10-5 A) B) C) D)
27 Unite 4: ACID BASE EQUILIBRIUM Calculate the ph of. 0.5 M solution of NH 4 Cl? Kb ( NH 3 ) = 1.75 X 10-5 A) B) C) D)
28 Unite 4: ACID BASE EQUILIBRIUM Calculate the ph of. 0.5 M solution of NH 4 Cl? Kb ( NH 3 ) = 1.75 X 10-5 A) ph = 2.1 B) ph = 4.8 C) ph= 7 D) ph = 9.6
29 Unite 4: ACID BASE EQUILIBRIUM Calculate the ph of 0.2 M solution of Ba(OH)2 A) poh = 12.1 B) poh = 8.9 C) poh = 5.5 D) poh = 0.4
30 Unite 4: ACID BASE EQUILIBRIUM Calculate the ph of 0.2 M solution of Ba(OH)2 A) 9.6 B) 11.6 C) 13.6
31 Unite 4: ACID BASE EQUILIBRIUM ph of salt ( strong acid+ weak base) A) ph=7 B) C) D)
32 Unite 4: ACID BASE EQUILIBRIUM ph of salt ( strong acid+ strong base) A) ph=7 B) C) poh = log k w C s k a ph of salt ( strong base + weak acid) A) ph=7 B) C) poh = log k w C s k a )
33 Unite 5: BUFFER SOLUTIONS ph of CH 3 COOH /CH 3 COONa A) poh = pk b + log C s C b B)pH = pk a + log C s C a C)pOH = log k w C s k a
34 Unite 5: BUFFER SOLUTIONS ph of NH 4 OH / NH 4 Cl A) poh = pk b + log C s C b B) ph = pk a + log C s C a C) poh = log k w C s k a
35 Unite 5: BUFFER SOLUTIONS Calculate the ph of a solution containing 0.1 M Na 2 CO 3 and o.2 M NaHCO 3? Ka2 (H 2 CO 3 ) = 4.7 X A) ph= 5 B) ph= 8 C) ph=10
36 Unite 5: BUFFER SOLUTIONS Calculate the ph of a solution resulting from adding 4 ml of 0.2 M of NaOH solution to 16 ml of a buffer solution containing 0.1 M CH 3 COOH ( K a = 1.8X10-5 ) and 0.2 M CH 3 COONa The number of mmol CH 3 COOH A) 0.8 mmol B) 1.6 mmol C) 4 mmol
37 Unite 5: BUFFER SOLUTIONS Calculate the ph of a solution resulting from adding 4 ml of 0.2 M of NaOH solution to 16 ml of a buffer solution containing 0.1 M CH 3 COOH ( K a = 1.8X10-5 ) and 0.2 M CH 3 COONa The number of mmol NaOH A) 0.8 mmol B) 1.6 mmol C) 4 mmol
38 Unite 5: BUFFER SOLUTIONS Calculate the ph of a solution resulting from adding 4 ml of 0.2 M of NaOH solution to 16 ml of a buffer solution containing 0.1 M CH 3 COOH ( K a = 1.8X10-5 ) and 0.2 M CH 3 COONa The number of mmol CH 3 COONa A) 0.8 mmol B) 1.6 mmol C) 4 mmol
39 Unite 5: BUFFER SOLUTIONS Calculate the ph of a solution resulting from adding 4 ml of 0.2 M of NaOH solution to 16 ml of a buffer solution containing 0.1 M CH 3 COOH ( K a = 1.8X10-5 ) and 0.2 M CH 3 COONa The ph of a solution A) ph = 2 B) ph= 5.4 C) ph= 9.4
40 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS In the case of titrating 10 ml solution of 0.1 M of Cl - ( in the conical flask ) by 0.2 M solution of Ag + ( in the burette ), calculate Veq.p of Ag + solution A)2.5 ml B)5 ml C)10 ml
41 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS In the case of titrating 10 ml of 0.1 M of H 2 SO 4 solution by 0.2 M of NaOH solution, calculate Veq.p of NaOH solution A)2.5 ml B)5 ml C)10 ml
42 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS Endpoint is A) The same (equal) equivalence point B) Before equivalence point C) After equivalence point
43 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS 300 mg of a Na 2 C 2 O 4 ( mw = 134 ) 95 %w/w pure reagent was transferred to a titration conical flask. After adding acid solution and a suitable indicator, C 2 O 4 2- was titrated with KMnO4 unknown solution according to the following titration reaction equation : 2 KMnO C 2 O H Mn CO2 + 8H2O If the volume of KMnO 4 solution at the equivalent point was 34 ml, calculate the molarity of KMnO 4 solution, the pure wieght of Na 2 C 2 O 4 A) 300 mg B) 285 mg C) mg
44 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS 300 mg of a Na 2 C 2 O 4 ( mw = 134 ) 95 %w/w pure reagent was transferred to a titration conical flask. After adding acid solution and a suitable indicator, C 2 O 4 2- was titrated with KMnO4 unknown solution according to the following titration reaction equation : 2 KMnO C 2 O H Mn CO2 + 8H2O If the volume of KMnO 4 solution at the equivalent point was 34 ml, calculate the molarity of KMnO 4 solution, the moles of Na 2 C 2 O 4 A) 5.61 mol B) 2.12 mol C) 0.85 mol
45 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS 300 mg of a Na 2 C 2 O 4 ( mw = 134 ) 95 %w/w pure reagent was transferred to a titration conical flask. After adding acid solution and a suitable indicator, C 2 O 4 2- was titrated with KMnO4 unknown solution according to the following titration reaction equation : 2 KMnO C 2 O H Mn CO2 + 8H2O If the volume of KMnO 4 solution at the equivalent point was 34 ml, calculate the molarity of KMnO 4 solution, the moles of KMnO 4 A) 5.61 mol B) 2.12 mol C) 0.85 mol
46 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS 750 mg of a sample containing iron ( aw = ) was transferred into a titration conical flask and all iron was converted to Fe 3+, then unknown excess of KI was added and the following reaction was occurred 2 I Fe 3+ I2 + 2 Fe 2+ The iodine I2 formed was titrated with M of Na2S2O3 solution using starch as indicator according to the following reaction equation : mmol of S 4 O 6 2- A) B) 1.39 C) 4.77 S 4 O I 2 S 2 O I -
47 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS 750 mg of a sample containing iron ( aw = ) was transferred into a titration conical flask and all iron was converted to Fe 3+, then unknown excess of KI was added and the following reaction was occurred 2 I Fe 3+ I2 + 2 Fe 2+ The iodine I2 formed was titrated with M of Na2S2O3 solution using starch as indicator according to the following reaction equation : S 4 O I 2 S 2 O I - mmol of I 2 A) B) 1.39 C) 0.69
48 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS 750 mg of a sample containing iron ( aw = ) was transferred into a titration conical flask and all iron was converted to Fe 3+, then unknown excess of KI was added and the following reaction was occurred 2 I Fe 3+ I2 + 2 Fe 2+ The iodine I2 formed was titrated with M of Na2S2O3 solution using starch as indicator according to the following reaction equation : S 4 O I 2 S 2 O I - mmol of Fe 3+ A) B) 1.39 C) 0.69
49 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS g of a sample containing Cr 6+ ( aw = 52 ) is dissolved in a conical flask. An excess of 1.5 g of Fe 2+ ( aw = ) are added which is oxidized to Fe 3+ during its reduction of Cr 6+ to Cr 3+. The excess Fe 2+ is titrated with 0.05 N of K 2 Cr 2 O 7 and was found to require 10 ml at the equivalent point. Calculate the %w/w percentage of Cr 6+ in the sample. m.eq. of K 2 Cr 2 O 7 A) 0.05 m.eq B) 0.5 m.eq C) 1.5 m.eq
50 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS g of a sample containing Cr 6+ ( aw = 52 ) is dissolved in a conical flask. An excess of 1.5 g of Fe 2+ ( aw = ) are added which is oxidized to Fe 3+ during its reduction of Cr 6+ to Cr 3+. The excess Fe 2+ is titrated with 0.05 N of K 2 Cr 2 O 7 and was found to require 10 ml at the equivalent point. Calculate the %w/w percentage of Cr 6+ in the sample. m.eq. of Fe 2+ (excess) which reacted with K 2 Cr 2 O 7 A) 0.05 m.eq B) 0.5 m.eq C) 1.5 m.eq
51 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS g of a sample containing Cr 6+ ( aw = 52 ) is dissolved in a conical flask. An excess of 1.5 g of Fe 2+ ( aw = ) are added which is oxidized to Fe 3+ during its reduction of Cr 6+ to Cr 3+. The excess Fe 2+ is titrated with 0.05 N of K 2 Cr 2 O 7 and was found to require 10 ml at the equivalent point. Calculate the %w/w percentage of Cr 6+ in the sample. m.eq. of Fe 2+ total A) 0.05 m.eq B) 0.78 m.eq C) m.eq
52 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS g of a sample containing Cr 6+ ( aw = 52 ) is dissolved in a conical flask. An excess of 1.5 g of Fe 2+ ( aw = ) are added which is oxidized to Fe 3+ during its reduction of Cr 6+ to Cr 3+. The excess Fe 2+ is titrated with 0.05 N of K 2 Cr 2 O 7 and was found to require 10 ml at the equivalent point. Calculate the %w/w percentage of Cr 6+ in the sample. m.eq. of Fe 2+ which reacted with Cr 6+ A) 0.05 m.eq B) 0.78 m.eq C) m.eq
53 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS g of a sample containing Cr 6+ ( aw = 52 ) is dissolved in a conical flask. An excess of 1.5 g of Fe 2+ ( aw = ) are added which is oxidized to Fe 3+ during its reduction of Cr 6+ to Cr 3+. The excess Fe 2+ is titrated with 0.05 N of K 2 Cr 2 O 7 and was found to require 10 ml at the equivalent point. Calculate the %w/w percentage of Cr 6+ in the sample. m.eq. of Cr 6+ A) 0.05 m.eq B) 0.78 m.eq C) m.eq
54 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS g of a sample containing Cr 6+ ( aw = 52 ) is dissolved in a conical flask. An excess of 1.5 g of Fe 2+ ( aw = ) are added which is oxidized to Fe 3+ during its reduction of Cr 6+ to Cr 3+. The excess Fe 2+ is titrated with 0.05 N of K 2 Cr 2 O 7 and was found to require 10 ml at the equivalent point. Calculate the %w/w percentage of Cr 6+ in the sample. mass of Cr 6+ A) mg B) mg C) mg
55 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS g of a sample containing Cr 6+ ( aw = 52 ) is dissolved in a conical flask. An excess of 1.5 g of Fe 2+ ( aw = ) are added which is oxidized to Fe 3+ during its reduction of Cr 6+ to Cr 3+. The excess Fe 2+ is titrated with 0.05 N of K 2 Cr 2 O 7 and was found to require 10 ml at the equivalent point. Calculate the %w/w percentage of Cr 6+ in the sample. A) % B) % C) 25.60%
56 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS 10 ml of 0.1 M of analyte A were titrated with 0.2 M of titrant B according to the following titration reaction equation : A + 2 B 3 E, Keq = 1x10 10 Calculate the molar concentration of each [A], [B] and [E] in the conical flask after the following additions of the titrant B solution : (1) 5 ml (2) 10 ml (3) 15 ml calculate the volume of the titrant B solution at the equivalent point A) 5 ml B) 2.5mL C) 10 ml
57 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS 10 ml of 0.1 M of analyte A were titrated with 0.2 M of titrant B according to the following titration reaction equation : A + 2 B 3 E, Keq = 1x10 10 Calculate the molar concentration of each [A], [B] and [E] in the conical flask after the following additions of the titrant B solution : (1) 3 ml (2) 10 ml (3) 15 ml calculate the molar concentration of [E] A) 1 ml B) 0.07mL C) 0.1 ml
58 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS 10 ml of 0.1 M of analyte A were titrated with 0.2 M of titrant B according to the following titration reaction equation : A + 2 B 3 E, Keq = 1x10 10 Calculate the molar concentration of each [A], [B] and [E] in the conical flask after the following additions of the titrant B solution : (1) 3 ml (2) 10 ml (3) 15 ml calculate the molar concentration of [A] A) 1 M B) 0.03 M C) 0.05 M
59 Unite 6: INTRODUCTION TO VOLUMETRIC ANALYSIS 10 ml of 0.1 M of analyte A were titrated with 0.2 M of titrant B according to the following titration reaction equation : A + 2 B 3 E, Keq = 1x10 10 Calculate the molar concentration of each [A], [B] and [E] in the conical flask after the following additions of the titrant B solution : (1) 3 ml (2) 10 ml (3) 15 ml calculate the molar concentration of [B] A) 0.3 M B) 1.8x10-6 M C) 8.3x10-7
60 Unit7: ACID BASE TITRATION CURVES If 10 ml of 0.1 M CH 3 COOH ((Ka = 1.8 x 10-5 ) is titrated with 0.2 M NaOH solution, calculate the ph of the titration solution in the conical flask after the addition of the following volumes of NaOH solution : V.eq.p A) 5 B) 10 C) 15
61 Unit7: ACID BASE TITRATION CURVES If 10 ml of 0.1 M CH 3 COOH ((Ka = 1.8 x 10-5 ) is titrated with 0.2 M NaOH solution, calculate the ph of the titration solution in the conical flask after the addition of the following volumes of NaOH solution : 3mL A) 2.91 B) 4.96 C) 11.51
62 Unit7: ACID BASE TITRATION CURVES If 10 ml of 0.1 M CH 3 COOH ((Ka = 1.8 x 10-5 ) is titrated with 0.2 M NaOH solution, calculate the ph of the titration solution in the conical flask after the addition of the following volumes of NaOH solution : 5mL A) 5.2 B) 8.8 C) 11.51
63 Unit7: ACID BASE TITRATION CURVES If 10 ml of 0.1 M CH 3 COOH ((Ka = 1.8 x 10-5 ) is titrated with 0.2 M NaOH solution, calculate the ph of the titration solution in the conical flask after the addition of the following volumes of NaOH solution : 10mL A) 2.91 B) 1.30 C) 12.71
64 Unit7: ACID BASE TITRATION CURVES A) if pka = 3, so, sharper the region of the curve near the equivalent point B) if pka = 7, so, sharper the region of the curve near the equivalent point C) if pka = 10, so, sharper the region of the curve near the equivalent point
65 Unite 8: ACID BASE TITRATION INDICATORS At a low ph, the human can distinguish the acidic color if: A) B) C)
66 Unite 8: ACID BASE TITRATION INDICATORS At a high ph, the human can distinguish the acidic color if: A) B) C)
67 Unite 8: ACID BASE TITRATION INDICATORS At low ph ( acidic), The ph of indicator is : A) ph In = pk HIn -1 B) ph In = pk HIn +1 C) pk HIn = ph In -1
68 Unite 8: ACID BASE TITRATION INDICATORS The Indicator s Range is: A) pk HIn = pk HIn ±1 B) pk HIn = ph In ±1 C) ph In = pk HIn ±1
69 Unite 8: ACID BASE TITRATION INDICATORS How do the following indicators behave with the titration of 10 ml of 0.05 M of the weak acid HA (K a = 1 X 10-5 ) with 0.1 M NaOH : (1) pk In = 7 (2) pk In = 9 (3) pk In = 11 Calculate V eq.p. A) 2.5 ml B) 5mL C) 7.5mL
70 Unite 8: ACID BASE TITRATION INDICATORS How do the following indicators behave with the titration of 10 ml of 0.05 M of the weak acid HA (K a = 1 X 10-5 ) with 0.1 M NaOH : (1) pk In = 7 (2) pk In = 9 (3) pk In = 11 Calculate the ph at eq.p. A) 4.7 ml B) 9.3mL C) 12.5mL
71 Unite 8: ACID BASE TITRATION INDICATORS How do the following indicators behave with the titration of 10 ml of 0.05 M of the weak acid HA (K a = 1 X 10-5 ) with 0.1 M NaOH : (1) pk In = 7 (2) pk In = 9 (3) pk In = 11 The suitable indicatore is A) ) pk In = 7 B) pk In = 9 C) pk In = 11
72 Unite 8: ACID BASE TITRATION INDICATORS How do the following indicators behave with the titration of 20 ml of 0.1 M of NH 3 ( K b 2 X 10-5 ) with 0.4 M HCl : (1) pk In = 3 (2) pk In = 5 (3) pk In = 7 Calculate V eq.p. A) 2.5 ml B) 5mL C) 7.5mL
73 Unite 8: ACID BASE TITRATION INDICATORS How do the following indicators behave with the titration of 20 ml of 0.1 M of NH 3 ( K b 2x 10-5 ) with 0.4 M HCl : (1) pk In = 3 (2) pk In = 5 (3) pk In = 7 Calculate the ph at eq.p. A) 11.9 B) 9.8 C) 5.2
74 Unite 8: ACID BASE TITRATION INDICATORS How do the following indicators behave with the titration of 20 ml of 0.1 M of NH 3 ( K b 2x 10-5 ) with 0.4 M HCl : (1) pk In = 3 (2) pk In = 5 (3) pk In = 7 The suitable indicatore is A) pk In = 3 B) pk In = 5 C) pk In = 7
75 Unite 9: COMPLEX FORMATION TITRATION Example of unidentate A) H 2 NCH 2 CH 2 NH 2 B) Br -, F -, I -, CN-, RNH2, RCOO C) EDTA
76 Unite 9: COMPLEX FORMATION TITRATION The charge of complex Fe 3+, Fe(CN) 6 A) 3+ B) 3- C) 6- D) 6+
77 Unite 9: COMPLEX FORMATION TITRATION Calculate the molar concentration of [CN - ] in a solution of Cu( CN ) 4 2- (Kd = 5.2 X ) prepared by dissolving 0.05 mole of the complex in water and completing the volume to one liter A) 2.52x B) 5.63x10-2 C) 32.2x10-28
Lecture #11-Buffers and Titrations The Common Ion Effect
Lecture #11-Buffers and Titrations The Common Ion Effect The Common Ion Effect Shift in position of an equilibrium caused by the addition of an ion taking part in the reaction HA(aq) + H2O(l) A - (aq)
INTRODUCTION TO VOLUMETRIC ANALYSIS
Objectives Subjects At the end of this unit the student is expected to be able to : 1- Answer questions such as : what is volumetric analysis?, Titration?, equivalent point?, end point?, primary standard?,
Titration a solution of known concentration, called a standard solution
Acid-Base Titrations Titration is a form of analysis in which we measure the volume of material of known concentration sufficient to react with the substance being analyzed. Titration a solution of known
Acid-Base Titration Solution Key
Key CH3NH2(aq) H2O(l) CH3NH3 (aq) OH - (aq) Kb = 4.38 x 10-4 In aqueous solution of methylamine at 25 C, the hydroxide ion concentration is 1.50 x 10-3 M. In answering the following, assume that temperature
Preparation of different buffer solutions
Preparation of different buffer solutions 1 - Buffers: - All biochemical reactions occur under strict conditions of the concentration of hydrogen ion. - Biological life cannot withstand large changes in
Understanding the shapes of acid-base titration curves AP Chemistry
Understanding the shapes of acidbase titration curves AP Chemistry Neutralization Reactions go to Completion Every acidbase reaction produces another acid and another base. A neutralization reaction is
Analyte: The substance whose concentration is not known in a titration. Usually the analyte is in the flask or beaker beneath the burette.
Key Worksheet 15 Acids & Base Equilibria: Acid Base Titrations Objectives To be able to calculate the ph, poh, and concentrations of all species present at any point of an acid base titration. Vocabulary
CHM 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
2] What is the difference between the end point and equivalence point for a monobasicmonoacid
![2] What is the difference between the end point and equivalence point for a monobasicmonoacid](/thumbs/79/79657977.jpg "2] What is the difference between the end point and equivalence point for a monobasicmonoacid")
4 Titrations modified October 9, 2013 1] A solution of 0.100 M AgNO 3 is used to titrate a 100.00 ml solution of 0.100 M KCl. The K sp of AgCl is 1.8e-11 a) What is pag if 50.00 ml of the titrant is added
AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS
AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS THE COMMON ION EFFECT The common ion effect occurs when the addition of an ion already present in the system causes the equilibrium to shift away
The Common Ion Effect
Chapter 17 ACID BASE EQUILIBRIA (Part I) Dr. Al Saadi 1 17.1 The Common Ion Effect A phenomenon known as the common ion effect states that: When a compound containing an ion in common with an already dissolved
STIOCHIOMETRY Single answer type questions: 21. SO -2
STIOCHIOMETRY Single answer type questions: 1. SO - + S - + I (Calculated amount) X + I -1. Then X in this reaction is a) SO - b) IO -1 c) SO - d) IO -1. A certain quantity of Ammonium chloride is boiled
Buffer solutions Strong acids and bases dissociate completely and change the ph of a solution drastically. Buffers are solutions that resist changes i
18.3 ph Curves Buffer solutions Strong acids and bases dissociate completely and change the ph of a solution drastically. Buffers are solutions that resist changes in ph even when acids and bases are added
Chapter 15. Acid-Base Equilibria
Chapter 15 Acid-Base Equilibria The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion already involved in the equilibrium
Questions #4-5 The following two questions refer to the following system: A 1.0L solution contains 0.25M HF and 0.60M NaF (Ka for HF = 7.2 x 10-4 ).
Multiple Choice 1) A solution contains 0.250 M HA (K a = 1.0 x 10-6 ) and 0.45 M NaA. What is the ph after 0.10 mole of HCl is added to 1.00L of this solution? a. 3.17 b. 3.23 c. 6.00 d. 10.77 e. 10.83
Review of Chemistry 11
Review of Chemistry 11 HCl C 3 H 8 SO 2 NH 4 Cl KOH H 2 SO 4 H 2 O AgNO 3 PbSO 4 H 3 PO 4 Ca(OH) 2 Al(OH) 3 P 2 O 5 Ba(OH) 2 CH 3 COOH 1. Classify the above as ionic or covalent by making two lists. Describe
= ) = )
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
A 95 g/mol B 102 /mol C 117 g/mol D 126 g/mol E 152 g/mol
Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
1) Write the Brønsted-Lowry reaction for weak acid HCN reacting with H 2 O.
1) Write the Brønsted-Lowry reaction for weak acid HCN reacting with H O. HCN + H O º H O + + CN ) Write the Brønsted-Lowry reaction for weak base NH reacting with H O NH + H O º OH + NH + ) Using the
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
Steward Fall 08. Moles of atoms/ions in a substance. Number of atoms/ions in a substance. MgCl 2(aq) + 2 AgNO 3(aq) 2 AgCl (s) + Mg(NO 3 ) 2(aq)
Dealing with chemical stoichiometry Steward Fall 08 of Not including volumetric stoichiometry of Chapter 6.0x10 A 6.0x10 Mol/mol ratio from balanced equation B 6.0x10 6.0x10 s, Equations, and Moles: II
Acid-Base Equilibria (Chapter 10.) Problems: 2,3,6,13,16,18,21,30,31,33
Acid-Base Equilibria (Chapter 10.) Problems: 2,3,6,13,16,18,21,30,31,33 Review acid-base theory and titrations. For all titrations, at the equivalence point, the two reactants have completely reacted with
Applications of Aqueous Equilibria Chapter 15. Titration Curves & Indicators Sections 4-5
Applications of Aqueous Equilibria Chapter 15 Titration Curves & Indicators Sections 45 Strong Acid vs. Strong Base Titration Titrate 50.0 ml of 0.200 M HNO 3 with 0.100 M NaOH What is the ph when no NaOH
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
Lecture 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.
Kotz 7 th ed. Section 18.3, pp
Lecture 15 Acid/base reactions. Equilibria in aqueous solutions. Titrations Kotz 7 th ed. Section 18.3, pp.821-832. In a titration a solution of accurately known concentration is added gradually added
Web Appendix 6. Volumetric Calculations Using Normality and Equivalent Weight A6-1 THE DEFINITIONS OF EQUIVALENT AND MILLIEQUIVALENT
Web Appendix 6 Volumetric Calculations Using Normality and Equivalent Weight The normality of a solution expresses the number of equivalents of solute contained in 1 L of solution or the number of milliequivalents
Chemical Equilibrium. Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B
Chemical Equilibrium Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B The point reached in a reversible reaction where the rate of the forward reaction (product formation,
CHAPTER 7 Acid Base Equilibria
1 CHAPTER 7 Acid Base Equilibria Learning Objectives Acid base theories Acid base equilibria in water Weak acids and bases Salts of weak acids and bases Buffers Logarithmic concentration diagrams 2 ACID
ACID-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
Acid-Base Equilibria. And the beat goes on Buffer solutions Titrations
Acid-Base Equilibria And the beat goes on Buffer solutions Titrations 1 Common Ion Effect The shift in equilibrium due to addition of a compound having an ion in common with the dissolved substance. 2
TYPES OF CHEMICAL REACTIONS
TYPES OF CHEMICAL REACTIONS Precipitation Reactions Compounds Soluble Ionic Compounds 1. Group 1A cations and NH 4 + 2. Nitrates (NO 3 ) Acetates (CH 3 COO ) Chlorates (ClO 3 ) Perchlorates (ClO 4 ) Solubility
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 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
Equilibri acido-base ed equilibri di solubilità. Capitolo 16
Equilibri acido-base ed equilibri di solubilità Capitolo 16 The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.
Chemical Equilibrium
Chemical Equilibrium Many reactions are reversible, i.e. they can occur in either direction. A + B AB or AB A + B The point reached in a reversible reaction where the rate of the forward reaction (product
Applications of Aqueous Equilibria. Chapter 18
Applications of Aqueous Equilibria Chapter 18 What we learn from Chap 18 This chapter is the third in the three-chapter sequence about equilibrium, this one building upon the core principles raised in
Grade A buffer: is a solution that resists changes in its ph upon small additions of acid or base.sq1
Chapter 15 Lesson Plan Grade 12 402. The presence of a common ion decreases the dissociation. BQ1 Calculate the ph of 0.10M CH 3 COOH. Ka = 1.8 10-5. [H + ] = = ( )( ) = 1.34 10-3 M ph = 2.87 Calculate
IB Chemistry ABS Introduction An acid was initially considered a substance that would produce H + ions in water.
IB Chemistry ABS Introduction An acid was initially considered a substance that would produce H + ions in water. The Brønsted-Lowry definition of an acid is a species that can donate an H + ion to any
A buffer is a an aqueous solution formed from a weak conjugate acid-base pair that resists ph change upon the addition of another acid or base.
1 A buffer is a an aqueous solution formed from a weak conjugate acid-base pair that resists ph change upon the addition of another acid or base. after addition of H 3 O + equal concentrations of weak
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
OCR A Chemistry A-Level Module 5 - Physical Chemistry & Transition Elements
OCR A Chemistry A-Level Module 5 - Physical Chemistry & Transition Elements Acids and Bases Notes and Example Calculations Answers given at the end of the booklet ph Acid Calculations Strong Acids To calculate
Lecture 12. Acid/base reactions. Equilibria in aqueous solutions.
Lecture 12 Acid/base reactions. Equilibria in aqueous solutions. Titrations Kotz 7 th ed. Section 18.3, pp.821-832. In a titration a solution of accurately known concentration is added gradually added
CHEM 1412 Answers to Practice Problems Chapters 15, 16, & 17
CHEM 1412 Answers to Practice Problems Chapters 15, 16, & 17 1. Definitions can be found in the end-of-chapter reviews and in the glossary at the end of the textbook! 2. Conjugate Base Conjugate Acid Compound
ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor
ph calculations MUDr. Jan Pláteník, PhD Brønsted-Lowry concept of acids and bases Acid is a proton donor Base is a proton acceptor HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) Acid Base Conjugate acid Conjugate
Normality Primer. Normality (N) = equivalents/liter = eq/l Examples: 1.0 M HCl = 1.0 N HCl 1.0 M H 2 S = 2.0 N H 2 S
Normality Primer Introduction Normality, equivalents and equivalent weight are all related terms typically used in titrations when the titration reaction is unknown or just not used. Consequently, definitions
-a base contains an OH group and ionizes in solutions to produce OH - ions: Neutralization: Hydrogen ions (H + ) in solution form
NOTES Acids, Bases & Salts Arrhenius Theory of Acids & Bases: an acid contains hydrogen and ionizes in solutions to produce H+ ions: a base contains an OH group and ionizes in solutions to produce OH ions:
Acid-Base Equilibria and Solubility Equilibria Chapter 17
PowerPoint Lecture Presentation by J. David Robertson University of Missouri Acid-Base Equilibria and Solubility Equilibria Chapter 17 The common ion effect is the shift in equilibrium caused by the addition
Chemistry 12 APRIL Course Code = CH. Student Instructions
MINISTRY USE ONLY MINISTRY USE ONLY Place Personal Education Number (PEN) here. Place Personal Education Number (PEN) here. MINISTRY USE ONLY Chemistry 12 2002 Ministry of Education APRIL 2002 Course Code
Chemistry 12 UNIT 4 ACIDS AND BASES
Chemistry 12 UNIT 4 ACIDS AND BASES PACKAGE #6 TITRATION -allows you to react measured amounts of a solution with a known volume of another solution until an equivalence point is reached. Recall from Indicators
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
mohd faisol mansor/chemistry form 4/chapter 7 CHAPTER 7 ACIDS AND BASES HCl (g) H 2 O H + (aq) + Cl - (aq) NaOH(s) H 2 O Na + (aq) + OH - (aq)
CHAPTER 7 ACIDS AND BASES Arrhenius Theory An acid is a chemical compound that produces hydrogen ions, H + or hydroxonium ions H3O + when dissolve in water. A base defined as a chemical substance that
15.0 g Fe O 2 mol Fe 55.8 g mol Fe = g
CHAPTER Practice Questions.1 1 Mg, O, H and Cl (on each side).. BaCl (aq) + Al (SO ) (aq) BaSO (s) + AlCl (aq).5 0.15 mol 106 g mol 1 = 1. g 15.0 g Fe O mol Fe 55.8 g mol Fe = 10.9 g 1 159.7 g mol FeO
Chemistry 3202 Pre-Public Examination May 2012 Name:
Chemistry 3202 Pre-Public Examination May 2012 Name: Section A: Multiple Choice This section contains 40 multiple choice covering concepts from the entire course. Please answer all multiple choice items
Student Worksheet for Buffers, Ksp, and Titrations
Student Worksheet for Attempt to work the following practice problems after working through the sample problems in the videos. Answers are given on the last page(s). Relevant Equations Keq= [C]c [D] d
Stoichiometry ( ) ( )
Stoichiometry Outline 1. Molar Calculations 2. Limiting Reactants 3. Empirical and Molecular Formula Calculations Review 1. Molar Calculations ( ) ( ) ( ) 6.02 x 10 23 particles (atoms or molecules) /
Dilutions 4/8/2013. Steps involved in preparing solutions from pure solids. Steps involved in preparing solutions from pure solids
Steps involved in preparing solutions from pure solids Steps involved in preparing solutions from pure solids Calculate the amount of solid required Weigh out the solid Place in an appropriate volumetric
Chemistry 265 December Exam 2011 Smith-Palmer
1 Chemistry 265 December Exam 2011 Smith-Palmer NAME: [1] 1. Define an anode [1] Define a cathode [2] What is the E o for the following reaction: Ag + + Cu Ag (s) + Cu 2+ Ag + + e - Ag (s) E o = 0.799
Acids and bases, ph and buffers. Dr. Mamoun Ahram Lecture 2
Acids and bases, ph and buffers Dr. Mamoun Ahram Lecture 2 ACIDS AND BASES Acids versus bases Acid: a substance that produces H+ when dissolved in water (e.g., HCl, H2SO4) Base: a substance that produces
Analytical Chemistry Lecture III by/ Dr. Ekhlas Q. J. BUFFER SOLUTIONS
Analytical Chemistry Lecture III by/ Dr. Ekhlas Q. J. BUFFER SOLUTIONS Buffer solutions Definition Solutions which resist changes in ph when small quantities of acid or alkali are added. a solution that
CH 4 AP. Reactions in Aqueous Solutions
CH 4 AP Reactions in Aqueous Solutions Water Aqueous means dissolved in H 2 O Moderates the Earth s temperature because of high specific heat H-bonds cause strong cohesive and adhesive properties Polar,
Acid-Base Titrations
Acid-Base Titrations The Titration One of the most important lab procedures involving acids and bases is the titration. A titration is an analytical procedure that allows for the measurement of the amount
Chapter 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
CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY
CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY 1 Chapter Three Goals 1. Chemical Equations. Calculations Based on Chemical Equations. Percent Yields from Chemical Reactions 4. The Limiting Reactant
The Solved Problems in Analytical Chemistry
The Solved Problems in Analytical Chemistry Dr. Mohammed Zaboon Al- saidi 2017-2018 Moles Q1: Calculate the number of mole of ammonia, NH 3, required to produce 2.5 mol of Cu(NH 3 ) 4 SO 4 according to
School of Chemistry, University of KwaZulu-Natal, Howard College Campus, Durban. CHEM191 Tutorial 1: Buffers
School of Chemistry, University of KwaZulu-Natal, Howard College Campus, Durban CHEM191 Tutorial 1: Buffers Preparing a Buffer 1. How many moles of NH 4 Cl must be added to 1.0 L of 0.05 M NH 3 to form
Classes at: - Topic: Ionic Equilibrium
PHYSICAL CHEMISTRY by: SHAILENDRA KR Classes at: - SCIENCE TUTORIALS; Opp Khuda Baksh Library, Ashok Rajpath, Patna PIN POINT STUDY CIRCLE; House No 5A/65, Opp Mahual Kothi, Alpana Market, Patna Topic:
Lecture 5. Percent Composition. etc. Professor Hicks General Chemistry II (CHE132) Percent Composition. (aka percent by mass) 100 g.
Lecture 5 Professor Hicks General Chemistry II (CHE132) Percent Composition (aka percent by mass) % by mass component 1 = mass component 1 mass sample 100% sample component 1 100 g sample component 1 component
Quiz name: Equilibria + Acids/Bases
Name: Quiz name: Equilibria + Acids/Bases Date: 1. 2. At 450 C, 2.0 moles each of H 2(g), I 2(g), and HI are combined in a 1.0 L rigid container. The value of K c at 450 C is 50. Which of the following
Unit 3 Chemistry - Volumetric Analysis
Unit 3 Chemistry Volumetric Analysis Volumetric analysis is a quantitative chemical analysis used to determine the unknown concentration of one reactant [the analyte] by measuring the volume of another
Unit 4a: Solution Stoichiometry Last revised: October 19, 2011 If you are not part of the solution you are the precipitate.
1 Unit 4a: Solution Stoichiometry Last revised: October 19, 2011 If you are not part of the solution you are the precipitate. You should be able to: Vocabulary of water solubility Differentiate between
Le Chatlier's principle can be used to decide whether the above equilibrium will be shifted left or right
Problems, Chapter 17 (with solutions) NOTE: Unless otherwise stated, assume T = 25. C in all problems) 1) In which of these solutions will HNO2 ionize less than it does in pure water? a) 0.10 M NaCl b)
ANSWER KEY CHEMISTRY F14O4 FIRST EXAM 2/16/00 PROFESSOR J. MORROW EACH QUESTION IS WORTH 1O POINTS O. 16.
discard 1 2 ANSWER KEY CHEMISTRY F14O4 FIRST EXAM 2/16/00 PROFESSOR J. MORROW PRINT NAME, LAST: FIRST: I.D.# : EACH QUESTION IS WORTH 1O POINTS 1. 7. 13. 2. 8. 14. 3. 9. 15. 4. 1O. 16. 5. 11. 17. 6. 12.
Chapter 8: Applications of Aqueous Equilibria
Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations
Chem 1A Dr. White Fall Handout 4
Chem 1A Dr. White Fall 2014 1 Handout 4 4.4 Types of Chemical Reactions (Overview) A. Non-Redox Rxns B. Oxidation-Reduction (Redox) reactions 4.6. Describing Chemical Reactions in Solution A. Molecular
Chem 115 POGIL Worksheet - Week #6 - Answers Oxidation Numbers, Redox Reactions, Solution Concentration, Titrations, First Law, and Enthalpy
Chem 115 POGIL Worksheet - Week #6 - Answers Oxidation Numbers, Redox Reactions, Solution Concentration, Titrations, First Law, and Enthalpy Key Questions, Exercises, and Problems 1. Assign the oxidation
Chapter 4 Chemical Quantities and Aqueous Reactions
Chapter 4 Chemical Quantities and Aqueous Reactions Reaction Stoichiometry the numerical relationships between chemical amounts in a reaction is called stoichiometry the coefficients in a balanced chemical
1) What is the volume of a tank that can hold Kg of methanol whose density is 0.788g/cm 3?
1) Convert the following 1) 125 g to Kg 6) 26.9 dm 3 to cm 3 11) 1.8µL to cm 3 16) 4.8 lb to Kg 21) 23 F to K 2) 21.3 Km to cm 7) 18.2 ml to cm 3 12) 2.45 L to µm 3 17) 1.2 m to inches 22) 180 ºC to K
2/4/2016. Chapter 15. Chemistry: Atoms First Julia Burdge & Jason Overby. Acid-Base Equilibria and Solubility Equilibria The Common Ion Effect
Chemistry: Atoms First Julia Burdge & Jason Overby 17 Acid-Base Equilibria and Solubility Equilibria Chapter 15 Acid-Base Equilibria and Solubility Equilibria Kent L. McCorkle Cosumnes River College Sacramento,
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
Titration of a weak acid with strong base
Titration of a weak acid with strong base - Objectives: - To study titration curves. - Determine the pka value of a weak acid. - Reinforce the understanding of buffers. - Titration Curves: - Titration
Homework: 14, 16, 21, 23, 27, 29, 39, 43, 48, 49, 51, 53, 55, 57, 59, 67, 69, 71, 77, 81, 85, 91, 93, 97, 99, 104b, 105, 107
Homework: 14, 16, 21, 23, 27, 29, 39, 43, 48, 49, 51, 53, 55, 57, 59, 67, 69, 71, 77, 81, 85, 91, 93, 97, 99, 104b, 105, 107 Chapter 15 Applications of Aqueous Equilibria (mainly acid/base & solubility)
PHYSICAL CHEMISTRY. by: SHAILENDRA KR. Meq. Approach. Classes at: - Topic: Mole Concept
PHYSICAL CHEMISTRY by: SHAILENDRA KR. asses at: - SCIENCE TUTORIALS; Opp. Khuda Baksh Library, Ashok Rajpath, Patna PIN POINT STUDY CIRCLE; House No. 5A/65, Opp. Mahual Kothi, Alpana Market, Patna Topic:
Chemistry 12. Resource Exam B. Exam Booklet
Chemistry 12 Resource Exam B Exam Booklet Contents: 21 pages Examination: 2 hours 50 multiple-choice questions in the Exam Booklet Additional Time Permitted: 60 minutes Province of British Columbia PART
Chapter 18 Problem Set Solutions S.E. Van Bramer 2/22/96
Chapter 18 Problem Set Solutions S.E. Van Bramer 2/22/96 1. Benzoic Acid (C6H5COOH) is a solid. 15.000 grams is dissolved in 200.00 ml of distilled water. a) Benzioic acid is a weak acid that will dissociate
5.1.3 Acids, Bases and Buffers
5..3 Acids, Bases and Buffers BronstedLowry Definition of Acid Base behaviour A BronstedLowry acid is defined as a substance that can donate a proton. A BronstedLowry base is defined as a substance that
CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.5 (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH(aq) H (aq) CH 3 COO (aq) Initial (): 0.40 0.00 0.00
Chemical Reaction Defn: Chemical Reaction: when starting chemical species form different chemicals.
Chemistry 11 Notes on Chemical Reactions Chemical Reaction Defn: Chemical Reaction: when starting chemical species form different chemicals. Evidence to indicate that a chemical reaction has occurred:
*In every acid-base reaction, equilibrium favors transfer of a proton from the stronger acid to the stronger base.
16.2 Bronsted-Lowry Acids and Bases An acid is a substance that can transfer a proton to another substance. A base is a substance that can accept a proton. A proton is a hydrogen ion, H +. Proton transfer
Problem Set III Stoichiometry - Solutions
Chem 121 Problem set III Solutions - 1 Problem Set III Stoichiometry - Solutions 1. 2. 3. molecular mass of ethane = 2(12.011) + 6(1.008) = 30.07 g 4. molecular mass of aniline = 6(12.011) + 7(1.008) +
Chapter 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
16.3 Weak Acids Weak Bases Titration
16.3 Weak Acids Weak Bases Titration Titration of Weak Acid with Strong Base Titration of Base Acid with Strong Acid Dr. Fred Omega Garces Chemistry 201 Miramar College 1 Weak Acids Weak Bases Titration
Formular weight of K 2 Cr 2 O 7 = 2( ) + 2( ) + 7( ) = (7 S.F.) (1 pt/each factor)
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 39.0983 ± 0.0003, Cr 51.9961
Institute for Competitive Examinations
INSTITUTE FOR COMETITIVE EXAMINATIONS DELHI CENTRES: E-6, Defence Colony New Delhi - 11004. h.: 4614, 46144. B-1/6, Main Nazafgarh Road, Janakpuri, New Delhi - 110058. h.: 011-557111/1/1/14. NOIDA CENTRE:
( 1 ) Concept of acid / base
Section 6.2 Ionic Equilibrium Unit 628 ( 1 ) Concept of acid / base The best definition of acids and bases is that proposed by T.M. Lowry and also, independently by J.N. Bronsted in 1923. BronstedLowry
Chapter 4: Chemical Quantities and Aqueous Reactions
Chapter 4: Chemical Quantities and Aqueous Reactions C (s) + O 2 (g) CO 2 (g) CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 0 (g) 2 C 8 H 18 (g) + 25 O 2 (g) 16 CO 2 (g) + 18 H 2 0 (g) Stoichiometry Calculations
CHAPTER 8 CHEMICAL EQUILIBRIUM SHORT QUESTION WITH ANSWERS Q.1 What is weak electrolyte? A compound which is only partially ionized in aqueous solution is called as weak electrolyte.e.g CH 3 COOH(Acetic
Copyright 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
Lecture 0403 Displacement Reactions, Molarity, Stoichiometric Calcs.
Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Lecture 0403 Displacement, Molarity, Stoichiometric Calcs. John D. Bookstaver St. Charles Community
Chapter 4 Reactions in Aqueous Solution
Chapter 4 Reactions in Aqueous Solution Homework Chapter 4 11, 15, 21, 23, 27, 29, 35, 41, 45, 47, 51, 55, 57, 61, 63, 73, 75, 81, 85 1 2 Chapter Objectives Solution To understand the nature of ionic substances
CHEM 1413 Chapter 4 Homework Questions TEXTBOOK HOMEWORK
CHEM 1413 Chapter 4 Homework Questions TEXTBOOK HOMEWORK Chapter 3 3.68 Calculate each of the following quantities: (a) Mass (g) of solute in 185.8 ml of 0.267 M calcium acetate (b) Molarity of 500. ml