Chapter 8
(8.2)
Write a sentence that shows how the following pairs of terms relate to one another.
a) Homogeneous and Heterogeneous
b) Solute, Solvent and Solution
c) Dilute and Concentrated
d) Surfactant and Detergent
e) Miscible and Immiscible
f) Saturated, Unsaturated and Super-saturated
a) In a Homogenous mixture, there is only one phase (distinct form) visible, meanwhile, in a heterogenous, there are two or more phases visible.
b) A solution is a homogenous mixture of a solute in a solvent.
A solute is a substance that there is less of.
A solvent is a substance that there is a greater quantity of.
c) A concentrated solution has a relatively high quantity of solute compared to the volume of the solution.
A dilute solution has a relatively low quantity of solute compared to the volume of Solution.
For example, coffee: strong coffee is concentrated, and weak coffee is dilute.
d) A surfactant is a substance that acts on the surface of a liquid, reducing surface tension. This would allow a polar solvent to mix with a non-polar liquid such as oil (mainly for cleaning.)
A detergent is a specific type of surfactant.
e) Miscible liquids mix to form solutions.
Immiscible liquids do not mix at all.
f) A saturated solution is a solution containing the maximum concentration of one or more dissolved solutes in a solvent
An unsaturated solution contains less solute than the solution is capable of dissolving.
Supersaturated solutions contain more dissolved solute than saturated solutions and dissolve more solute than the solution can at a given temperature.
(9.1)
True or false?
Ions that do not actively participate in a reaction are called spectator ions.
True!
(10.1)
A substance feels slippery on your skin and turns blue in the presence of litmus. Which of the following is most likely to be true?
(a) The substance does not dissolve in water.
(b) The substance is a poor conductor of electricity.
(c) The substance has a sour taste.
(d) The substance reacts with fats.
d) is the answer
It is a base:
(a) The substance does not dissolve in water: This is unlikely because bases are generally soluble in water (e.g., sodium hydroxide, potassium hydroxide).
(b) The substance is a poor conductor of electricity: Bases, especially strong bases, typically dissociate in water to form ions, making them good conductors of electricity.
(c) The substance has a sour taste: Sour taste is characteristic of acids, not bases.
(8.3)
Describe what occurs when an ionic compound dissolves within water: NaCl
Once the ionic compound is submerged and dissolved in water, the water molecules will move and reorient themselves. The negative (oxygen) end of each molecule is attracted to the nearest positively charged sodium ion meanwhile the positive (hydrogen) end is attracted to the nearest negatively charged chlorine ion. If the water-ion attraction does pull the ions away from the crystal, it dissolves. If not, it remains undissolved. For a substance to dissolve, the solute-solvent attractions must beat the molecular forces within.
(9.2)
Due to sewage contamination, a pond begins to lose many of its fish species. How might installing a fountain in the pond water with oxygen improve the situation?
The sewage contamination depletes the oxygen in the pond. A fountain would help to mix the water with air, which would allow more oxygen to dissolve in the water. A higher oxygen concentration can help the fish species survive.
(10.2)
Equal volumes of 1.0 mol/L solutions of hydrochloric acid, HCl(aq), and sodium hydroxide, NaOH(aq), are combined. Which of the following is true?
(a) The solution contains a dissolved ionic compound.
(b) Both anions are present in the solution.
(c) The resulting solution will be acidic.
(d) The resulting solution will be basic.
a) is the right answer.
This is because:
(b) Both anions are present in the solution: In the final solution, the main anion present is the chloride ion (Cl⁻). The hydroxide ion (OH⁻) reacts with the hydrogen ion (H⁺) to form water, so OH⁻ is not present.
(c) The resulting solution will be acidic: Since equal volumes and concentrations of a strong acid (HCl) and a strong base (NaOH) neutralize each other, the resulting solution will be neutral, not acidic.
(d) The resulting solution will be basic: For the same reason as (c), the resulting solution will be neutral, not basic.
(8.1)
What special kind of molecular structure does water have? List three unique characteristics of water and explain how each characteristic is related to water’s special structure.
Water has a very polar structure due to its shape. As a result, its structure and polarity allow for hydrogen bonding between water molecules. This hydrogen bonding holds water molecules together:
Results in high surface tension, high melting + boiling points, and expansion when it freezes.
(9.2)
Arrange the steps in the water treatment process in the correct sequence:
Filtration
Fluoridation
Aeration
Disinfection
Ammoniation
Collection
Removal of debris
Softening
Post-Chlorination
Coagulation, Flocculation, + Sedimentation
a) Explain the process and purpose of softening. Include a chemical equation in your explanation.
The following is arranged in order:
1. Collection (Removal of larger particles)
2. Coagulation, Flocculation, + Sedimentation (Removal of smaller particles)
3. Filtration (Removal of fine biological/physical impurities)
4. Disinfection (Use of Chlorine/Ozone/etc)
5. Aeration (Improvement of taste/colour)
6. Softening (Removal of Calcium/Magnesium ions)
7. Fluoridation (Addition of small quantity of fluorine)
8. Post-Chlorination (Kills remaining micro-organisms)
9. Ammoniation (Stabilizes Chlorine in water)
a) Certain communities have hard water, which contains a high concentration of calcium/magnesium ions. This is not only detrimental to health, but it also clogs water pipes and causes soap to precipitate.
10.3
A friend who has a summer job in a lab needs to perform a titration to find the concentration of an unknown solution. Write a general step-by-step procedure for your friend to follow.
Step 1. Rinse the burette with water and then with the titrant solution several times.
Step 2. Prepare the burette with the titrant solution.
Step 3. Pipette the required volume of unknown solution into a clean flask.
Step 4. Add an appropriate indicator to the flask. Place a sheet of white paper under the flask to make colour changes more visible.
Step 5. Record the initial burette volume.
Step 6. Add titrant to the flask quickly at first. As you near the endpoint, add the titrant drop-by-drop, swirling the flask after every drop.
Step 7. Use a wash bottle to rinse the sides of the flask to make sure that all titrants have reacted.
Step 8. Stop titrating when you see a permanent colour change. This is the endpoint. Record the
final burette volume.
Step 9. Repeat steps 2 to 8 until you have at least three trials with results that are similar.
(8.7)
Water is added to 0.500 L of 2.500 mol/L hydrochloric acid to make a final volume of 4.250 L. What is the concentration of the resulting solution?
GRASS
Given:
C1 = 2.500 mol/L
V1 = 0.500 L
V2 = 4.250 L
Required:
The concentration for the diluted solution, C2
Analysis:
C1V1 = C2V2
Solution:
Step 1. Substitute into concentration equation
C2 = C1V1/V2
= (3.0 mol/L)(4.5 L)/20.0 L
C2 = 0.68 mol/L
Statement:
Therefore, the concentration of the diluted solution is 0.68 mol/L
(9.1/10.2)
A strong acid, H2SO4(aq) is completely neutralized with a strong base, KOH(aq)
a) Predict the products of the neutralization reaction
b) Write the balanced equation
c) Write the total ionic equation
d) Write the net ionic equation
e) What can you draw from this acid-base neutralization from the net ionic equation you wrote?
a)
The products of the neutralization reaction are water and an ionic compound.
b)
H2SO4(aq) + 2KOH(aq) → 2H2O(l) + K2SO4(aq)
c)
2H+(aq) + SO42–(aq) + 2K+(aq) + 2OH–(aq) → 2 H2O(l) + SO42–(aq) + 2K+(aq)
d)
2H+(aq) + 2OH–(aq) → 2H2O(l)
e)
The net ionic equation shows that any acid-base neutralization reaction can be reduced to an interaction between hydrogen ions and hydroxide ions to form water.
(10.2)
A 1.0 mL sample of 1.0 mol/L hydrochloric acid is diluted with water to make a solution whose volume is 1.0 L
a) Write the ionization equation for hydrogen chloride in water to form hydrochloric acid.
b) What is the concentration of H+ ions in the original solution?
c) What is the concentration of H1 ions in the diluted solution?
a)
HCl(aq) → H+(aq) + Cl‒(aq)
b)
The concentration of hydrogen ions in the original solution is the same as the concentration of hydrochloric acid, which is 1.0 mol/L
c)
GRASS
Given:
C1 = 1.0 mol/L
V1 = 1.0 mL
V2 = 1.0L
Required:
Concentration of diluted solution (C2)
Analysis:
C1V1 = C2V2
Solution:
Convert the volumes into litres
V1 = 1.0 mL x (1 L/1000 mL)
Rearrange into the following equation and solve:
C2 = C1V1/V2
C2 = ((1.0 mol/L)(0.001L)/1.0 L)
C2 = 0.001 mol/L
Statement:
Therefore, the concentration of the H+ ions in the diluted solution is 0.001 mol/L
(8.5)
A student prepares a supersaturated solution by dissolving 60 g of ammonium chloride, NH4Cl(s), in 100 g of water at 70 °C. She then gradually cools the solution to 50 °C. Finally, she adds a small seed crystal to the solution. Refer to Figure 4 in Section 8.5 to predict how much ammonium chloride crystallizes out of the solution, leaving a saturated solution.
The solubility of ammonium chloride, NH4Cl(s), at 50C, is 50 g/100 g H2O. Originally,60 g of ammonium chloride was dissolved in 100 g of water. Therefore, the amount of ammonium chloride that crystallizes out of solution is:
mNH4Cl = 60 g - 50 g
mNH4Cl = 10 g
Ten grams of ammonium chloride crystallizes out of solution, leaving a saturated solution of 50 g/100 g H2O.
(9.1)
A solution is suspected of containing:
Strontium ions, Sr2+(aq), Manganese ions, Mn2+(aq), or both.
a) Plan a qualitative analysis to identify the ions. Describe your plan.
b) Draw a flow chart to represent your qualitative analysis.
a) A qualitative analysis for identifying the ions would involve a series of steps that would separate them.
Step 1.
Add a sodium hydroxide solution, NaOH, to precipitate manganese ions but not strontium ions. Be sure to add an excess of sodium hydroxide to make sure all of the manganese precipitates. If a precipitate forms, the solution contains manganese ions. If a precipitate does not form, there were no manganese ions.
Step 2.
To precipitate any strontium ions, add sodium sulphate. Strontium sulphate will precipitate out of the solution.
b) NOTE: Images require premium on this site, so a Google Doc link is provided:
https://docs.google.com/document/d/1RFYaQEG9F7lFnrsHka6JP3vw5y6OsR5gyUhB99GfpuY/edit?usp=sharing
(10.3)
An 18 mL sample of hydrochloric acid, HCl(aq), in a flask was titrated with a primary standard solution of sodium carbonate, Na2CO3(aq). Methyl red was used as an indicator. The primary standard solution was prepared by dissolving 0.53 g of sodium carbonate in enough water to make 100 mL of solution. In a single trial of the titration, the initial volume reading on the burette was 0.21 mL and the final volume reading was 26.23 mL.
a) What volume of primary standard solution was used in this trial?
b) What amount of Sodium Carbonate reacted with the acid, during the trial?
c) What was the concentration of the Hydrochloric Acid solution?
a)
26.23 mL – 0.21 mL = 26.02 mL
The volume of the primary standard solution used in this trial was 26.02 mL.
b)
GRASS
Given:
Mass of Sodium Carbonate, mNa2CO3 = 0.53 g
The molar mass of Sodium Carbonate, MM = 105.99 g/mol
Volume of Standard Solution, VNa2CO3 (standard) = 100 mL
Volume of Titrated Solution, VNa2CO3 (titrated) = 26.02 mL
Required:
Amount of Sodium Carbonate in Titrated Solution, nNaCO3
Solution:
Step 1.
Convert the volume of solutions used to litres.
VNa2CO3 (standard) = 100 ml x (1 L/ 1000 mL)
VNa2CO3 (standard) = 0.100 L
VNa2CO3 (titrated) = 26.02 mL x (1 L/ 1000 mL)
VNa2CO3 (titrated) = 0.02602 L
Step 2.
Convert the mass of the Sodium Carbonate to the amount in 100 mL of solution.
nNaCO3 = 0.53 g x (1 mol/105.99 g)
nNaCO3 = 5.00 x 10-3 mol
Step 3.
Use the concentration equation to determine the concentration of Sodium Carbonate in the Standard Solution.
CNa2CO3 (titrated) = nNaCO3/VNa2CO3 (standard)
= 5.00 x 10-3 mol/ 0.100 L
CNa2CO3 (titrated) = 1.3 x 10-2 mol/L
Step 4.
Use the concentration equation to calculate the amount of sodium carbonate in the titrated solution with the new volume of the titrated Sodium Carbonate.
nNaCO3 = (CNa2CO3 (titrated))(VNa2CO3 (titrated))
= (5.00 x 10-2 mol/1 L) x (0.02602 L)
nNaCO3 = 1.3 x 10-3 mol
Statement:
Therefore, the amount of Sodium Carbonate that reacted with acid was 1.3 x 10-3 mol.
c)
Given:
Amount of Sodium Carbonate that reacted with the acid, nNaCO3 = 1.3 x 10-3 mol
Volume of acid, 18 mL
Required:
Concentration of the Hydrochloric Acid solution, CHCL
Analysis:
C = n/V
Solution:
Step 1.
Convert the volume of solution to litres.
VHCl = 18 mL x (1 L/1000 mL)
VHCl = 0.018 L
Step 2.
Write the balanced equation for the reaction
Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
Step 3.
Use the mole ratio in the balanced equation in order to determine the amount in the acid.
nHCl = 1.3 x 10-3 mol x (2 mol/1 mol)
nHCl = 2.6 x 10-3 mol
Step 4.
Using the concentration equation, determine the concentration of the acid with the new values.
CHCl = nHCl/VHCl
= 2.6 x 10-3 mol/0.018 L
CHCl = 0.14 mol/L
Statement:
Therefore, the concentration of the Hydrochloric acid is 0.14 mol/L