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chemistry form 4 - Acids n bases (concentration)

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chemistry form 4 - Acids n bases (concentration)

  1. 1. Concentration of Acids and Alkalis
  2. 2.  Definition: - Quantity of solute in a given volume of solution which is usually 1 dm3 (text book) - The mass (in grams) or the number of moles of solute dissolved in a solvent to form 1.0 dm3 (1ooo cm3 ) of solution (reference book)
  3. 3.  Hence, the concentration of a solution can be defined in two ways: Concentration = Mass of solute dissolved (g) (g/dm3) Volume of solution (dm3) Concentration = Number of moles of solute (mol) (mol/dm3) Volume of solution (dm3)
  4. 4.  Number of moles of solute that are present in 1 dm3 of solution Molarity = Concentration (g/ dm3) (mol/dm3) Molar mass (g / mol) = Number of moles of solute (mol) Volume of solution (dm³)
  5. 5.  However, the concentration unit that is widely used by chemists is molarity (mol/dm³) or molar concentration (M)  The two units of concentration can be inter-converted as below: Molarity (mol/dm³) Concnetration (g/dm³) Relationship between molarity and concentration X Molar mass ÷ Molar mass
  6. 6.  Molarity = Number of moles of solute (mol) Volume of solution (dm³)  From the above equation, we can calculate the number of moles of solute in a given volume of solution.  Number of moles of solute = Molarity x volume of solution  In short,  n= Number of moles of solute (mol)  M= Molarity of solution (mol dm¯³)  V= Volume of solution (dm³) n = MV
  7. 7.  5 g of copper (II) sulphate dissolved in water to form 500 cm³ solution. Calculate the concentration of copper (II) sulphate solution in g/dm³.  What is the mass of sodium carbonate required to dissolve in water to prepare a 200 cm³ solution and contains 50 g dm¯³.  Calculate the number of moles of ammonia in 150 cm³ of 2 mol dm¯³ aqueous ammonia.  A 250 cm³ solution contains 0.4 moles of nitric acid. Calculate the molarity of the nitric acid.
  8. 8.  4.0 g of sodium carbonate powder, Na₂CO₃, is dissolved in water and made up to 250 cm³. what is the molarity of the sodium carbonate solution? [r.a.m.: C=12, O=16, Na=23]  Dilute hydrochloric acid used in school laboratories usually has a concentration of 2.0 mol dm¯³. calculate the mass of hydrogen chloride in 250 cm³ of the hydrochloric acid? [r.a.m.: H=1, Cl=35.5]
  9. 9.  Standard solution – a solution in which its concentration is accurately known (tx book) / a solution with a known concentration (ref book)  Volumetric flask (standard flask) – apparatus with a known volume such as 100 cm³, 200 cm³, 250 cm³, 500 cm³ and 1000 cm³. - Used to prepare standard solution. - Can measure the volume of a liquid accurately, up to one decimal point.
  10. 10.  Steps involved in the preparation of a standard solution: a) Calculate the mass (m g) of the chemical required to prepare v cm³ of solution where v is the volume of the volumetric flask. b) Weight out the exact mass (m g) of the chemical accurately in a weighting bottle using an electronic balance. c) Dissolve m g of the chemical in a small amount of distilled water. d) Transfer the dissolved chemical into the volumetric flask. e) Add enough water until the graduation mark.
  11. 11.  Dilution method Dilution – process of diluting a concentrated solution by adding a solvent such as water to obtain a more dilute solution. When a solution is diluted, the volume of solvent increases but the number of moles of solute remain s constant. Hence, the concentration of the solution decreases.
  12. 12. If a solution with volume of V₁ cm³ and molarity of M₁ mol dm¯³ is diluted to become V₂ cm³, the new concentration of the diluted solution, M₂ mol dm¯³ can be determined by: Number of moles of = M₁V₁ solute before dilution 1000 Number of moles of = M₂V₂ solute after dilution 1000
  13. 13. However, the number of moles of solute before dilution is the same as the number of moles of solute after dilution. M₁V₁ = M₂V₂ ------- ------- OR M₁V₁ =M₂V₂ 1000 1000 M₁ - molarity of solution before water is added V₁ - volume of solution before water is added M₂ - molarity of solution after water is added V₂ - volume of solution after water is added
  14. 14.  pH value of an acid or an alkali depends on 2 factors: a) degree of dissociation b) molarity or concentration
  15. 15.  At the same concentration, the pH value of an acid or an alkali depends on the degree of dissociation. a) Degree of dissociation of an acid increases, the pH value of the acid is decreases b) Degree of dissociation of an alkali increases, the pH value of the alkali is increases
  16. 16.  For an acid or alkali, its pH value depends on the molarity of the solution. a) Molarity of an acid increases, the pH value of the acid is decreases b) Molarity of an alkali increases, the pH value of the alkali is increases

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