chemistry hp unit 7 solutions learning targets …7+packet...1 chemistry hp unit 7 – solutions...
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Chemistry HP Unit 7 – Solutions
Learning Targets (Your exam at the end of Unit 6 will assess the following:)
7. Solutions
7-1. Define a solution and give examples.
7-2. Define solute and solvent and determine the solute and solvent for a given solution
7-3. Define molarity using a mathematical equation and give the appropriate units.
7-4. Perform calculations involving molarity/solution concentration.
7-5. Perform calculations involving solution dilution.
7-6. Define dissociation and describe the interactions between water and ionic compounds that lead to dissociation.
7-7. Write dissociation equations for ionic compounds and represent dissociation equations using a diagram showing ions and
water molecules.
7-8. Calculate the concentration of ions in a solution from the concentration of an ionic compound.
7-9. Calculate the concentration of ions resulting from mixing two solutions by applying solution dilution and dissociation
equations.
7-10. Compare unsaturated and saturated solutions.
7-11. Determine if a compound is soluble or insoluble in water using a solubility table.
7-12. Define precipitate in terms of solubility.
7-13. Describe the effect of temperature on solubility. Determine the solubility of a compound at a given temperature from a
solubility curve.
7-14. Write the formula equation, complete ionic equation, and net ionic equation for a precipitation reaction, giving the
appropriate state for each substance.
7-15. Develop a procedure to selectively precipitate an ion from solution on the basis of solubility rules. 7-16. Perform stoichiometric calculations involving molarity of solutions.
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7-1. Define a solution and give examples.
7-2. Define solute and solvent and determine the solute and solvent for a given solution.
Solution Definitions
A solution is a homogeneous mixture of substances.
A mixture is a physical combination of two or more substances. Substances that are physically combined have not reacted chemically with each other. Since they are physically combined, they can be physically separated. Examples of mixtures are air, which is a mixture of nitrogen, oxygen and other gases, and a pile of sugar and salt. Each of these mixtures contains more than one substance physically combined with each other. Air is composed of a mixture of gases, while each gas retains its own properties. Oxygen is still oxygen, mixed with nitrogen, carbon dioxide and other gases. This is a good thing, because if oxygen reacted chemically with any of the other gases in the atmosphere, we wouldn’t have oxygen to breathe.
A homogeneous mixture refers to a mixture that is the same throughout. Air is a homogeneous mixture of gases because the gases are in the same phase and have the same uniform appearance.
Every solution consists of a solute and solvent. A solute is the substance being dissolved, and a solvent is the dissolving medium. When sweetening hot tea with sugar, the sugar is the solute, and the hot tea is the solvent. If both the solute and solvent are in the same phase — meaning they are both solids or liquids or gases — the solute refers to the substance in the lesser amount.
Water is known as the “universal solvent,” because many solutes can be dissolved in water.
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7-3. Define molarity using a mathematical equation and give the appropriate units.
7-4. Perform calculations involving molarity/solution concentration.
Molarity
Solution concentration is most often measured as molarity, which is defined as the number of moles solute per liter of solution.
moleMolarity
Volume
The unit of molarity is moles/liter, which is also referred to as a molar (M).
Sample Problem. What is the molarity of a solution in which 4.67 moles of Li2SO3 are dissolved to make 2.00 liters of solution?
Sample Problem. What is the molarity of a solution made by dissolving 20.0 g of NaOH to make 50.0 mL of solution?
Sample Problem. How many grams of H2SO4 would be needed to make 750.0 mL of a 2.00 M solution?
147 g H2SO4
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7-4. Perform calculations involving molarity/solution concentration.
Sample Problem. Determine the moles of solute needed to prepare 6.20 L of a 4.00 M Na2O solution.
24.8 mol
Sample Problem. Calculate the concentration of a 5.0 L solution that contains 9.03 x 1024 molecules of H2SO4.
3.0 M
Preparing a Molar Solution
To prepare a solution of desired molarity, place the required mass of dry reagent in a container. Dissolve in a small
amount of water (less than the final volume) and add water to reach the final volume. It is important to note that the
volume when calculating molarity refers to the volume of the entire solution, consisting of both the original dry reagent
and the added water.
Sample Problem. Describe how to prepare 100 mL of 0.400 M Na2CO3 solution.
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Worksheet 7-1 (Learning Targets 7-3 and 7-4)
Give the equation used to solve each problem and answer with the appropriate units.
(1) Calculate the concentration of a 1.5 L solution that contains 0.24 mol HCl. _________________
(2) Calculate the molarity of a 600 mL solution that contains 0.750 mol NH4Cl. _________________
(3) How many moles of AlCl3 are contained in 0.25 L of 2.0 M solution? _________________
(4) What is the volume of a 5.0 M solution of HNO3 that contains 2.0 mol? _________________
(5) How many moles of Cu(NO3)2 are contained in 400 mL of 0.90 M solution? _________________
(6) What is the volume of 0.20 M solution of MgCl2 that contains 0.30 mol? _________________
(7) Calculate the molarity of a 0.25 L solution that contains 17 g of NaNO3. _________________
(8) Calculate the concentration of a 5.0 L solution that contains 9.03x1024 molecules of NaOH. _________________
(9) What is the mass of KI present in 200 mL of 2.00 M solution? _________________
(10) How many molecules of H2SO4 are present in 125 mL of 0.400 M solution? _________________
(11) Describe how to prepare 0.200 L of 0.250 M CaCl2 solution.
(12) Describe how to prepare 100 mL of 0.400 M Al(NO3)3 solution.
Answers: (1) 0.16 M (2) 1.25 M (3) 0.50 mol (4) 0.40 L (5) 0.36 mol (6) 1.5 L (7) 0.80 M (8) 3.0 M (9) 66.4 g KI (10)
3.01x1022 molecules (11) Weigh 5.55 g of CaCl2. Dissolve in a small amount of water (less than 0.200 L). Add water UP TO
a total volume of exactly 0.200 L of solution. (12) Weigh 8.52 g of Al(NO3)3. Dissolve in a small amount of water (less than
100 mL). Add water UP TO a total volume of exactly 100 mL of solution.
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7-5. Perform calculations involving solution dilution.
Solutions of desired molarity can be prepared by diluting solutions of higher concentration. Dilution volumes and concentrations can be calculated using the following equation:
1 1 2 2M V M V
Where: M1 =molarity of the first, more concentrated, solution V1 = volume to take from the more concentrated solution M2 = molarity of the diluted solution V2 = volume of the diluted solution
Sample Problem Concentrated H2SO4 is 18.0 M. What volume of 18.0 M solution is needed to make 2.00 L of 1.00 M H2SO4 solution?
Sample Problem You have 50.0 mL of a 0.40 M NaCl solution and dilute it to 1000. mL. What is the concentration of NaCl in the new solution?
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Worksheet 7-2 (Learning Target 7-5: Perform calculations involving solution dilution.)
(1) 1.5 L of a 1.0 M solution of HCl is diluted with water to a final volume of 2.5 L. Calculate the final concentration.
_________
(2) 125 mL of a 0.400 M solution of NaOH is diluted to give a solution with a concentration of 0.250 M. Calculate the final
volume. What volume of water was added to the solution? _________
volume added: _________
(3) A 0.20 M solution of copper (II) sulfate is diluted to give 800 mL of 0.15 M solution. What was the initial volume of
solution? What volume of water was added to the solution?
_________
volume added: _________
(4) 2.0 L of a sulfuric acid solution is diluted to give 10 L of solution with a final concentration of 0.40 M. Calculate to
initial concentration.
_________
(5) 100 mL of 0.15 M potassium permanganate solution is diluted to a final concentration of 0.020 M. Calculate the final
volume of the solution. What volume of water was added to the solution?
_________
volume added: _________
(6) Calculate the final concentration if 250 mL of 0.30 M solution of nitric acid is diluted by adding 500 mL of water.
_________
(7) 2.5 L of solution contains 170 g of sodium nitrate. What is the initial concentration of the solution? If the solution is
diluted to give a final volume of 4.0 L, what is the final concentration?
_________
_________
(8) 200 mL of solution contains 10.7 g of potassium iodate. Calculate the initial concentration of the solution. Calculate
the final concentration if 300 mL of water are added to the solution.
_________
_________
(9) 400 mL of a solution of iron (III) nitrate of an unknown initial concentration is diluted to give 500 mL of solution with
a final concentration of 0.600 M. Calculate the initial concentration. What mass of iron (III) nitrate is found in the
solution?
_________
mass: _________
(10) 250 mL of water are added to 150 mL of a solution of Na3PO4 with an unknown concentration. The final
concentration of the solution is 0.120 M. Calculate the initial concentration. What mass of Na3PO4 found in the solution?
_________
mass: _________
Answers: (1) 0.60 M (2) 0.200 L, 0.075 L added (3) 0.60 L, 0.20 L added (4) 2.0 M (5) 0.75 L, 0.65 L added (6) 0.10 M (7) 0.80 M, 0.50 M (8) 0.250 M, 0.100 M (9) 0.750
M, 72.6 g (10) 0.320 M, 7.87 g
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7-6. Define dissociation and describe the interactions between water and ionic compounds that lead to dissociation.
Solvation of Ionic Compounds
Most ionic compounds, when added to water, dissolve in water by separating into their constituent ions. The degree of separation of an ionic compound into its constituent ions is a function of its solubility in water. The more soluble an ionic compound, the more it will dissociate into its ions. This process of an ionic compound dissolving in water is called Solvation.
Solvation is defined as the interaction on the molecular level between solute and solvent (water). As an ionic compound dissolves in water, the positive ion (cation) of the ionic compound interacts with the negative oxygen end of the water molecule and the negative ion (anion) of the ionic compound interacts with the positive hydrogen end of the water molecule. These interactions result in dissociation: the separation of an ionic compound into its constituent ions. The ions then undergo hydration: they become surrounded by water molecules.
Here is an image of how a sodium cation (positive ion) gets surrounded by the negative oxygen ends of water molecules:
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7-7. Write dissociation equations for ionic compounds and represent dissociation equations using a diagram showing ions and water
molecules.
Writing Dissociation Equations
An ionic compound soluble in water will separate into its constituent ions when added to water. Here is a particle diagram of the dissociation of NaCl when dissolved in water.
When NaCl is dissolved in water, it undergoes the process of solvation, separating into its Na+ and Cl- ions while still retaining the identity and properties of NaCl.
Similarly, when CaCl2 is dissolved in water, it separates into one Ca2+ and two Cl- ions.
When an ionic compound containing a polyatomic ion dissociates, the polyatomic anion remains intact.
Sample Problem. Write the dissociation equation for copper (II) phosphate.
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7-8. Calculate the concentration of ions in a solution from the concentration of an ionic compound.
Dissociation and Ion Concentration
To determine the concentration of ions dissolved in solution after an ionic compound is introduced to water, first write a dissociation equation. To find the concentration of each ion in solution, multiply the concentration of the solution by the coefficient of that ion in solution. For example, to determine the concentration of each ion in a 0.100 M solution of NaCl, we first write the dissociation equation: NaCl —> Na+ + Cl-
NaCl Na+ Cl-
Initial 0.100 M 0 M 0 M
Change - 0.100 M + 1(0.100 M) + 1(0.100 M)
Final 0 M 0.100 M 0.100 M
0.100 M solution of NaCl contains 0.100 M Na+ and 0.100 M Cl-.
Sample Problem Determine the concentration of ions in a solution of 1.0 M calcium nitrate.
Sample Problem Write the dissociation equation for 0.30 M ammonium sulfate solution, and calculate the concentration of each ion present in solution.
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Worksheet 7-3 (Learning Targets 7-7 and 7-8)
(1) Write dissociation equations for the following compounds.
(a) HI
(f) KCN
(b) CaBr2
(g) Cr2(SO4)3
(c) MgSO4
(h) (NH4)2SO4
(d) Al(NO3)3
(i) sodium carbonate
(e) CuCl2
(j) cobalt (II) chloride
(2) Write a dissociation equation for each the following compounds. Calculate the concentration of each ion in solution.
(a) 2.0 M NaF [ ] = ______ M
[ ] = ______ M
(b) 0.50 M BaI2 [ ] = ______ M
[ ] = ______ M
(c) 0.40 M K3PO4 [ ] = ______ M
[ ] = ______ M
(d) 0.080 M CrCl3 [ ] = ______ M
[ ] = ______ M
(e) 0.10 M Ni(NO3)2 [ ] = ______ M
[ ] = ______ M
(f) 2.0 x 10–3 M ZnBr2 [ ] = ______ M
[ ] = ______ M
(g) 0.050 M K2Cr2O7 [ ] = ______ M
[ ] = ______ M
(h) 0.30 M Al2(SO4)3 [ ] = ______ M
[ ] = ______ M
Answers: (1) (a) HI → H+ + I− (b) CaBr2 → Ca2+ + 2Br– (c) MgSO4 → Mg2+ + SO4 2− (d) Al(NO3)3 → Al3+ + 3NO3 − (e) CuCl2 → Cu2+ + 2Cl− (f) KCN → K+ + CN− (g) Cr2(SO4)3 → 2Cr3+ + 3SO4 2− (h) (NH4)2SO4 → 2NH4 + + SO4 2− (i) Na2CO3 → 2Na+ + CO3 2− (j) CoCl2 → Co2+ + 2Cl− (2) (a) NaF → Na+ + F– [Na+ ] = 2.0 M [F− ] = 2.0 M (b) BaI2 → Ba2+ + 2I– [Ba2+] = 0.50 M [I– ] = 1.0 M (c) K3PO4 → 3K+ + PO4 3– [K+ ] = 1.2 M [PO4 3–] = 0.40 M (d) CrCl3 → Cr3+ + 3Cl− [Cr3+] = 0.080 M [Cl− ] = 0.24 M (e) Ni(NO3)2 → Ni2+ + 2NO3 − [Ni2+] = 0.10 M [NO3
− ] = 0.20 M (f) ZnBr2 → Zn2+ + 2Br– [Zn2+] = 2.0x10–3 M [Br– ] = 4.0x10–3 M (g) K2Cr2O7 → 2K+ + Cr2O7 2– [K+ ] = 0.10 M [Cr2O7 2–] = 0.050 M (h) Al2(SO4)3 → 2Al3+ + 3SO4 2− [Al3+] = 0.60 M [SO4 2− ] = 0.90 M
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7-9. Calculate the concentration of ions resulting from mixing two solutions by applying solution dilution and dissociation equations.
Ion Concentration in Mixtures
When two solutions are mixed, the concentration of each ion is diluted as a result of the added volume. The initial concentration of each ion is first calculated from the dissociation equation. After two solutions are mixed, though, the total volume of the mixture increases, which lowers the concentration of the ions in solution. The same number of moles of ion are present, but the increased volume results in a lower molarity according to the molarity equation:
1 1 2 2M V M V
The final concentration of each ion can then be calculated using the solution dilution equation. The final volume is the total added volume of both solutions.
Sample Problem Calculate the concentration of each ion resulting from mixing 50.0 mL of 0.10 M potassium iodide with 30.0 mL of 0.35 M magnesium chloride, given that no reaction occurs.
Sample Problem Calculate the concentration of each ion resulting from mixing 2.5 L of 0.72 M ammonium chloride with 1.5 L of 0.80 M ammonium sulfate, given that no reaction occurs.
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Worksheet 7-4 (Learning Target 7-9. Calculate the concentration of ions resulting from mixing two solutions by applying
solution dilution and dissociation equations.)
Calculate the concentration of each ion resulting from mixing the following solutions, given that no reaction occurs.
(a) 4.0 L of 1.2 M HBr mixed with 8.0 L of 2.4 M NaCl
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
(b) 2.0 L of 0.25 M KNO3 mixed with 3.0 L of 0.45 M SrI2
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
(c) 150 mL of 0.80 M CaCl2 mixed with 50 mL of 0.60 M Mg(NO3)2
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
(d) 0.40 L of 0.35 M NaBr mixed with 0.60 L of 0.20 M Na2CO3
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
(e) 400 mL of 0.18 M FeCl3 mixed with 200 mL of 0.72 M KCl
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
(f) 20 mL of 0.15 M Al(NO3)3 mixed with 10 mL of 0.30 M Pb(NO3)2
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
[ ] = ______ M
Answers: (a) [H+ ] = 0.40 M [Br– ] = 0.40 M [Na+ ] = 1.6 M [Cl– ] = 1.6 M (b) [K+ ] = 0.10 M [NO3 – ] = 0.10 M [Sr2+] = 0.27 M
[I– ] = 0.54 M (c) [Ca2+] = 0.60 M [Cl– ] = 1.2 M [Mg2+] = 0.15 M [NO3 – ] = 0.30 M (d) [Na+ ] = 0.38 M [Br– ] = 0.14 M [CO3 2–]
= 0.12 M (e) [Fe3+] = 0.12 M [Cl– ] = 0.60 M [K+ ] = 0.24 M (f) [Al3+] = 0.10 M [NO3 – ] = 0.50 M [Pb2+] = 0.10 M
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7-11. Determine if a compound is soluble or insoluble in water using a solubility table.
7-12. Define precipitate in terms of solubility.
Solubility Definitions
Solubility: Solubility is defined as the amount of solute that will dissolve in a solvent at a given temperature to form a
saturated solution.
Saturated: A solution is said to be saturated when no more solute can be dissolved in it.
Unsaturated: A solution is said to be unsaturated when still more solute can be dissolved in it.
Substances in which more than 0.10 moles of a solute dissolves in 1 L of water are said to be soluble in water.
Substances in which less than 0.10 moles of solute dissolves in 1 L of water are said to be insoluble in water. When
compounds are insoluble in solution, they form a precipitate.
Precipitate: a solid compound that does not readily dissolve in that solvent.
Solubility data for various combinations of anions and cations is available on solubility tables, which is included on your
Formula Sheet.
Note: in general, the solubility of a compound increases with increased temperature.
Sample Problem Classify the following compounds as soluble or insoluble in water (a) PbI2 (b) MgS (c) Al(OH)3 (d) Li2CO3 (e) CuCl2
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7-14. Write the formula equation, complete ionic equation, and net ionic equation for a precipitation reaction, giving the
appropriate state for each substance.
Net Ionic Equations
There are 3 types of equations:
1. Complete Molecular Equation
2. Complete Ionic Equation
3. Net Ionic Equation
Sample Reaction:
Barium chloride solution reacts with sodium sulfate solution to make solid barium sulfate and aqueous sodium chloride.
Complete Molecular Equation
Complete Ionic Equation
Net Ionic Equation
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Worksheet 7-5 (Learning Targets 7-11, 7-12 and 7-14).
(1) Determine whether the following compounds are soluble or insoluble in water.
(a) CaF2 (b) sodium hydroxide
(c) iron (II) carbonate (d) PbBr2 (e) Ca(OH)2
(f) silver chloride (g) ammonium phosphate (h) K2S (i) CuI2 (j) calcium sulfate
(2) Write the complete molecular equation, complete ionic equations, and net ionic equation for the following reactions.
(a) silver nitrate and sodium chloride
(b) MgSO4 + KF →
(c) FeCl3 + NaOH →
(d) ZnBr2 + K3PO4 →
(e) aluminum sulfate and calcium chloride
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Worksheet 7-5 (con’t) (Learning Targets 7-11, 7-12 and 7-14).
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7-13. Describe the effect of temperature on solubility. Determine the solubility of a compound at a given temperature
from a solubility curve.
Factors Affecting Solubility
The solubility of a solute in a solvent is affected by:
1. Temperature
2. Surface area of the solute
3. Pressure (gases ONLY)
1. Temperature
Temperature plays a very important role in affecting solubility. In general, solubility of a solute in a solvent increases
with increasing temperature. For example, it’s much easier to dissolve sugar (solute) in hot tea (solvent), than in
cold tea.
Whereas the solubility of most solutes increases with temperature, the solubility of gases DECREASES with
increasing temperature.
2. Surface Area
Surface area refers to the area of solute in contact with the solvent. Larger surface area means that the solute
particles are in greater contact with the solvent. This increased contact causes an increase in solubility. For
example, granulated sugar has more surface area than a sugar cube, making the former more soluble than the
latter.
3. Pressure (Gases ONLY)
Pressure affects the solubility of gases only. As pressure of a gas increases, so does its solubility.
For example, when you open a can of soda, you can hear and see the bubbles coming out of solution. Opening the
can releases the pressure, and the gas comes out of solution.
Another example concerns deep sea divers. Deep sea divers need to be careful of diving too deep. The pressure of
O2 and N2 increases with increasing depth. If too much N2 gets dissolved into the blood, the diver may become
disoriented. Further, if the diver ascends too rapidly, the N2 bubbles up out of the blood much like gas comes out of
an open can of soda, threatening blood vessels and the heart.
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7-13. Describe the effect of temperature on solubility. Determine the solubility of a compound at a given temperature
from a solubility curve.
Reading a Solubility Curve
The following is a solubility curve. Each curve on the graph represents a certain solute. The y-axis reads grams of solute
that can be dissolved in 100 g H2O, and the x-axis reads temperature in ᵒC.
Sample Problem What is the solubility of NaNO3 at 50 °C
Saturated Solution of NaNO3 at 50 °C: Unsaturated Solution of NaNO3 at 50 °C:
Supersaturated Solution of NaNO3 at 50 °C:
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Worksheet 7-6 (Learning Target 7-13. Describe the effect of temperature on solubility. Determine the solubility of a
compound at a given temperature from a solubility curve.
Use the given solubility curve to answer the questions below.
(1) In general, what happens to the solubility of a salt when as temperature increases? Which salts are exceptions
to this trend?
(2) Write the chemical name of each salt beside the formula on the solubility curve.
(3) Which salt has a solubility of approximately 105 g (per 100 g H2O at 70 ᵒC)?
(4) Which salt has a solubility of approximately 80 g (per 100 g H2O at 30 ᵒC)?
(5) What is the solubility (in g solute/100 g H2O) of KBr at 60 ᵒC?
(6) What is the solubility (in g solute/100 g H2O) of Ce2(SO4)3 at 20 ᵒC?
(7) What is the solubility (in g solute/100 g H2O) of NH4I at 10 ᵒC?
(8) What is the solubility (in g solute/100 g H2O) of Pb(NO3)2 at 80 ᵒC?
(9) What is the solubility (in g solute/100 g H2O) of RbClO3 at 90 ᵒC?
(10) What is the solubility (in g solute/100 g H2O) of BaS at 50 ᵒC?
(11) What is the solubility (in g solute/100 g H2O) of NiBr2 at 40 ᵒC?
(12) What is the solubility (in g solute/100 g H2O) of Li2SO4 at 100 ᵒC?
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7-15. Develop a procedure to selectively precipitate an ion from solution on the basis of solubility rules.
Selective Precipitation
Selective precipitation is a procedure in which ions are separated from one another in solution based on their differing
solubility. An ion must be found that forms a precipitate with only ONE of the ions present in the solution and not with
any other. In order to add the ion to the solution, it must be added with an “escort” ion. An escort ion can be chosen if it
is soluble with the ion being added.
Good positive escort ions that are soluble with all anions include the
alkali ions
hydrogen
ammonium
Good negative escort ions that are soluble with all cations include
nitrate
chlorate
hypochlorite
perchlorate
acetate
Once the precipitate is formed, it can be filtered from the solution and the next ion(s) can be precipitated. There are
many possible variations to a selective precipitation procedure.
Sample Problem. A solution contains Ca2+ and Cu+ ions. Describe a procedure to separate each of the ions from solution.
Ca2+ Cu+
Cl- No ppt ppt
F- ppt X
(1) Add NaCl. CuCl precipitates out. Filter.
(2) Add NaF. CaF2 precipitates out. Filter.
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7-15. Develop a procedure to selectively precipitate an ion from solution on the basis of solubility rules.
Sample Problem. A solution contains ions Ba2+, Pb2+, and Fe2+. Describe a procedure to separate each of the ions from
solution.
Ba2+ Pb2+ Fe2+
Sample Problem. A solution contains Br–, F–, and S2– ions. Describe a procedure to separate each of the ions from
solution.
Br- F- S2-
Worksheet 7-7 (Learning Target 7.13)
(1) Describe a procedure to separate each of these ions from solution.
(a) Cl− and F− (d) SO4 2− , OH– , and S2−
(b) Ag+ and Ca2+ (e) Ba2+, Ag+, and Ca2+
(c) Cu+, Mg2+, and Sr2+
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7-16. Perform stoichiometric calculations involving molarity of solutions.
Solution Stoichiometry
Convert everything to moles…
Sample Problem. 100 mL of 0.300 M barium chloride reacts with 200 mL of sodium sulfate.
(a) What concentration of sodium sulfate solution is required?
(b) What would be the mass of each of the products?
+ +
moles (mol)
mass (g)
volume (L)
Molarity (M)
molar mass (g/mol)
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Worksheet 7-8 (Learning Target 7.16)
(1) A solution of magnesium chloride is reacted with a solution of sodium fluoride to produce magnesium fluoride and
sodium chloride. Write a balanced chemical equation for this reaction.
(a) If 0.150 L of 0.400 M magnesium chloride is reacted with 0.200 L of sodium fluoride, what concentration of sodium
fluoride is required? What would be the mass of each of the products?
(b) If 100 mL of 0.500 M sodium fluoride is reacted with a 0.200 M magnesium chloride, what volume of magnesium
chloride is required? What would be the mass of each of the products?
(2) Sodium carbonate solution is reacted with iron (III) chloride solution to produce sodium chloride and iron (III)
carbonate. Write a balanced chemical equation for this reaction.
(a) If 0.200 L of 0.600 M sodium carbonate is reacted with 0.250 L of iron (III) chloride, what concentration of iron (III)
chloride is required? What would be the mass of each of the products? If the reaction actually yields 12.6 g of sodium
chloride, what is the percent yield of the reaction? What mass of iron (III) carbonate would actually be produced?
(b) If 400 mL of 0.250 M iron (III) chloride is reacted with a 0.300 M sodium carbonate, what volume of sodium
carbonate is required? What would be the mass of each of the products? If the reaction has an 80.0 % yield, what mass
would actually be obtained for each of the products?
(3) Mercury (II) nitrate is reacted with sodium iodide. Write a balanced chemical equation for this reaction.
(a) If 75.0 mL of 0.100 M mercury (II) nitrate solution is reacted with 80.0 mL of 0.150 M sodium iodide solution, which
reactant is limiting and which is excess? What is the mass of each of the products?
(b) If 40.0 mL of 0.200 M mercury (II) nitrate solution is reacted with 50.0 mL of 0.400 M sodium iodide solution, which
reactant is limiting and which is excess? What is the mass of each of the products?
(4) Silver nitrate is reacted with aluminum chloride. Write a balanced chemical equation for this reaction.
If 200 mL of 0.450 M silver nitrate solution is reacted with 125 mL of 0.160 M aluminum chloride solution, which
reactant is limiting and which is excess? What is the mass of each of the products? If the percent yield for the reaction is
70.0%, what mass would actually be obtained for each of the products? Write the formula equation, complete ionic
equation, and net ionic equation for the reaction. What are the spectator ions for this reaction?
Answers:
(1) MgCl2 + 2NaF → MgF2 + 2NaCl
(a) [NaF] = 0.600 M, 3.74 g MgF2 and 7.01 g NaCl (b) 0.125 L MgCl2, 1.56 g MgF2 and 2.92 g NaCl (2) 3Na2CO3 + 2FeCl3 → 6NaCl + Fe2(CO3)3 (a) [FeCl3] = 0.320 M, 14.0 g NaCl and 11.7 g Fe2(CO3)3, 90.0%, 10.5 g Fe2(CO3)3 actually produced (b) 0.500 L Na2CO3, 17.5 g NaCl and 14.6 g Fe2(CO3)3, 14.0 g NaCl and 11.7 g Fe2(CO3)3 actually produced (3) Hg(NO3)2 + 2NaI → HgI2 + 2NaNO3 (a) limiting: NaI, excess: Hg(NO3)2, 2.73 g HgI2 and 1.02 g NaNO3 (b) limiting: Hg(NO3)2, excess: NaI, 3.64 g HgI2 and 1.36 g NaNO3 (4) 3AgNO3 + AlCl3 → 3AgCl + Al(NO3)3 limiting: AlCl3, excess: AgNO3, 8.60 g AgCl and 4.26 g Al(NO3)3 6.02 g AgCl and 2.98 g Al(NO3)3 actually produced formula equation: 3AgNO3 (aq) +AlCl3 (aq) → 3AgCl (s) + Al(NO3)3 (aq) complete ionic equation: 3Ag+(aq) + 3NO3– (aq) + Al3+ (aq) + 3Cl– (aq) → 3AgCl (s) + Al3+ (aq) + 3NO3
– (aq) net ionic equation: Ag+ (aq) + Cl– (aq) →AgCl (s) (REDUCE COEFFICIENTS) spectator ions: Al3+ and NO3
–
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Worksheet 7-9: Solutions Review
I. Solution Concentration and Solution Dilution (1) Calculate the concentration of a 4.0 L solution that contains 2.4 mol of NaCl. (2) How many moles of magnesium chloride are contained in 500 mL of a 0.40 M solution? What mass of magnesium chloride is present in the solution? (3) 1.2 L of 0.10 M CaI2 solution is diluted to a final volume of 3.0 L. Calculate the final concentration. Determine the mass of CaI2 present in the solution. (4) 2.5 L of 0.80 M potassium chloride solution is diluted to give a solution with a final concentration of 0.50 M. Calculate the final volume of the solution. What volume of water was added to dilute the solution? II. Dissociation (1) Write dissociation equations for the following compounds. Calculate the concentration of each ion in solution. (a) 0.30 M K2SO4 (b) 2.5x10–3 M Na3PO4 (2) Calculate the concentration of each ion resulting from mixing the following solutions, given that no reaction occurs. (a) 2.0 L of 0.10 M HCl mixed with 3.0 L of 0.15 M Ca(C2H3O2)2 (b) 300 mL of 0.015 M MgCl2 mixed with 600 mL of 0.018 M Mg(NO3)2 III. Solubility (1) Determine whether the following compounds are soluble or insoluble) in water. (a) SrSO4 (b) K3PO4 (c) ZnS (d) CuCl2 (2) Write the formula equation, complete ionic equations, and net ionic equation for the following reactions. (a) sodium fluoride and strontium nitrate (b) K2CO3 + AlCl3
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IV. Solution Stoichiometry (1) Barium chloride reacts with sodium sulfate. Give the balanced chemical equation for this reaction. If 125 mL of 0.40 M barium chloride solution is reacted with 200 mL of sodium sulfate solution, what molarity of sodium sulfate solution is required? What would be the mass of each of the products? (2) Silver nitrate is reacted with calcium chloride. Give the balanced chemical equation for this reaction. If 50.0 mL of 0.150 M silver nitrate solution is reacted with 40.0 mL of 0.120 M calcium chloride solution, which reactant is limiting and which is excess? What would be the mass of each of the products? Give the net ionic equation for the reaction. Answers: I. Solution Concentration and Solution Dilution (1) 0.60 M (2) 0.20 mol, 19 g (3) 0.040 M, 35 g (4) 4.0 L, 1.5 L added II. Dissociation (1) (a) K2SO4 → 2K+ + SO4 2–, [K+ ] = 0.60 M, [SO4 2–] = 0.30 M (b) Na3PO4 → 3Na+ + PO4 3–, [Na+ ] = 7.5x10–3 M, [PO4 3–] = 2.5x10–3 M (2) (a) [H+ ] = 0.040 M, [Cl– ] = 0.040 M, [Ca2+] = 0.090 M, [C2H3O2 – ] = 0.18 M (b) [Mg2+] = 0.017 M, [Cl– ] = 0.010 M, [NO3 – ] = 0.024 M III. Solubility (1) (a) insoluble (b) soluble (c) insoluble (d) soluble (2) (a) formula equation: 2NaF (aq) + Sr(NO3)2 (aq) → 2NaNO3 (aq) + SrF2 (s) complete ionic equation: 2Na+ (aq) + 2F– (aq) + Sr2+ (aq) + 2NO3 – (aq) → 2Na+ (aq) + 2NO3 – (aq) + SrF2 (s) net ionic equation: Sr2+ (aq) + 2F– (aq) → SrF2 (s) (b) formula equation: 3K2CO3 (aq) + 2AlCl3 (aq) → 6KCl (aq) + Al2(CO3)3 (s) complete ionic equation: 6K+ (aq) + 3CO3 2– (aq) + 2Al3+ (aq) + 6Cl– (aq) → 6K+ (aq) + 6Cl– (aq) + Al2(CO3)3 (s) net ionic equation: 2Al3+ (aq) + 3CO3 2– (aq)→ Al2(CO3)3 (s) IV. Solution Stoichiometry (1) BaCl2 + Na2SO4 → BaSO4 + 2NaCl [Na2SO4] = 0.25 M, 12 g BaSO4, 5.8 g NaCl (2) 2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2 lim: AgNO3, ex: CaCl2 1.08 g AgCl, 0.615 g Ca(NO3)2 net ionic equation: Ag+ + Cl– → AgCl (REDUCE COEFFICIENTS!)