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Topic Acid-Base ChemistryAll problems and solutions as
pdf here.Learning objectives resp. question areas:
- Ab SB-01 : definitions of acids, behaviour of acids
- from SB-10 : corresponding acid-base chemistry pairs
- from SB-20 : statements to the acid strength, pKs
- from SB-25 : degree of dissociation a acid
- from SB-40 : pH-value - dilutions
- From SB-60 : Mixtures - Neutralization
- From SB-70 : Titrations
- This forum has 29 topics, 29 replies, and was last updated 4 days, 22 hours ago by .
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- Behaviour of an acid or base (SB-03)Formulate reactions for the following given substances. the abbreviation LM stands for solvent.
It is not a question of judging whether The required reactions possible are or not, but that is why, that z.B. At a required acidic reaction the given particles a H+ is emitted and the solvents this and then is absorbed.- (a) HBr, acid reaction, LM: water
- (b) HF, base reaction, LM: water
- (c) HF, acid reaction, LM: water
- (d) HF, acid reaction, LM: NH3
- (e) H2O, acid reaction, LM: water
- (f) CH3COOH, acid reaction, LM: water
- (g) CH3OH, base reaction, LM: NH3
- (h) CH3OH, acid reaction, LM: water
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- Konjugiert_2 (SB-11)Indicate following acid-base chemistry-pairs more accurately ('x is conjoined acid / base of y and so on'). Specifically holds it thus, at the exercise a) or b) each all 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)Complete the following table with the first example solved.
Should be it for a substance z.B. No conjoined base to give (z.B. F-), and then should the cell contents are marked with an 'X'.Other articles of clothing. Base Other articles of clothing. Acid H2O OH– H3O+ HS– F– H2PO–4 PO-34 H3PO4 NH3 NO–2 - 1
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- Conjugated, various examples 2 (SB-13)Enter the following reactions. Order the participating substances in corresponding acid – bases 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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- Acids in water (SB-14)The following acids are dissolved in water. Es of which: the adopted, that the acids in full deprotonizing. Write the corresponding ions analogue to exercise (a).{(a) Acetic acid: }CH3COOH + H2O <=> CH3COO^- + H3O^+{(b) Acidic acid:}{(c) Folic acid: }{(d) Sulphuric acid: }{(e) Salpetic acid: }{(e) Carbonic acid: }
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- Drawing various acids (SB-15)Draw the Lewis structures of the following acids.{(a) Acetic acid: }{(b) Acidic acid:}{(c) Folic acid: }{(d) Sulphuric acid: }{(e) Salpetic acid: }{(e) Carbonic acid: }
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- Acid strength (SB-22)Given acetic acid:CH3COOH{: } H-C<|H><`|H>-C<`||O>-O-H
Which H's are acid at the acetic acid: those who H’s at the same time, carbon or the H at the same time, oxygen? - 1
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- Statements, which ones are correct (SB-25)It is a watered, 1 molar solution a unprotonated heavy acid manufacture ('HX'). Which of the following statements is correct? Reason for the choice.
- The acid is dissolved by 0.1%
- The acid is dissolved at 1%
- The acid is dissolved by 10%
- The acid is 100% dissolved
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- 4 days, 22 hours ago
- Frequency of particles (SB-26)What kind of particles are present in an aqueous 1 M H2SO4 solution? In what relative quantities are they present? Make a ranking, start with the most common particle.
pKs(H2SO4) is -3, pks(HSO–4) = 1.92.
Note: Es [translation unavailable] only following particles before: H2O, H3O+, HSO–4, SO2-4, (almost) no H2SO4, there sulphuric acid a strong acid is - 1
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- Interpretation of pKs value, 1 (SB-27)The pKs(H2SO4) is -3, pks(HSO–4) = 1.92.
Which of the following statements is correct? And almost more importantly, why?- 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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- 4 days, 22 hours ago
- Production of a specific concentration (SB-40)Describe the manufacture of the following solutions in the laboratory: the solvent is always distilled water.
Adopted the substances be as salts of which: and solving itself fully in water.- (a) 1 litre of 0.3 M NaOH solution
- (b) 1 litre of 0.4 M HCL solution
- (c) 25 ml 0.5 M NaOH solution
- d) 30 ml 0.6 M H2SO4
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- Calculate various values (pH etc) (SB-44)Calculate / add 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 basic, specify area (SB-45)Complete the gaps in the table with the values < 7, = 7, > 7, < 10-7, = 10-7 and > 10-7
name and address Acidic solution Neutral solution Basic solution c(H3O+ mol/l) pH c(OH– mol/l) pOH - 1
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- Concentrations of hydroxonium etc (SB-46)Given be a solution A with pH = 7 and a solution B with pH = 8. Assessment (correct – wrong) of the following statements:
- (a) The H3O+-Konzentration of A is ten times larger than that of solution of B
- (b) The H3O+-Konzentration of A is equal to the OH– concentration of A
- (c) The H3O+-Konzentration of B is ten times smaller than the c(OH–) of solution of B
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- pH calculations, 1 (SB-47)Calculate the pH of the following solutions if the solvent is always pure water:
- 1 g hydrochloric acid (HCl) dissolved in a litres water
- 1 kg of hydrochloric acid (HCl) dissolved in one cubic meter of water
- 1 g NaOH, dissolved in 2 litres of water
- 81 mg HBr in 0.1 litres of water
- 1 grams sulphuric acid in 1 m3 water. Assumption: complete deprotonation of the sulphuric acid.
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- pH calculations, 2 (SB-48)Calculate the mass at H3O+ and OH– in …
- a) … one litre of water. pH = 6
- (b)… in Lake Constance. The pH was 7
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- Calculations of pH, dilutions (SB-49)Calculate the pH of the following solutions if the solvent is always pure water:
- A 0.03 molar HCl solution is diluted a hundred times.
Calculate the pH before and after dilution - 20 ml 0.05 M HCl is diluted with 100 ml distilled water.
Calculate the pH before and after dilution - 50 ml of 0.2 mo of NaOH solution is given 100 ml of distilled water.
Calculate the pH before and after dilution
- A 0.03 molar HCl solution is diluted a hundred times.
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- Neutralizations (SB-60)Form the neutralization reactions in ion type and form the corresponding salts, analogous to exercise a).{(a) Sodium hydroxide + hydrochloric acid }Na^+ + OH^- + H3O^+ + Cl^- -> NaCl{([translation unavailable])} + 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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- Calculations of pH, mixtures (SB-61)Calculate the pH of the following mixtures.
- For 80 ml of a 0.05 M solution of hydrochloric acid, 100 ml of a 0.01 M solution of sodium nitrate is given.
- For 60 ml of a 0.015 KOH solution, 30 ml of 0.2 M saline solution is given.
- 10 ml 0.2 M saltic acid solution is first diluted tenfold and then replaced with
1.0 g Ca(OH)2.
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- pH calculations, 4 (SB-62)One liter of two solutions A and B with the same starting pH of 4.8 shall be given. A is a hydrochloric acid solution, B is an acidic acid solution.
Now one liter of water is added to each solution.- Calculate the new pH of solution A and B
- In the case of a solution that has a greater change in pH,
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- Calculation titration 1 (SB-70)Titration of an acid (20 ml, unknown concentration) with a base (0.1 M) is required up to the equivalence point of 30 ml of the base. Calculate the concentrations of the acid.
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