Topics Covered:
Review of Chemistry 11
Soluble and Low Solubility Defined
Deciding if a substance is soluble
Equilibrium Equations for Low Solubility Compounds
Calculating Solubility
Precipitation
Soluble and Low Solubility Defined
Deciding if a substance is soluble
Equilibrium Equations for Low Solubility Compounds
Calculating Solubility
Precipitation
Using
Trial Ksp (T.I.P) to Determine if a Precipitate Forms
Water Pollution and Hardness Removal
Changing the Solubility of A Salt
Water Pollution and Hardness Removal
Changing the Solubility of A Salt
Review of Chemistry 11:
Solution: homogeneous mixture ⇒ looks like a pure substance
Solute: dissolved in solvent ⇒ lesser part of mixture
Solvent: dissolves solute ⇒ greater part of mixture
Saturated Solution: maximum possible amount of solute dissolved
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Solution: homogeneous mixture ⇒ looks like a pure substance
Solute: dissolved in solvent ⇒ lesser part of mixture
Solvent: dissolves solute ⇒ greater part of mixture
Saturated Solution: maximum possible amount of solute dissolved
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Soluble and Low Solubility Defined
Some substances are soluble ⇒ dissolve completely ⇒ can form > 0.10M [solution]
Many substances don't appear to
dissolve BUT they are a little
soluble, these substances are described as having
LOW SOLUBILITY ⇒ [saturated solution] ≤ 0.10mol/L
An equilibrium occurs between
undissolved solid and solution
rate DISSOLVING = rate CRYSTALLIZATION (or precipitation)
Deciding If a
Substance is Soluble
How to use this table:
1) Look at the ionic compound you are interested in and split it up into its ions
2) Find that pair of ions on the table - sometimes your ion may be part of "All others" so be careful.
Things to be careful about: Cu+ ion vs Cu2+ ion, know your Alkali ions (if you cannot remember them look at the first row on the table!), be aware that ALL nitrates are soluble.
Examples
1) NaCl ions are Na+ and Cl- since Na+ is an alkali ion and all compounds containing alkali ions are soluble ⇒ NaCl is soluble
2) AgCl ions are Ag+ and Cl- find Ag+ with Cl- on table ⇒ AgCl is low solubility
3) AgNO3 ions are Ag+ and NO3- since all nitrates are soluble ⇒ AgNO3 is soluble
4) MgCO3 ions are Mg2+ and CO32- find Mg2+ with CO32- on the table ⇒ MgCO3 is low solubility
5) CuBr ions are Cu+ and Br- find Cu+ with Br- on table ⇒ CuBr is low solubility
6) CuBr2 ions are Cu2+ and Br- find Cu2+ with Br- on the table ⇒ CuBr2 is soluble
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The equilibrium equation ALWAYS has the solid on the REACTANTS side of the equilibrium equation. The FORWARD direction shows the solid dissolving, the REVERSE direction shows the solid precipitating.
Examples:
1) Equilibrium equation for a saturated solution of AgCl:
AgCl(s) ↔ Ag+(aq) + Cl-(aq)
2) Equilibrium equation for a saturated solution of Pb3(PO4)2:
Pb3(PO4)2(s) ↔ 3Pb2+(aq) + 2PO43+(aq)
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Example 1: 0.51g of CuBr dissolves in 3.5L of water. Calculate the molar solubility.
Example 2: [CuCl2] = 0.51M Calculate the concentration of this solution in g/L.
Example 3: What concentration of oxide ions are there in a 0.050M aluminium oxide solution?
Formula for aluminium oxide is Al2O3
Equation for dissociation of Al2O3 into ions is: Al2O3→ 2Al3+(aq) + 3O2-(aq)
[O2-(aq)] = 3× [Al2O3] = 3× 0.050 = 0.15M
Example 4: [Li2O] = 1.3M
a) Calculate [Li+] in moles/L
b) Calculate the concentration of Li+ in grams/L
When 2 solutions mix, sometimes a DOUBLE REPLACEMENT reaction occurs and a precipitate is formed.
Precipitates are usually observed as because they make the solutions turn cloudy.
Precipitation can be expressed using three different types of equation - Balanced equation, Total Ionic Equation, and Net Ionic Equation.
Balanced Equation - this is the sort of equation you learned about in junior science and chemistry 11:
Total Ionic Equation - this equation splits all the soluble (aq) species into their ions
Net Ionic Equation - this equation ONLY uses the ions involved in the precipitation, all other ions are considered to be spectator ions. In this example, Pb2+(aq) and 2I-(aq) are the ions that form the precipitate PbI2(s) , the K+(aq) and NO3-(aq) ions remain in solution at the start and end of the reaction, so they are spectator ions.
How To Determine If Solid forms:
1) Write a balanced equation for the possible double replacement reaction ⇒ positive ions switch places
2) Check each product on solubility table to see if its LOW SOLUBILITY (s) or SOLUBLE (aq)
3) Write TOTAL and/or NET ionic equation if asked.
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Separating Cations - this is done by adding anions such that only ONE cation is precipitated at a time.
Example: Separating a mixture of Ag+, Cu2+, and Mg2+
Looking at the data table on page 4 of the data booklet, you can see that Ag+ will precipitate with most of the anions listed. This means that Ag+ should be precipitated first because it is easiest to remove using either Cl−, Br−, or I−. Now we have to separate Cu2+ from Mg2+, looking at page 4 of the data booklet Cu2+ precipitates with S2−, but Mg2+ is soluble with S2− ⇒ they can be separated. If we were doing this as a lab, a schematic of what you would do looks like:
Now you would normally not want to draw out a schematic to answer a separating mixtures of ions question, so the schematic can be summarized as follows:
1) Add NaCl to give Cl− to ppt Ag+ as AgCl. FILTER.
2) Add Na2S to give S2− to ppt Cu2+ as CuS. FILTER.
3) Add Na2CO3 to give CO32− to ppt Mg2+ as MgCO3. FILTER.
Separating Anions - this is done by adding cations such that only ONE anion is precipitated at a time.
Example: Separating a mixture of Cl−, OH−, and SO32−
Looking at the data table on page 4 of the data booklet, you can see that SO32− will precipitate with most of the cations listed. This means that SO32− should be precipitated first because it is easiest to remove, using Sr2+ will allow you to do this without precipitating any of the other ions. Now we have to separate Cl− from OH−, looking at page 4 of the data booklet Ca2+ precipitates with OH−, but is soluble with Cl− ⇒ OH− and Cl− can be separated. If we were doing this as a lab, a schematic of what you would do looks like:
Now you would normally not want to draw out a schematic to answer a separating mixtures of ions question, so the schematic can be summarized as follows:
1) Add Sr(NO3)2 to give Sr2+ to ppt SO32− as SrSO3. FILTER.
2) Add Ca(NO3)2 to give Ca2+ to ppt OH− as Ca(OH)2. FILTER.
3) Add Pb(NO3)2 to give Pb2+ to ppt Cl− as PbCl2. FILTER.
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Use the table on page 4
of your data booklet a copy is provided below:
How to use this table:
1) Look at the ionic compound you are interested in and split it up into its ions
2) Find that pair of ions on the table - sometimes your ion may be part of "All others" so be careful.
Things to be careful about: Cu+ ion vs Cu2+ ion, know your Alkali ions (if you cannot remember them look at the first row on the table!), be aware that ALL nitrates are soluble.
Examples
1) NaCl ions are Na+ and Cl- since Na+ is an alkali ion and all compounds containing alkali ions are soluble ⇒ NaCl is soluble
2) AgCl ions are Ag+ and Cl- find Ag+ with Cl- on table ⇒ AgCl is low solubility
3) AgNO3 ions are Ag+ and NO3- since all nitrates are soluble ⇒ AgNO3 is soluble
4) MgCO3 ions are Mg2+ and CO32- find Mg2+ with CO32- on the table ⇒ MgCO3 is low solubility
5) CuBr ions are Cu+ and Br- find Cu+ with Br- on table ⇒ CuBr is low solubility
6) CuBr2 ions are Cu2+ and Br- find Cu2+ with Br- on the table ⇒ CuBr2 is soluble
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The equilibrium equation ALWAYS has the solid on the REACTANTS side of the equilibrium equation. The FORWARD direction shows the solid dissolving, the REVERSE direction shows the solid precipitating.
Examples:
1) Equilibrium equation for a saturated solution of AgCl:
AgCl(s) ↔ Ag+(aq) + Cl-(aq)
2) Equilibrium equation for a saturated solution of Pb3(PO4)2:
Pb3(PO4)2(s) ↔ 3Pb2+(aq) + 2PO43+(aq)
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Example 1: 0.51g of CuBr dissolves in 3.5L of water. Calculate the molar solubility.
Example 2: [CuCl2] = 0.51M Calculate the concentration of this solution in g/L.
Example 3: What concentration of oxide ions are there in a 0.050M aluminium oxide solution?
Formula for aluminium oxide is Al2O3
Equation for dissociation of Al2O3 into ions is: Al2O3→ 2Al3+(aq) + 3O2-(aq)
[O2-(aq)] = 3× [Al2O3] = 3× 0.050 = 0.15M
Example 4: [Li2O] = 1.3M
a) Calculate [Li+] in moles/L
b) Calculate the concentration of Li+ in grams/L
a) 2 Li+
in Li2O ⇒ [Li+] = 2×
1.3 = 2.6M
b)
(top)b)
When 2 solutions mix, sometimes a DOUBLE REPLACEMENT reaction occurs and a precipitate is formed.
Precipitates are usually observed as because they make the solutions turn cloudy.
Precipitation can be expressed using three different types of equation - Balanced equation, Total Ionic Equation, and Net Ionic Equation.
Balanced Equation - this is the sort of equation you learned about in junior science and chemistry 11:
2KI(aq) + Pb(NO3)2(aq)
→ PbI2(s) + 2KNO3(aq)
Total Ionic Equation - this equation splits all the soluble (aq) species into their ions
2K+(aq) + 2I-(aq) + Pb2+(aq) + 2NO3-(aq) → PbI2(s) + 2K+ + 2NO3-(aq)
Net Ionic Equation - this equation ONLY uses the ions involved in the precipitation, all other ions are considered to be spectator ions. In this example, Pb2+(aq) and 2I-(aq) are the ions that form the precipitate PbI2(s) , the K+(aq) and NO3-(aq) ions remain in solution at the start and end of the reaction, so they are spectator ions.
Pb2+(aq) +
2I-(aq) → PbI2(s)
How To Determine If Solid forms:
1) Write a balanced equation for the possible double replacement reaction ⇒ positive ions switch places
2) Check each product on solubility table to see if its LOW SOLUBILITY (s) or SOLUBLE (aq)
3) Write TOTAL and/or NET ionic equation if asked.
(top)
Separating Cations - this is done by adding anions such that only ONE cation is precipitated at a time.
Example: Separating a mixture of Ag+, Cu2+, and Mg2+
Looking at the data table on page 4 of the data booklet, you can see that Ag+ will precipitate with most of the anions listed. This means that Ag+ should be precipitated first because it is easiest to remove using either Cl−, Br−, or I−. Now we have to separate Cu2+ from Mg2+, looking at page 4 of the data booklet Cu2+ precipitates with S2−, but Mg2+ is soluble with S2− ⇒ they can be separated. If we were doing this as a lab, a schematic of what you would do looks like:
Now you would normally not want to draw out a schematic to answer a separating mixtures of ions question, so the schematic can be summarized as follows:
1) Add NaCl to give Cl− to ppt Ag+ as AgCl. FILTER.
2) Add Na2S to give S2− to ppt Cu2+ as CuS. FILTER.
3) Add Na2CO3 to give CO32− to ppt Mg2+ as MgCO3. FILTER.
Separating Anions - this is done by adding cations such that only ONE anion is precipitated at a time.
Example: Separating a mixture of Cl−, OH−, and SO32−
Looking at the data table on page 4 of the data booklet, you can see that SO32− will precipitate with most of the cations listed. This means that SO32− should be precipitated first because it is easiest to remove, using Sr2+ will allow you to do this without precipitating any of the other ions. Now we have to separate Cl− from OH−, looking at page 4 of the data booklet Ca2+ precipitates with OH−, but is soluble with Cl− ⇒ OH− and Cl− can be separated. If we were doing this as a lab, a schematic of what you would do looks like:
Now you would normally not want to draw out a schematic to answer a separating mixtures of ions question, so the schematic can be summarized as follows:
1) Add Sr(NO3)2 to give Sr2+ to ppt SO32− as SrSO3. FILTER.
2) Add Ca(NO3)2 to give Ca2+ to ppt OH− as Ca(OH)2. FILTER.
3) Add Pb(NO3)2 to give Pb2+ to ppt Cl− as PbCl2. FILTER.
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The
Solubility
Product Ksp
This is a special case of Keq
that applies to the equilibrium between a low solubility salt and the
small amount of dissolved ions that are produced. Remember that solids
are not included in the Keq expression, and this is true for
Ksp as well.
Examples:
1) AgCl(s) ↔ Ag+(aq) + Cl−(aq) Ksp = [Ag+][Cl−]
2) Ag3PO4(s) ↔ 3Ag+(aq) + PO43−(aq) Ksp = [Ag+]3[PO43−]
3) Al2O3 ↔ 2Al3+(aq) + 3O2−(aq) Ksp = [Al3+]2[O2−]3
The larger the value of Ksp the greater the solubility of the salt.
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Examples:
1) Ksp for CuS is 6.0×10−37. Calculate the concentration of a saturated solution of CuS.
Equm: CuS(s) ↔ Cu2+(aq) + S2−(aq)
Ksp expression is: Ksp = [Cu2+][S2−]
Let solubility of CuS = x mol/L ⇒ [Cu2+] = x and [S2−] = x
∴ Ksp = x⋅x = 6.0×10−37
⇒x2 = 6.0×10−37
square root both sides
x = √(6.0×10−37)
⇒ x = 7.7×10−19M
2) Calculate the concentration of a saturated solution of silver chromate, Ag2CrO4.
In this case the Ksp value is not given in the question ⇒ you will have to look it up on page 5 of the data booklet.
According to the data table:
Ksp for silver chromate = 1.1×10−12
Ksp = [Ag+]2[CrO42−]
Let x = [Ag2CrO4] ⇒ [Ag+] = 2x and [CrO42−] = x
⇒ Ksp = (2x)2⋅x = 1.1×10−12
⇒ (2x)2⋅x = 1.1×10−12
⇒ 4x3 = 1.1×10−12
divide both sides by 4
x3 = 2.75×10-13
cube root both sides
x = ∛(2.75×10-13) = 6.50×10−5
∴ the concentration of a saturated solution of silver chromate = 6.50×10−5M
3) The Ksp for Al2O3 = 1.9×10−18 calculate the concentration of a saturated solution.
Al2O3 ↔ 2Al3+(aq) + 3O2−(aq)
Let x = solubility of Al2O3 ⇒ [Al3+] = 2x and [O2−] = 3x
Ksp = [Al3+]2[O2−]3
⇒ Ksp = (2x)2⋅(3x)3
108x5 = 1.9×10−18
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Examples:
1) AgCl(s) ↔ Ag+(aq) + Cl−(aq) Ksp = [Ag+][Cl−]
2) Ag3PO4(s) ↔ 3Ag+(aq) + PO43−(aq) Ksp = [Ag+]3[PO43−]
3) Al2O3 ↔ 2Al3+(aq) + 3O2−(aq) Ksp = [Al3+]2[O2−]3
The larger the value of Ksp the greater the solubility of the salt.
(top)
Examples:
1) Ksp for CuS is 6.0×10−37. Calculate the concentration of a saturated solution of CuS.
Equm: CuS(s) ↔ Cu2+(aq) + S2−(aq)
Ksp expression is: Ksp = [Cu2+][S2−]
Let solubility of CuS = x mol/L ⇒ [Cu2+] = x and [S2−] = x
∴ Ksp = x⋅x = 6.0×10−37
⇒x2 = 6.0×10−37
square root both sides
x = √(6.0×10−37)
⇒ x = 7.7×10−19M
2) Calculate the concentration of a saturated solution of silver chromate, Ag2CrO4.
In this case the Ksp value is not given in the question ⇒ you will have to look it up on page 5 of the data booklet.
According to the data table:
Ksp for silver chromate = 1.1×10−12
Ksp = [Ag+]2[CrO42−]
Let x = [Ag2CrO4] ⇒ [Ag+] = 2x and [CrO42−] = x
⇒ Ksp = (2x)2⋅x = 1.1×10−12
⇒ (2x)2⋅x = 1.1×10−12
⇒ 4x3 = 1.1×10−12
divide both sides by 4
x3 = 2.75×10-13
cube root both sides
x = ∛(2.75×10-13) = 6.50×10−5
∴ the concentration of a saturated solution of silver chromate = 6.50×10−5M
3) The Ksp for Al2O3 = 1.9×10−18 calculate the concentration of a saturated solution.
Al2O3 ↔ 2Al3+(aq) + 3O2−(aq)
Let x = solubility of Al2O3 ⇒ [Al3+] = 2x and [O2−] = 3x
Ksp = [Al3+]2[O2−]3
⇒ Ksp = (2x)2⋅(3x)3
108x5 = 1.9×10−18
(top)
Using
Trial Ksp (T.I.P) to Determine if a Precipitate Forms
Trial Ion Product (T.I.P.) is used
to determine if a solution is saturated or not, this is useful for
determining whether a mixture of ions will precipitate.
T.I.P. is calculated by substituting given values for ion concentration into the Ksp expression. The value obtained is then compared to the Ksp
If
TIP = Ksp then you are at equilibrium ⇒ you have a saturated solution and precipitate is just starting to form.
TIP < Ksp then not enough ions to form precipitate ⇒ no ppt forms
TIP > Ksp more than enough ions to form precipitate ⇒ ppt forms
Water Pollution and Hardness Removal
Changing the Solubility of A Salt
T.I.P. is calculated by substituting given values for ion concentration into the Ksp expression. The value obtained is then compared to the Ksp
If
TIP = Ksp then you are at equilibrium ⇒ you have a saturated solution and precipitate is just starting to form.
TIP < Ksp then not enough ions to form precipitate ⇒ no ppt forms
TIP > Ksp more than enough ions to form precipitate ⇒ ppt forms
Remember that when you mix two solutions together the total volume is changed, so you will need to do dilution calculations to find the concentration of the ions in the mixture before reaction occurs − this is theoretical, but still important. Another thing to consider is the concept of mixing equal volumes of two solutions − this has a similar effect to mixing 1.0L with 1.0L so the concentration of the ions is halved.
Dilution calculations are carried out using the theory that moles = VM and since the moles
before and after dilution do not change then: MiVi = MfVf.
This equation can be rearranged to determine the concentration of the
ions in the mixture.
Sample
Dilution Calculation
10.0mL of 6.0M NaOH is diluted to 45.0mL. What is the
concentration of the diluted solution?
Examples of using T.I.P.
1) 10.0mL of 0.10M AgNO3
is mixed with 10.0mL of 0.10M NaCl. Will a precipitate of AgCl form?
[Ag+]in
= 0.10M
[Cl−]in
= 0.10M
Ksp
= [Ag+][Cl−] =
1.8×10−10
⇒ T.I.P.
= [Ag+][Cl−] = 0.050
× 0.050 = 2.5×10−3
T.I.P.
> Ksp ⇒ ppt forms
2) 5.0mL of 0.010M lead (II) nitrate is mixed with 20.0mL of
0.0010M NaCl. Will a precipitate of PbCl2 form?
Ksp
= [Pb2+][Cl−] = 1.2×10−5
T.I.P. = (0.0020)×(0.00080)2 = 1.28×10−9
TIP < Ksp ⇒ no ppt forms
(top)
T.I.P. = (0.0020)×(0.00080)2 = 1.28×10−9
TIP < Ksp ⇒ no ppt forms
(top)
Water Pollution and Hardness Removal
Heavy Metal Removal:
Most heavy metal ions precipitate when OH is added. Adding OH− is relatively safe provided it is done in a controlled manner without adding any excess. Remember too much OH− will effect the pH of a system which can be just as harmful to organisms as the heavy metal you are trying to remove.
Most heavy metal ions precipitate when OH is added. Adding OH− is relatively safe provided it is done in a controlled manner without adding any excess. Remember too much OH− will effect the pH of a system which can be just as harmful to organisms as the heavy metal you are trying to remove.
Example: A copper mine must ensure that the concentration of Cu2+
in its effluent is no greater than 1.2×10−6. The Cu2+
is precipitated using OH−. What is the minimum concentration
of OH− required to reduce [Cu2+] to acceptable
levels? Ksp for Cu(OH)2 = 1.4×10−28
Equilibrium:
Cu(OH)2(s) ↔ Cu2+(aq) + 2OH−(aq)
Hardness:
Hard water contains Ca2+ and/or Mg2+ because the river or reservoir supplying the water runs through an area where limestone is present.
There are two types of hard water:
Permanently Hard Water only contains Ca2+ and/or Mg2+ there are no HCO3−
Hardness can only be removed from permanently hard water by adding a source of CO32−. The most common substance used to remove hardness is Washing Soda which is the commercial name for Na2CO3
Temporarily Hard Water contains HCO3− as well as Ca2+ and/or Mg2+
Temporarily hard water will lose "hardness" when heated because the HCO3− breaks down to form carbonate ions:
Hard water contains Ca2+ and/or Mg2+ because the river or reservoir supplying the water runs through an area where limestone is present.
There are two types of hard water:
Permanently Hard Water only contains Ca2+ and/or Mg2+ there are no HCO3−
Hardness can only be removed from permanently hard water by adding a source of CO32−. The most common substance used to remove hardness is Washing Soda which is the commercial name for Na2CO3
Temporarily Hard Water contains HCO3− as well as Ca2+ and/or Mg2+
Temporarily hard water will lose "hardness" when heated because the HCO3− breaks down to form carbonate ions:
2HCO3− → CO32− + CO2 + H2O
since Ca2+
and Mg2+ form a ppt with the CO32−
the hardness is removed when the water is boiled. This results in
a build up of Kettle Scale on anything used to heat temporarily hard
water such as Kettles, hot water pipes, boilers, dishwashers, washing
machines, etc. This causes a problem when heating the water
because the scale acts as an insulator, which means the cost to heat
the water is greater. Another problem caused by water hardness is
that it reacts with soap to form soap scum. Soap Scum is calcium
stearate and/or magnesium stearate, which are sticky substances that
tend to float on the surface of the water. This is a huge problem
not only for taking a shower or bath, but for doing laundry because you
need twice as much detergent if you are not using a water
softner. This is why Washing Soda Na2CO3
is commonly available. In areas where water is hard there is
often a huge industry set up providing water softeners and filters.
Changing the Solubility of A Salt
The solubility of a low solubility
compound can be manipulated using Le Chatelier's principle to shift the
equilibrium in the direction desired. The equilibrium can be
shifted left by adding one of the ions on the products side of the
equilibrium, this is known as the Common Ion Effect. The
equilibrium can be shifted right by removing one of the ion on the
product side of the equilibrium by precipitating it out using another
ion that is not part of the equilibrium.
Common Ion Effect
Common Ion Effect
AgCl(s)
↔ Ag+(aq) + Cl−(aq)
If NaCl is added to this
equilibrium then [ Cl−] increases
which causes the equilibrium to shift LEFT. Since there are now
fewer dissolved ions the compound has been made LESS soluble.
Increasing the solubility
If NaOH is added to this equilibrium then the OH−
ions will cause the small amount of Ag+
to precipitate forming AgOH, this causes the [ Ag+]
to decrease, and the equilibrium will shift right. Since this
causes more ions to be dissolved the compound has been made MORE
soluble.
Example b:
(top)
Increasing the solubility
Example a:
AgCl(s) ↔ Ag+(aq) + Cl−(aq)Example b:
CuSO3(s) ↔ Cu2+(aq) + SO32−(aq)
If MgNO3 is added the SO32−
will precipitate as MgSO3
which causes the concentration of SO32−
to decrease, and the equilibrium will shift right. Since
this
causes more ions to be dissolved the compound has been made MORE
soluble.(top)
You should now be able to do ALL
the unit 3 questions from Hebden. Don't worry about Titration
questions they are just stoichiometry questions! More information
about titrations in the next two units!