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Acid Base Equilibria
• A Bronsted-Lowry acid is defined as a substance that can donate
a proton.
• A Bronsted-Lowry base is defined as a substance that can
accept a proton.
• Each acid is linked to a conjugate base on the other side of the
equation
Calculating pH
Where [H+] is the concentration of hydrogen ions in the
solution
Calculating pH of strong acids
• The concentration of hydrogen ions in a monoprotic strong acid will be the
same as the concentration of the acidFor HCl and HNO3 the [H+(aq)] will
be the same as the
• original concentration of the acid.
• For 0.1M HCl the pH will be –log[0.1] =1.00.
Finding [H+] from pH
• [H+] = 1 x 10-pH
Ionic Product for water
• This equilibrium has the following equilibrium expression
• Because [H2O (l)] is much bigger than theconcentrations of the ions,
we assume its value is constant and make a new constant Kw
• At 25oC the value of Kw for all aqueous solutions is 1x10-14 mol2dm-6
• The Kw expression can be used to calculate [H+ (aq)] ions if we know the
[OH- (aq)] ions and vice versa
• Finding pH of pure water.
Calculating pH of Strong Base
• For bases we are normally given the concentration of the hydroxide ion.
• To work out the pH we need to work out [H+(aq)] using the kw expression.
• We need know
• The concentration of the OH- ions in solution
• The ionization expression for the ionization of water Kw = [H+] [OH–]
• As Kw = [H+] [OH–]
• [H+] =Kw/[OH–]
Strong base (a quicker method)
A quick way to get the same answer is to:■■ find –log10 [OH–] (here –log10 [OH–] =
–log10 (0.0500)= 1.3).
■■ subtract this value from 14 (in this example
14 – 1.3 = 12.7).
This works because –log10 [H+] – log10 [OH–] = 14.
Weak Acids
using the acid dissociation constant
• Weak acids only slightly dissociate when dissolved in water, giving an
equilibrium mixture.
• The units of Ka are determined in the same way as for Kc
• For the dissociation of a monobasic acid the units are mol dm–3.
• ■■ A high value for Ka (for example, 40 mol dm–3) indicates that the position of
equilibrium lies to the right. The acid is almost completely ionised.
• ■■ A low value for Ka (for example, 1.0 × 10–4 mol dm–3) indicates that the
position of equilibrium lies to the left.
• The acid is only slightly ionised and exists mainly as HA molecules and
comparatively few H+ and A– ions.
•pKa = –log10Ka
pH Calculations involving
neutralisation Reactions
• These can be quite complex calculations working out the pH of a
partially neutralised acid or the pH of the solution if too
• much alkali has been added and has gone past neutralisation. The
method differs if the acid is strong or weak for the partially
neutralised case.
Examples
1) 45cm3 of 1M HCl is reacted with 30cm3 of 0.65M NaOH. What will
be the pH of the resulting mixture?
2) 2)55cm3 of 0.5M CH3CO2H is reacted with 25cm3 of 0.35M NaOH.
What will be the pH of the resulting mixture?
• CH3CO2H+ NaOH CH3CO2Na + H2O ka is 1.7 x 10-5mol dm-3
Indicators and acid–base
titrations
• indicators are used to detect the end-point in acid–alkali titrations
• indicators such as litmus to test whether a substance is
acidic or alkaline.
The colour change in phenolphthalein is due to small differences in
the structure of its molecule when hydrogen ions or hydroxide ions
are added
• An acid–base indicator is a dye or mixture of dyes that changes
colour over a specific pH range. In simple terms, many indicators
can be considered as weak acids in which the acid (HIn) and its
conjugate base (In–) have different colours.
• ■■ Adding an acid to this indicator solution shifts the position
• of equilibrium to the left. There are now more molecules of colour A.
• ■■ Adding an alkali shifts the position of equilibrium to the right. There are
now more ions of colour B.
• ■■ The colour of the indicator depends on the relative concentrations of
HIn and In–. The colour of the indicator
• during a titration depends on the concentration of H+ ions present.
The red petals of pelargonium (geranium)
contain the dye pelargonidin. Hydrogen ions or
hydroxide ions can make small changes in its
molecular structure to produce different
colours.
Titration curves
• •Measure initial pH of the acid
• • Add alkali in small amounts noting the volume added
• • Stir mixture to equalise the pH
• • Measure and record the pH to 1 dp
• • When approaching endpoint add in smaller volumes of alkali
• •Add until alkali in excess
• Calibrate meter first by measuring known pH of a buffer solution.
This is necessary because pH meters can lose accuracy on
storage Can improve accuracy by maintaining constant
temperature
Titration curves
• There are 4 main types of curve
•1. Strong acid and strong base
•2. Weak acid and strong base
•3. Strong acid and weak base
•4. Weak acid and weak base
The Key points to sketching a curve
• :
• Initial and final pH
• Volume at neutralisation
• General Shape (pH at neutralisation
• At the equivalence point (when 25.0 mL of NaOH solution has been
added), the neutralization is complete: only a salt remains in solution (NaCl),
and the pH of the solution is 7.00.
• Adding more NaOH produces a rapid increase in pH, but eventually the
pH levels off at a value of about 13.30, the pH of 0.20 M NaOHNaOH .
• The titration of either a strong acid with a strong base or a strong
base with a strong acid produces an S-shaped curve. The curve is
somewhat asymmetrical because the steady increase in the
volume of the solution during the titration causes the solution to
become more dilute. Due to the leveling effect, the shape of the
curve for a titration involving a strong acid and a strong base
depends on only the concentrations of the acid and base, not their
identities.
• In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a
weak base depends dramatically on the identity of the acid or the base and the
corresponding Ka or Kb.
• the pH also changes much more gradually around the eq
uivalence point in the titration of a weak acid or a weak base.
• [H+] of a solution of a weak acid (HA) is not equal to the concentration of the acid but
depends on both its pKa and its concentration.
• Because only a fraction of a weak acid dissociates,
• [H+] is less than [HA] . Thus the pH of a solution of a weak acid is greater than
the pH of a solution of a strong acid of the same concentration.
• Before any base is added, the pH of the acetic acid solution is
greater than the pH of the HCl solution, and the pH changes more
rapidly during the first part of the titration. Note also that the pH of
the acetic acid solution at the equivalence point is greater than
7.00. That is, at the equivalence point, the solution is basic.
• In addition, the change in pH around the equivalence point is only about
half as large as for the HCl titration; the magnitude of the pH change at the
equivalence point depends on the pKa of the acid being titrated. Above the
equivalence point, however, the two curves are identical. Once the acid has
been neutralized, the pH of the solution is controlled only by the amount of
excess NaOH present, regardless of whether the acid is weak or strong.
Choosing an Indicator
• Indicators can be considered as weak acids.
• The acid must have a different colour to its conjugate . An indicator
changes colour from HIn to In- over a narrow range.
• Different indicators change colours over a different ranges
• The end-point of a titration is reached when [HIn] =[In-].
• To choose a correct indicator for a titration oneshould pick an indicator
whose end-point coincideswith the equivalence point for the titration.
phenolphthalein
• Only use phenolphthalein in titrations with strong bases but not
weak bases-
• Colour change: colourless acid pink alkali
methyl orange
• Use methyl orange with titrations with strong acids but not weak
acids
• Colour change: redacid yellow alkali (orange end
point)
BUFFER
• A Buffer solution is one where the pH does not change
significantly if small amounts of acid or alkali are added to it.
• An acidic buffer solution is made from a weak acid and a salt of that
weak acid ( made from reacting the weak acid with a strong base).
• An acidic buffer solution is simply one which has a pH less than 7.
• Example : ethanoic acid and sodium ethanoateCH3CO2H (aq) and
CH3CO2- Na+
• A basic buffer solution is made from a weak base and a salt of that
weak base ( made from reacting the weak base with a strong acid).
• Example :ammonia and ammonium chloride NH3 and NH4+Cl-
• An alkaline buffer solution has a pH greater than 7.
How Buffer solutions work
Calculating the pH of buffer solutions
• We still use the weak acids dissociation expression
rearranged
• The salt content can be added in several ways: a salt solution could
be added to the acid or some solid salt added.
• A buffer can also be made by partially neutralising a weak acid with
alkali and therefore producing salt.
• We also assume the Initial concentration of the acid has remained
constant, because amount that has dissociated or reacted is small
If a buffer is made by adding sodium
hydroxide to partially neutralise a weak
acid then follow the method below
• Uses of buffer solutions
• Buffer solutions play an important part in many industrial processes,
including electroplating, the manufacture ofdyes and in the treatment of
leather. They are also used to make sure that pH meters record the
correct pH.
• Many animals depend on buffers to keep a constant pH in various parts
of their bodies. In humans, the pH of the blood is kept between 7.35 and
7.45 by a number of different buffers in the blood:
• ■■ hydrogencarbonate ions, HCO3–
• ■■ haemoglobin and plasma proteins
• ■■ dihydrogenphosphate (H2PO4 –) and hydrogenphosphate (HPO4 2–)
ions
• ■■ the position of this equilibrium shifts to the left
• ■■ H+ ions combine with HCO3 – ions to form carbon dioxide and water until equilibrium
is restored
• ■■ this reduces the concentration of hydrogen ions in the blood and helps keep the pH
constant.
• If the H+ ion concentration decreases:
• ■■ the position of this equilibrium shifts to the right
• ■■ some carbon dioxide and water combine to form H+ and HCO3
• – ions until equilibrium is restored
• ■■ this increases the concentration of hydrogen ions in the blood and helps keep the pH
constant.
SALT HYDROLYSIS
• Salts of strong acids and strong bases form neutral solutions eg sodium
chloride
• Salts of weak acids and strong bases form alkaline
solutions in water. Eg Sodium ethanoate –it fully ionizes in water
• Most of H+ are removed in the reaction CH3COOH is weak acid. This
[H+] becomes less and the pH becomes larger than 7 . The solution is
alkaline.
• Salts of strong acids & weak bases form acidic solutions in water.
• ammonium sulphate
(NH4 )2SO4
Solubility product
• An equilibrium is established when an undissolved ionic compound
is in contact with a saturated solution of its ions.
• The ions move from the solid to the saturated solution at the same
rate as they move from the solution to the solid
• For any solid, the concentration of the solid phase remains
constant and can be combined with the value of Kc.
DEFNIITION OF SLUBILITY PRODUCT
• Solubility product is the product of the concentrations of each ion in
a saturated solution of a sparingly soluble salt at 298 K, raised to
the power of their relative concentrations.
LIMITATIONS OF SOLUBILITY
CONSTANT
1 Only applies only to sparingly soluble ionic substances at
2 Solubility constant is only constant at a particular temperature. Temperature
usually coated is 298k
Calculating solubility product question
The common ion effect
• The common ion effect is the reduction in the solubility of a
dissolved salt achieved by adding a solution of a compound which
has an ion in common with the dissolved salt.
• when we add a solution of sodium chloride to a saturated solution of silver chloride and silver
chloride precipitates.
• In a saturated solution of silver chloride in water, we have the following equilibrium:
• AgCl(s) Cl–(aq) + Ag+(aq)
• We now add a solution of sodium chloride:
• ■■ the chloride ion is common to both sodium chloride and silver chloride
• ■■ the added chloride ions shift the position of equilibrium to the left
• ■■ silver chloride is precipitated.
• The addition of the common ion, Cl–, has reduced the solubility of the silver chloride because its
solubility product has been exceeded. When [Ag+] [Cl–] is greater than the Ksp for silver
chloride a precipitate will form.
Question
• 14 a Thallium(I) chloride is a salt that is sparingly soluble in water. When hydrochloric acid is
added to a saturated solution of thallium(I) chloride, a precipitate is formed. Explain why a
precipitate is formed.
• b Calcium sulfate is a sparingly soluble salt that can be made by mixing solutions containing
calcium and sulfate ions. A 0.001 00 mol dm–3 solution of aqueous calcium chloride, CaCl2, is
mixed with an equal volume of 0.001 00 mol dm–3 solution of aqueous sodium sulfate, Na2SO4.
• i Calculate the concentration of calcium and sulfate ions when equal volumes of these
• solutions of calcium chloride and sodium sulfate are mixed.
• ii Will a precipitate of calcium sulfate form?
• (Ksp of calcium sulfate = 2.0 × 10–5 mol2 dm–6)
Deciding whether precpitates will
form or not
Partition coefficient
Acid Base Equilibria ppt.pptx titrations
Acid Base Equilibria ppt.pptx titrations
Acid Base Equilibria ppt.pptx titrations

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Acid Base Equilibria ppt.pptx titrations

  • 2. • A Bronsted-Lowry acid is defined as a substance that can donate a proton. • A Bronsted-Lowry base is defined as a substance that can accept a proton. • Each acid is linked to a conjugate base on the other side of the equation
  • 3.
  • 4. Calculating pH Where [H+] is the concentration of hydrogen ions in the solution Calculating pH of strong acids
  • 5. • The concentration of hydrogen ions in a monoprotic strong acid will be the same as the concentration of the acidFor HCl and HNO3 the [H+(aq)] will be the same as the • original concentration of the acid. • For 0.1M HCl the pH will be –log[0.1] =1.00.
  • 6. Finding [H+] from pH • [H+] = 1 x 10-pH
  • 7.
  • 8. Ionic Product for water • This equilibrium has the following equilibrium expression
  • 9. • Because [H2O (l)] is much bigger than theconcentrations of the ions, we assume its value is constant and make a new constant Kw • At 25oC the value of Kw for all aqueous solutions is 1x10-14 mol2dm-6
  • 10. • The Kw expression can be used to calculate [H+ (aq)] ions if we know the [OH- (aq)] ions and vice versa • Finding pH of pure water.
  • 11.
  • 12. Calculating pH of Strong Base • For bases we are normally given the concentration of the hydroxide ion. • To work out the pH we need to work out [H+(aq)] using the kw expression. • We need know • The concentration of the OH- ions in solution • The ionization expression for the ionization of water Kw = [H+] [OH–] • As Kw = [H+] [OH–] • [H+] =Kw/[OH–]
  • 13.
  • 14. Strong base (a quicker method) A quick way to get the same answer is to:■■ find –log10 [OH–] (here –log10 [OH–] = –log10 (0.0500)= 1.3). ■■ subtract this value from 14 (in this example 14 – 1.3 = 12.7). This works because –log10 [H+] – log10 [OH–] = 14.
  • 15.
  • 16.
  • 17.
  • 18.
  • 19. Weak Acids using the acid dissociation constant • Weak acids only slightly dissociate when dissolved in water, giving an equilibrium mixture.
  • 20.
  • 21. • The units of Ka are determined in the same way as for Kc • For the dissociation of a monobasic acid the units are mol dm–3. • ■■ A high value for Ka (for example, 40 mol dm–3) indicates that the position of equilibrium lies to the right. The acid is almost completely ionised. • ■■ A low value for Ka (for example, 1.0 × 10–4 mol dm–3) indicates that the position of equilibrium lies to the left. • The acid is only slightly ionised and exists mainly as HA molecules and comparatively few H+ and A– ions.
  • 23.
  • 24.
  • 25.
  • 26. pH Calculations involving neutralisation Reactions • These can be quite complex calculations working out the pH of a partially neutralised acid or the pH of the solution if too • much alkali has been added and has gone past neutralisation. The method differs if the acid is strong or weak for the partially neutralised case.
  • 27.
  • 28. Examples 1) 45cm3 of 1M HCl is reacted with 30cm3 of 0.65M NaOH. What will be the pH of the resulting mixture? 2) 2)55cm3 of 0.5M CH3CO2H is reacted with 25cm3 of 0.35M NaOH. What will be the pH of the resulting mixture? • CH3CO2H+ NaOH CH3CO2Na + H2O ka is 1.7 x 10-5mol dm-3
  • 29. Indicators and acid–base titrations • indicators are used to detect the end-point in acid–alkali titrations • indicators such as litmus to test whether a substance is acidic or alkaline.
  • 30. The colour change in phenolphthalein is due to small differences in the structure of its molecule when hydrogen ions or hydroxide ions are added
  • 31. • An acid–base indicator is a dye or mixture of dyes that changes colour over a specific pH range. In simple terms, many indicators can be considered as weak acids in which the acid (HIn) and its conjugate base (In–) have different colours.
  • 32. • ■■ Adding an acid to this indicator solution shifts the position • of equilibrium to the left. There are now more molecules of colour A. • ■■ Adding an alkali shifts the position of equilibrium to the right. There are now more ions of colour B. • ■■ The colour of the indicator depends on the relative concentrations of HIn and In–. The colour of the indicator • during a titration depends on the concentration of H+ ions present.
  • 33. The red petals of pelargonium (geranium) contain the dye pelargonidin. Hydrogen ions or hydroxide ions can make small changes in its molecular structure to produce different colours.
  • 34.
  • 35. Titration curves • •Measure initial pH of the acid • • Add alkali in small amounts noting the volume added • • Stir mixture to equalise the pH • • Measure and record the pH to 1 dp • • When approaching endpoint add in smaller volumes of alkali • •Add until alkali in excess
  • 36. • Calibrate meter first by measuring known pH of a buffer solution. This is necessary because pH meters can lose accuracy on storage Can improve accuracy by maintaining constant temperature
  • 37. Titration curves • There are 4 main types of curve •1. Strong acid and strong base •2. Weak acid and strong base •3. Strong acid and weak base •4. Weak acid and weak base
  • 38.
  • 39. The Key points to sketching a curve • : • Initial and final pH • Volume at neutralisation • General Shape (pH at neutralisation
  • 40.
  • 41.
  • 42. • At the equivalence point (when 25.0 mL of NaOH solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. • Adding more NaOH produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M NaOHNaOH .
  • 43. • The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities.
  • 44.
  • 45. • In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding Ka or Kb. • the pH also changes much more gradually around the eq uivalence point in the titration of a weak acid or a weak base. • [H+] of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its pKa and its concentration. • Because only a fraction of a weak acid dissociates, • [H+] is less than [HA] . Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration.
  • 46.
  • 47.
  • 48.
  • 49. • Before any base is added, the pH of the acetic acid solution is greater than the pH of the HCl solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic.
  • 50. • In addition, the change in pH around the equivalence point is only about half as large as for the HCl titration; the magnitude of the pH change at the equivalence point depends on the pKa of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess NaOH present, regardless of whether the acid is weak or strong.
  • 51.
  • 52. Choosing an Indicator • Indicators can be considered as weak acids. • The acid must have a different colour to its conjugate . An indicator changes colour from HIn to In- over a narrow range. • Different indicators change colours over a different ranges • The end-point of a titration is reached when [HIn] =[In-]. • To choose a correct indicator for a titration oneshould pick an indicator whose end-point coincideswith the equivalence point for the titration.
  • 53.
  • 54.
  • 55. phenolphthalein • Only use phenolphthalein in titrations with strong bases but not weak bases- • Colour change: colourless acid pink alkali
  • 56. methyl orange • Use methyl orange with titrations with strong acids but not weak acids • Colour change: redacid yellow alkali (orange end point)
  • 57.
  • 58.
  • 59.
  • 60. BUFFER • A Buffer solution is one where the pH does not change significantly if small amounts of acid or alkali are added to it.
  • 61. • An acidic buffer solution is made from a weak acid and a salt of that weak acid ( made from reacting the weak acid with a strong base). • An acidic buffer solution is simply one which has a pH less than 7. • Example : ethanoic acid and sodium ethanoateCH3CO2H (aq) and CH3CO2- Na+
  • 62. • A basic buffer solution is made from a weak base and a salt of that weak base ( made from reacting the weak base with a strong acid). • Example :ammonia and ammonium chloride NH3 and NH4+Cl- • An alkaline buffer solution has a pH greater than 7.
  • 64.
  • 65.
  • 66.
  • 67.
  • 68. Calculating the pH of buffer solutions • We still use the weak acids dissociation expression rearranged
  • 69.
  • 70. • The salt content can be added in several ways: a salt solution could be added to the acid or some solid salt added. • A buffer can also be made by partially neutralising a weak acid with alkali and therefore producing salt. • We also assume the Initial concentration of the acid has remained constant, because amount that has dissociated or reacted is small
  • 71.
  • 72.
  • 73. If a buffer is made by adding sodium hydroxide to partially neutralise a weak acid then follow the method below
  • 74. • Uses of buffer solutions • Buffer solutions play an important part in many industrial processes, including electroplating, the manufacture ofdyes and in the treatment of leather. They are also used to make sure that pH meters record the correct pH. • Many animals depend on buffers to keep a constant pH in various parts of their bodies. In humans, the pH of the blood is kept between 7.35 and 7.45 by a number of different buffers in the blood:
  • 75. • ■■ hydrogencarbonate ions, HCO3– • ■■ haemoglobin and plasma proteins • ■■ dihydrogenphosphate (H2PO4 –) and hydrogenphosphate (HPO4 2–) ions
  • 76. • ■■ the position of this equilibrium shifts to the left • ■■ H+ ions combine with HCO3 – ions to form carbon dioxide and water until equilibrium is restored • ■■ this reduces the concentration of hydrogen ions in the blood and helps keep the pH constant. • If the H+ ion concentration decreases: • ■■ the position of this equilibrium shifts to the right • ■■ some carbon dioxide and water combine to form H+ and HCO3 • – ions until equilibrium is restored • ■■ this increases the concentration of hydrogen ions in the blood and helps keep the pH constant.
  • 77. SALT HYDROLYSIS • Salts of strong acids and strong bases form neutral solutions eg sodium chloride • Salts of weak acids and strong bases form alkaline solutions in water. Eg Sodium ethanoate –it fully ionizes in water
  • 78. • Most of H+ are removed in the reaction CH3COOH is weak acid. This [H+] becomes less and the pH becomes larger than 7 . The solution is alkaline. • Salts of strong acids & weak bases form acidic solutions in water. • ammonium sulphate (NH4 )2SO4
  • 79. Solubility product • An equilibrium is established when an undissolved ionic compound is in contact with a saturated solution of its ions. • The ions move from the solid to the saturated solution at the same rate as they move from the solution to the solid
  • 80.
  • 81. • For any solid, the concentration of the solid phase remains constant and can be combined with the value of Kc.
  • 82. DEFNIITION OF SLUBILITY PRODUCT • Solubility product is the product of the concentrations of each ion in a saturated solution of a sparingly soluble salt at 298 K, raised to the power of their relative concentrations.
  • 83. LIMITATIONS OF SOLUBILITY CONSTANT 1 Only applies only to sparingly soluble ionic substances at 2 Solubility constant is only constant at a particular temperature. Temperature usually coated is 298k
  • 85.
  • 86.
  • 87.
  • 88.
  • 89.
  • 90. The common ion effect • The common ion effect is the reduction in the solubility of a dissolved salt achieved by adding a solution of a compound which has an ion in common with the dissolved salt.
  • 91. • when we add a solution of sodium chloride to a saturated solution of silver chloride and silver chloride precipitates. • In a saturated solution of silver chloride in water, we have the following equilibrium: • AgCl(s) Cl–(aq) + Ag+(aq) • We now add a solution of sodium chloride: • ■■ the chloride ion is common to both sodium chloride and silver chloride • ■■ the added chloride ions shift the position of equilibrium to the left • ■■ silver chloride is precipitated. • The addition of the common ion, Cl–, has reduced the solubility of the silver chloride because its solubility product has been exceeded. When [Ag+] [Cl–] is greater than the Ksp for silver chloride a precipitate will form.
  • 92.
  • 93. Question • 14 a Thallium(I) chloride is a salt that is sparingly soluble in water. When hydrochloric acid is added to a saturated solution of thallium(I) chloride, a precipitate is formed. Explain why a precipitate is formed. • b Calcium sulfate is a sparingly soluble salt that can be made by mixing solutions containing calcium and sulfate ions. A 0.001 00 mol dm–3 solution of aqueous calcium chloride, CaCl2, is mixed with an equal volume of 0.001 00 mol dm–3 solution of aqueous sodium sulfate, Na2SO4. • i Calculate the concentration of calcium and sulfate ions when equal volumes of these • solutions of calcium chloride and sodium sulfate are mixed. • ii Will a precipitate of calcium sulfate form? • (Ksp of calcium sulfate = 2.0 × 10–5 mol2 dm–6)
  • 94. Deciding whether precpitates will form or not
  • 95.
  • 96.