Nun im Themenbereich Säure-Base_en
Topic Acid-Base ChemistryAll problems and solutions as
pdf here.Lernziele resp. Fragengebiete:
- From SB-01 : Definitions, What ? are Sof which:, Behaviour of Sof which:
- From SB-10 : Corresponding Sgold base Pairs
- From SB-20 : Statements to the SYour strength, pKs
- From SB-25 : Degree of dissociation One of them SYour
- From SB-40 : pH-Value - Dilutions
- From SB-60 : Mixtures - neutralization
- From SB-70 : Titrations
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- Behaviour One of them SYour or base (SB-03)Formulation reactions for the following: given Substances. The Abbreviation LM is for This LVegetable products.
Es Goes not That is why to to assess, whether or not The Required reactions possible are or not, but That is why, that z.B. In the case of One of them Required SYour reaction This given Particles a H+ is emitted and This LVegetable products This and then is absorbed.- a) HBr, SYour reaction, LM: Water
- b) HF, Baseline reaction, LM: Water
- c) HF, SYour reaction, LM: Water
- d) HF, SYour reaction, LM: NH3
- e) H2O, SYour reaction, LM: Water
- f) CH3COOH, SYour reaction, LM: Water
- g) CH3OH, Baseline reaction, LM: NH3
- h) CH3OH, SYour reaction, LM: Water
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- Konjugiert_2 (SB-11)Indicate the following: SGold base pairs more accurately (-of which: is conjoined Säure / Base of y etc'). Specifically Applicable It 's So what do you mean?, In the case of The exercise a) Responsibility. b) in each case All of them four Components with each other in Relationship to to set.
a) H2CO3 + 2 H2O ⇄ CO2-3 + 2 H3O+
b) NH3 + H2O ⇄ NH+4 + OH– - 1
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- Conjugated, Various Examples 1 (SB-12)Supplements the following: Table, Whereas This first Example solved is.
Should It 's for One Substance z.B. No conjoined base to give (z.B. F-), and then It should be The Cell content with a 'X' Notwithstanding the provisions will be.Other articles of clothing. base Other articles of clothing. SYour H2O OH– H3O+ HS– F– H2PO–4 PO-34 H3PO4 NH3 NO–2 - 1
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- Conjugated, Various Examples 2 (SB-13)Given are the following: reactions. Orderly The Participants Substances in Corresponding SGold base parts{a) } NH3 + HCl <=> NH4^+ + Cl^-{b) } HSO4^- + CN^- <=> HCN + SO4^2-{c) } H3O^+ + HS^- <=> H2S + H2O{d) } N2H4 + HSO4^- <=> N2H5^+ + SO4^2-{e) } H2O + NH2^- <=> NH3 + OH^-
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- Sof which: in Water (SB-14)The following: Sof which: are in Water solved. Es of which: The Adopted, that The Sof which: in full Deprotonizing. Write The This is the case for Ions by analogy to the exercise a) I ‘m going ..{a) Acetic acid: }CH3COOH + H2O <=> CH3COO^- + H3O^+{b) Hydrochloric acid:}{c) Fatty acids: }{d) Sulphuric acid: }{e) Hydrochloric acid: }{e) Carbonic acid: }
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- Different Sof which: drawing (SB-15)Drawing The Lewis structures The the following Sof which:.{a) Acetic acid: }{b) Hydrochloric acid:}{c) Fatty acids: }{d) Sulphuric acid: }{e) Hydrochloric acid: }{e) Carbonic acid: }
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- SYour strength (SB-22)Given be Acetic acid:CH3COOH{: } H-C<|H><`|H>-C<`||O>-O-H
What ? H's are Acid In the case of The Acetic acid: Those who H’s At the same time, Carbon or This H At the same time, Oxygen? - 1
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- 3 hours, 46 minutes ago
- Statements, which correctly (SB-25)Es will be One watered, 1 molar Lthe island One of them Unprotonated heavy SYour Manufacture ('HX'). What ? The the following Statements is correctly. Reasons The Selection.
- The SYour is to 0.1% dissolved
- The SYour is to 1% dissolved
- The SYour is to 10% dissolved
- The SYour is to 100% dissolved
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- Hfrequency The Particles (SB-26)What ? for Particles lying in One of them of a width of not more than 30 mm 1 M H2SO4-Lösung before? In which Relative Quantities are They of which:? Establishment One List of ranks, Starting with to the most common Particles.
pKs(H2SO4) be -3, pks(HSO–4) = 1.92.
Notes: Es Come on . It 's just … the following: Particles before: H2O, H3O+, HSO–4, SO2-4, (almost) not H2SO4, There you go. Sulphuric acid One heavy SYour is - 1
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- Interpretation pKs-Value, 1 (SB-27)Given be Sulphuric acid. The pKs(H2SO4) be -3, pks(HSO–4) = 1.92.
What ? The the following Statements is correctly? And then … almost more important: Why not??- c(H2SO4) > c(HSO–4 ) > c(SO2-4 )
- c(HSO–4) > c(H2SO4) > c(SO2-4)
- c(H2SO4) = c(HSO–4) > c(SO2-4)
- c(HSO–4) > c(SO2-4) > c(H2SO4)
- c(SO2-4) > c(HSO–4) > c(H2SO4)
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- Manufacture One of them specified Concentration (SB-40)Descriptions The Manufacture the following: LManufacture in which all the materials used are: In the Laboratory. This LVegetable products be Always Distilled Water.
Adopted The Substances are as salts of which: and Solving The Commission fully in Water.- a) 1 Litre One of them 0.3 M NaOH-Lthe island
- b) 1 Litre One of them 0.4 M HCL-Lthe island
- c) 25 Milligrams 0.5 M NaOH-Lthe island
- d) 30 Milligrams 0.6 M H2SO4
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- pH-Value 1 (SB-43)Calculate The pH One of them 3 molars Saltic acid solution.
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- Calculate Various Values (pH and so on) (SB-44)Calculate / Supplements The Values in the following: Table.
c(H3O+ mol/l) c(OH– mol/l) pH pOH 0.001 0.01 0.1 0.2 12 11 - 1
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- Acid, Neutral and Basically, Area specify (SB-45)Supplements The missing LOats and oats in The Table with The Values < 7, = 7, > 7, < 10-7, = 10-7 sowie > 10-7
Name and address Acid Lthe island Neutral Lthe island basic Lthe island c(H3O+ mol/l) pH c(OH– mol/l) pOH - 1
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- Concentrations Hydroxonium and so on (SB-46)Given be One Lthe island A with pH = 7 and One Lthe island B with pH = 8. Assessments (That 's right – wrong) the following: Statements:
- a) The H3O+-Konzentration of A is 10 times This is it. Large How The The Lthe island B
- b) The H3O+-Konzentration of A is Right now. The OH– Concentration of A
- c) The H3O+-Konzentration of B is 10 times small as The c(OH–) The Lthe island B
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- pH-calculations, 1 (SB-47)Calculate The pH-Values the following: LManufacture in which all the materials used are:, This LVegetable products be Always of a purity Water:
- 1 g Hydrochloric acid (HCl) solved in a Litre Water
- 1 kg Hydrochloric acid (HCl) solved in a Cubic metres Water
- 1 g NaOH, solved in a 2 Litre Water
- 81 mg HBr in 0.1 Litre Water
- 1 Grams Sulphuric acid in 1 m3 Water. Adopted: complete Deprotonation The Sulphuric acid.
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- pH-calculations, 2 (SB-48)Calculate The Weight to H3O+ and OH– in …
- a) … a Litre Water. pH = 6
- b) … In the Lake Constance. The pH be 7
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- pH-calculations, Dilutions (SB-49)Calculate The pH-Values the following: LManufacture in which all the materials used are:, This LVegetable products be Always of a purity Water:
- One of them. 0.03 molar HCl-Lthe island will be One hundred times diluted.
Calculate The pH before and after The Dilution - 20 Milligrams 0.05 M HCl will be with 100 Milligrams Distilled Water diluted.
Calculate The pH before and after The Dilution - To the 50 Milligrams One of them 0.2 molars NaOH-Lthe island will be 100 Milligrams Distilled Water given.
Calculate The pH before and after The Dilution
- One of them. 0.03 molar HCl-Lthe island will be One hundred times diluted.
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- Neutralizations (SB-60)Formulation The neutralization reactions in Ionic writing and Figure The This is the case for salts, by analogy to the exercise a).{a) Sodium hydroxide + Hydrochloric acid }Na^+ + OH^- + H3O^+ + Cl^- -> NaCl{(aq)} + 2H2O{b) Sodium hydroxide + Hydrochloric acid }{c) Sodium hydroxide + Carbonic acid }{d) Potassium hydroxide + Sulphuric acid }{e) Potassium hydroxide + Phosphoric acid }
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- pH-calculations, Mixtures (SB-61)Calculate The pH-Value the following: Mixtures.
- To the 80 Milligrams One of them 0.05 M Saltic acid solution will be 100 Milligrams One of them 0.01 M Sodium alloy solution given.
- To the 60 Milligrams One of them 0.015 KOH-Lthe island will be 30 Milligrams 0.2 M Saltic acid solution given.
- 10 Milligrams 0.2 M Saltic acid solution will be first. 10 times diluted and and then with
1.0 g Ca(OH)2 of which:.
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- pH-calculations, 4 (SB-62)Given be in each case a Litre of two LManufacture in which all the materials used are: A and B with to the equal to Starting-pH of 4.8. A be One Saltic acid solution, B One Acetic acid solution.
Well, you know what? will be to each Lthe island a Litre Water Admitted.- Calculate The new pH-Value The Lthe island A and B
- By which Lthe island has It's One Greater Changes of the pH-Values, Reasons
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- calculation titration 1 (SB-70)The titration One of them SYour (20 Milligrams, Unknown Concentration) with One of them base (0.1 M) Required to to the equivalence point 30 Milligrams The base. Calculate of which: The Concentrations The SYour.
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