CHEMISTRY 3ThiChung Le
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ThiChung Le
Unit 3 Assignment: Solutions, solubility, and acid-base chemistry
1.
For each of the following mixtures, explain whether it is a solution (homogeneous mixture) or a heterogeneous mixture. For the solutions only, identify the solute and the solvent. [5 marks]
a.
Ketchup: homogeneous, the solvent is water, and the solute is a combination of various ingredients in ketchup including tomatoes, vinegar, sugar, salt, and spices.
b.
Ice water: heterogenous, non-uniform composition.
c.
Orange juice: homogeneous,the solvent is water ,and the solute is orange ( which is include dissolved sugars, flavors, and other compounds present in the juice).
d.
Rainwater: homogeneous, the solvent is water, and the solute is sodium
dissolved which contains in rainwater include gases, minerals, and particulate matter from the atmosphere.
e.
14-karat gold in jewelry: homogeneous mixture. 2. Suppose to you eat a piece of candy and let it dissolve it in your mouth. Identify the solvent and solute for the resulting solution? [4 marks]
When you eat a piece of candy and it dissolves in your mouth, the solvent is saliva and the solute is the candy.
The solvent: -
Watery solution is Saliva which contains enzymes, electrolytes, and other substances.
-
The main purpose of this components is help to break down the candy become smaller pieces that can be dissolved and absorbed by the body.
-
In the process of lubricating the food and making it easier to swallow, Saliva is also play
an important role into it.
The solute:
-
There are various ingredients, including sugars, starches, fats and other flavorings that candy contains. -
These ingredients begin to dissolve and break down, when the candy comes into meet with saliva.
-
The dissolved candy components and the components of saliva mixture is the resulting of
the solution.
3. Write reactions to show the dissociation into ions of the following substances when they dissolve in water: [4 marks]
a.
Sodium phosphate: Na
3
PO
4
Na
3
PO
4
is a strong electrolyte and It dissociates completely into ions ( 3Na
+
and 1PO
4
3-
) when it dissolves in water.
Na
3
PO
4
(s) → 3Na
+
(aq)+ PO
4
3-
(aq)
b.
Ammonium hydroxide: NH
4
OH
NH
4
OH is d
issolves in water and i
t dissociates completely into ions ( 1NH
4
+
and 1OH
-
) NH
4
OH(aq) → NH
4
+
(aq) + OH
-
(aq)
4. Explain why water is known as the “universal solvent.” [4 marks]
Because of the polarity which make water become excellent solvent . In water, Oxygen atoms tend to attract more electrons than hydrogen atoms. Because of that, the molecule has a partially positive end and a partially negative end.
At the result, polar lysing molecules can be separated by water molecules.
The solute molecules is surrounded by the solvent molecules when the solute dissolves in
water.
Water molecules are generally small, so a lot of them can surround the molecules in solution
Polar covalent bonds have charged ends, but Ionic material split into charge ions.
This help water to easily dissolve such substances. At the result, ionic substances and polar covalent bonds are easily dissolved in water.
5. Use a solubility table to predict which of the following combinations of compounds would produce a precipitate. State what the precipitate will be if there will be one. (4 marks)
a.
strontium hydroxide and ammonium phosphate:
The products in this reaction are the result of the exchange of ion partners. That is, the strontium ion combine with the phosphate ion, and the ammonium ion combines with the hydroxide ion. The latter combination results in NH
4
OH, which is a soluble compound. Therefore, NH
4
+ and OH
−
remain dissolved in the solution. However, Sr
2+
and PO4
3−
combine to produce Sr
3
(PO4)
2
, which is an insoluble compound.
Therefore, The combination of the compounds results in the precipitation of Sr
(PO ) .
₃
₄ ₂
3Sr(OH)
2
+ 2(NH4)
3
PO
4
→
Sr
(PO
)
+ 6NH
₃
₄ ₂
4
OH
b.
silver nitrate and lithium acetate
In this reaction, the products formed result from the exchange of ion partners. That is, the silver ion combines with the acetate ion, and the lithium-ion combines with the nitrate ion. The former results in AgCH
3
COO while the latter results in LiNO
3
. Because both combinations lead to water-soluble compounds, no precipitation reaction will occur.
6.
Write balanced molecular, ionic, and net ionic equations for each of the following double displacement reactions. Assume all reactions occur in aqueous solution. Include states of matter in your balanced equation. [10 marks]
a.Sodium chloride and lead (II) nitrate
Molecular equation:
2NaCl(aq)+Pb(NO
3
)
2
(aq) →
2NaNO
3
(aq)+PbCl
2
(s)
Complete ionic equation:
2Na
+
(aq)+2Cl
-
(aq)+Pb
2+
+2NO
3
-
(aq) →
2Na
+
(aq)+2NO
3
-
(aq)
+PbCl
2
(s)
Net ionic equation:
2Cl
-
(aq)+ Pb
2+
(aq) → PbCl
2
(s)
b.
Sodium carbonate and iron (II) chloride
Molecular equation:
Na
2
CO
3
(aq)+FeCl
2
(aq) →
2NaCl(aq)+FeCO
3
(s)
Complete ionic equation:
2Na
+
(aq)+CO
3
2-
(aq)+Fe
2+
+2Cl
-
(aq) →
2Na
+
(aq)+2Cl
-
(aq)
+FeCO
3
(s)
Net ionic equation:
CO
3
2-
(aq)+ Fe
2+
(aq) → FeCO
3
(s)
7. A conductivity test of a dilute solution of nitrous acid (HNO
2
(aq) has very low electrical conductivity (the bulb on the tester glows dimly) and the pH is measured to be about 4. The same
concentration of a nitric acid (HNO
3
(aq) solution makes the conductivity tester’s bulb glow brightly and has a pH of less than 1. Explain the difference between these two acids. [4 marks]
The difference between nitrous acid (HNO
2
) and nitric acid (HNO
3
) lies in the number of oxygen
atoms attached to the nitrogen atom.
Nitrous acid has 2 oxygen atoms, while nitric acid has 3 oxygen atoms. As a result, nitrous acid is a weak acid and has a higher pH than nitric acid. This difference in acidity is also reflected in the conductivity of the solutions.
Since nitrous acid is a weaker acid, it has fewer ions in the solution, resulting in lower conductivity. On the other hand, nitric acid is a strong acid, resulting in more ions in the solution and thus higher conductivity.
8.
Identify the following household substances as either very acidic, slightly acidic, very basic, slightly basic, or neutral. [5 marks]
1.
drain cleaner, pH = 13
Very basic
2.
milk, pH = 6.8
neutral
3.
baking soda, pH = 10
Slightly basic
4.
coffee, pH = 5
Slightly acidic
5. muriatic acid, pH = 1
Very acidic
9. Explain the Arrhenius theory of acids in your own words. Write a reaction showing why hydrogen carbonate is an acid. [4 marks]
According to Arrhenius theory, an acids is a substance that dissociates to give H+ ions in an aqueous
medium. Example: HNO3- is an Arrhenius acid.
HCO
3
-
(aq) ⇌
H
+
(aq)+ CO
3
2-
(aq)
In this reaction, hydrogen carbonate (HCO3-) dissociates in water to produce a hydrogen ion (H+) and a carbonate ion (CO3^2-). The hydrogen ion (H+) released is responsible for the acidic behavior of hydrogen carbonate.
10. Determine the concentration as a percentage by volume if a 500 mL bottle of vinegar contains 40.0 mL of acetic acid. [2 marks]
We have: V
solute
= 40 ml ( acetic acid)
V
solution
= 500 ml ( vinegar)
C = ?
C = (V
solute / V
solution
) x 100%
C = ( 40 ml / 500 ml ) x 100%
C = 8 %
Therefore, the concentration as a percentage by volume if a 500 mL bottle of vinegar contains 40.0 mL of acetic acid is 8%.
11.Determine the concentration as a percentage weight by volume if a 250 mL aqueous sodium chloride solution contains 22.7 g of NaCl. [2 marks]
We have: m
solute
= 22.7 g ( NaCl )
V
solution
= 250 ml (aqueous sodium chloride )
C = ?
C = ( m
solute
/ V
solution
) x 100%
C = ( 22.7 g / 250 ml ) x 100%
C = 9.08 %
Therefore, he concentration as a percentage weight by volume if a 250 mL aqueous sodium chloride solution contains 22.7 g of NaCl is 9.08 %.
12. Calculate the molar concentration of a 6.70 L solution that contains 38.5 g of dissolved sodium hydrogen carbonate. [6 marks]
We have: mNaHCO3 = 38.5 g
MNaHCO3 = MNa+ MH+MC+3xMO MNaHCO3 = 23.00 g/mol+ 1 g/mol + 12.01g/mol + 3 x 15.999 g/mol
MNaHCO3 = 84.0 g/mol
So nNaHCO3 = mNaHCO3 / MNaHCO3
nNaHCO3 = 38.5( g) / 84.0 (g/mol)
nNaHCO3 = 0.458 mol
We have: Vsolution = 6.7 L
So C = n / V = 0.458 mol / 6.7 L C = 0.068 Therefore, the molar concentration is 0.068
13.You and your friend conduct an experiment in which you need to make 150 mL of a 0.250 M NaOH solution from an existing stock solution which has a concentration 6.25 M. What volume of concentration sodium hydroxide is required? [7 marks]
We have:
V2 = 150 ml = 0.15 L
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Related Questions
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Each value represents a different aqueous solution at 25 °C. Classify each solution as an acid, base, or neutral.
[OH−]=
3.0
×
10
−
6
[H+]=
8.1
×
10
−
13
pOH= 13.67
pOH= 7.00
[H+]=
5.3
×
10
−
3
[OH-]= 3.0x 10^-6 is a(n)
Question Blank 1 of 5
, [H+]= 8.1 x 10^-13 is a(n)
Question Blank 2 of 5
, pOH= 13.67 is a(n)
Question Blank 3 of 5
, pOH= 7.00 is a(n)
Question Blank 4 of 5
, [H+]= 5.3 x 10^-3 is a(n)
Question Blank 5 of 5
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The Patient:
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Under normal conditions this system functions to:
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