BCLN CHEMISTRY 11 - Rev. July/2014
Unit 6 ~ Learning Guide Name: ______________________________
Instructions:
Using a pencil, complete the following notes as you work through the related lessons. Show ALL
work as is explained in the lessons. You are required to have this package completed BEFORE
you write your unit test. Do your best and ask questions if you don’t understand anything!
Mixtures and Solutions:
1. In the table below list 3 differences between pure substances and mixtures
Pure Substances
Mixtures
Has constant physical properties
Has variable physical properties
Has constant chemical properties
Has variable chemical properties
Are homogeneous and so exhibit one phase
May be homogenous or heterogeneous with
multiple phases
Students may list other possibilities
2. A solid dissolved in a liquid is known as an amalgam while a solid dissolved in another solid is
known as what?
An alloy
3. What are four factors that affect solubility?
Molecular polarity, size of molecules, temperature, pressure
Molar concentration:
1. Write the long form and the short form of the concentration formula
𝐦𝐨𝐥𝐚𝐫𝐢𝐭𝐲 =
𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 𝐦𝐨𝐥𝐞𝐬
𝐥𝐢𝐭𝐫𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
Page 1 of 10
C=
𝐧
𝐕
BCLN CHEMISTRY 11 - Rev. July/2014
2. If 1.60 mol of table sugar (C12H22O11) is dissolved to give 750. mL of solution, what is the molar
concentration of the solution?
𝑪=
𝐧
𝐕
=
𝟏.𝟔𝟎 𝐦𝐨𝐥
.𝟕𝟓𝟎 𝐋
= 𝟐. 𝟏 𝐌
3. A solution of epsom salts may be prepared by dissolving 35.6 g of magnesium sulfate in
sufficient water to make 500.0 mL of solution. Determine the molar concentration of the epsom
salts solution.
Magnesium Sulfate = MgSO4
Molar Mass = 1 Mg + 1 S + 4 O = 1(24.31) + 1(32.06) + 4(16.00) = 120.37 g/mol
𝟑𝟓. 𝟔𝐠 𝐌𝐠𝐒𝐎𝟒 𝐱
𝑪=
𝐧
𝐕
=
𝟏 𝐦𝐨𝐥 𝐌𝐠𝐒𝐎𝟒
𝟏𝟐𝟎.𝟑𝟕𝐠 𝐌𝐠𝐒𝐎𝟒
𝟎.𝟐𝟗𝟔 𝐦𝐨𝐥
.𝟓𝟎𝟎 𝐋
= 𝟎. 𝟐𝟗𝟔 𝐦𝐨𝐥 𝐌𝐠𝐒𝐎𝟒
= 𝟎. 𝟓𝟗𝟐 𝐌 𝐌𝐠𝐒𝐎𝟒
4. Hydrogen peroxide, H2O2, forms a colorless solution. How many moles of H2O2 are found in
3.50 L of disinfectant solution that has a concentration of 0.450 M H2O2?
𝐧 = 𝐕 × 𝐂 = 𝟑. 𝟓𝟎 𝐋 𝐱
𝟎.𝟒𝟓𝟎 𝐦𝐨𝐥
𝟏𝐋
= 1.58 mol 𝐇𝟐 𝐎𝟐
5. What mass of nitre, KNO3•8H2O, is necessary to make 400. mL of a 0.0500 mol/L solution?
Molar Mass Nitre = 1 K + 1 N + 11 O + 16 H = 1(39.10) + 1(14.01) + 11(16.00) + 16(1.01)
Molar Mass Nitre = 245.27 g/mol
𝟎. 𝟒𝟎𝟎 𝐋 𝐊𝐍𝐎𝟑 𝟖𝐇𝟐 𝐎 𝐱
𝟎.𝟎𝟓𝟎𝟎 𝐦𝐨𝐥 𝐊𝐍𝐎𝟑 𝟖𝐇𝟐 𝐎
𝟏 𝐋 𝐊𝐍𝐎𝟑 𝟖𝐇𝟐 𝐎
= 𝟒. 𝟗𝟏 𝐠 𝐊𝐍𝐎𝟑 𝟖𝐇𝟐 𝐎
Page 2 of 10
𝐱
𝟐𝟒𝟓.𝟐𝟕𝐠 𝐊𝐍𝐎𝟑 𝟖𝐇𝟐 𝐎
𝟏 𝐦𝐨𝐥 𝐊𝐍𝐎𝟑 𝟖𝐇𝟐 𝐎
BCLN CHEMISTRY 11 - Rev. July/2014
6. Salt of lemon is a dilute solution of potassium binoxalate (KHC2O4H2O) often used to remove
ink stains. What volume of 1.25 M salt of lemon contains 6.00 mol of potassium binoxalate?
𝑽=
𝐧
𝐂
=
𝟔.𝟎𝟎 𝐦𝐨𝐥
𝟏.𝟐𝟓 𝐦𝐨𝐥/𝐋
= 𝟒. 𝟎𝟎 𝐋
7. Commonly called bleach, sodium hypochlorite (NaOCl) has many uses as a strong base and
cleaner. What volume (in mL) of 1.70 M NaOCl can be prepared from 54.7 g of solute?
Molar Mass NaOCl = 1 Na + 1 O + 1 Cl = 1(22.99) + 1(16.00) + 1(35.45) = 74.44 g/mol
𝟓𝟒. 𝟕 𝐍𝐚𝐎𝐂𝐥 𝐱
𝟏 𝐦𝐨𝐥 𝐍𝐚𝐎𝐂𝐥
𝟕𝟒.𝟒𝟒𝐠 𝐍𝐚𝐎𝐂𝐥
𝐱
𝟏 𝐋 𝐍𝐚𝐎𝐂𝐥
𝟏.𝟕𝟎 𝐦𝐨𝐥 𝐍𝐚𝐎𝐂𝐥
= 𝟎. 𝟒𝟑𝟐 𝐋 𝐨𝐫 𝟒𝟑𝟐 𝐦𝐋
Polarity in Molecules:
1. What is the definition of electronegativity?
The attraction that an atom has for a covalently bonded electron pair
2. Why is CCl4 not a polar molecule?
It is a symmetrical molecule which means all polar bonds cancel each other out
3. Fill in the following table. It is vital that you have this information memorized for your test!
Solute
dissolves in
Solvent
Non-polar
dissolves in
non-polar
Polar
dissolves in
polar
Ionic
dissolves in
polar
Page 3 of 10
BCLN CHEMISTRY 11 - Rev. July/2014
Dissociation and Ionic Equations:
1. Why do covalent (molecular) dissolve but not dissociate?
They are made of whole neutral molecules not ions.
2. For each of the following write i) a balanced chemical formula equation, ii) a total ionic equation
and iii) a net ionic equation. Remember if you are given the name of an ionic compound you
must balance the formula correctly!
a) Magnesium metal is placed in a solution of hydrochloric acid and reacts to form a magnesium
chloride solution and hydrogen gas
i)
Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g)
ii)
Mg(s) + 2 H+(aq) + 2 Cl-(aq)  Mg2+(aq) + 2 Cl-(aq) + H2(g)
iii)
Mg(s) + 2 H+(aq)  Mg2+(aq) + H2(g)
b) Zinc metal is placed in a silver nitrate solution and reacts to form silver metal and a zinc nitrate
solution.
i)
Zn(s) + 2AgNO3(aq)  2Ag(s) + Zn(NO3)2(aq)
ii)
Zn(s) + 2Ag+(aq) + 2NO3- (aq)  2Ag(s) + Zn2+(aq) + 2NO3-(aq)
iii)
Zn(s) + 2Ag+(aq)  2Ag(s) + Zn2+(aq)
c) Barium chloride solution is added to lead(II) nitrate solution in a double replacement reaction.
One lead containing precipitate forms while the other product remains in solution.
i)
BaCl2(aq) + Pb(NO3)2(aq)  PbCl2(s) + Ba(NO3)2(aq)
ii)
Ba2+(aq)+2Cl-(aq)+Pb2+(aq)+2NO3-(aq)  PbCl2(s)+Ba2+(aq)+2NO3-(aq)
iii)
2Cl-(aq) + Pb2+(aq)  PbCl2(s)
Page 4 of 10
BCLN CHEMISTRY 11 - Rev. July/2014
d) Sulfuric acid is added to sodium hydroxide. No precipitate forms from this neutralization
reaction.
i)
H2SO4(aq) + 2NaOH(aq)  2H2O(l) + Na2SO4(aq)
ii)
2H+(aq) + SO42-(aq) + 2Na+(aq) + 2OH-(aq)  2H2O(l) + 2Na+(aq) + SO42-(aq)
iii)
H+(aq) + OH-(aq)  H2O(l)
e) Aqueous chlorine is added to a solution of aluminum bromide in a single replacement reaction
which forms no precipitate and bromine liquid.
i)
3 Cl2(aq) + 2 AlBr3(aq)  2 AlCl3(aq) + 3 Br2(l)
ii)
3Cl2(aq) + 2Al3+(aq) + 6Br-(aq)  2Al3+(aq) + 6Cl-(aq) + 3Br2(l)
iii)
Cl2(aq) + 2Br-(aq)  2Cl-(aq) + Br2(l)
3. What is the concentration of each ion in a 5.50 M solution of aluminum sulfate? Write a
dissociation equation and put the correct concentration below each ion.
Dissociation equation: Al2(SO4)3(aq)  2 Al3+(aq) + 3 SO42-(aq)
Concentration:
5.5 M
Page 5 of 10
11.0 M
16.5 M
BCLN CHEMISTRY 11 - Rev. July/2014
4. What is the concentration of each ion formed when 94.5g of nickel (III) sulphate is dissolved into
850.0 mL of water? You will first need to calculate the concentration of the chemical then write a
dissociation equation and put the correct concentration below each ion.
Molar Mass Ni2(SO4)3 = 2 Ni + 3 S + 12 O = 2(58.69) + 3(32.06) + 12(16.00) = 405.56 g/mol
𝟗𝟒. 𝟓𝐠 𝐍𝐢𝟐 (𝐒𝐎𝟒 )𝟑 𝐱
𝑪=
𝐧
𝐕
=
𝟎.𝟐𝟑𝟑 𝐦𝐨𝐥
.𝟖𝟓𝟎𝟎 𝐋
𝟏 𝐦𝐨𝐥 𝐍𝐢𝟐 (𝐒𝐎𝟒 )𝟑
𝟒𝟎𝟓.𝟓𝟔𝐠 𝐍𝐢𝟐 (𝐒𝐎𝟒 )𝟑
= 𝟎. 𝟐𝟑𝟑 𝐦𝐨𝐥 𝐍𝐢𝟐 (𝐒𝐎𝟒 )𝟑
= 𝟎. 𝟐𝟕𝟒 𝐌 𝐍𝐢𝟐 (𝐒𝐎𝟒 )𝟑
Dissociation equation: Ni2(SO4)3(aq)  2 Ni3+(aq) + 3 SO42-(aq)
Concentration:
0.274 M
0.548 M
0.822 M
5. A 0.657 g sample of sodium hydroxide and a 1.62 g sample of strontium hydroxide are dissolved
in 100. mL of solution. Assuming that no reaction occurs, calculate the concentration of each ion
in the solution.
Molar Mass NaOH = 1 Na + 1 O + 1 H = 22.99 + 16.00 + 1.01 = 40.00 g/mol
Molar Mass Sr(OH)2 = 1 Sr + 2 O + 2 H = 1(87.62) + 2(16.00) + 2(1.01) = 121.64 g/mol
𝐦𝐨𝐥𝐬 𝐍𝐚𝐎𝐇 = 𝟎. 𝟔𝟓𝟕𝐠 𝐍𝐚𝐎𝐇 ×
𝐦𝐨𝐥𝐬 𝐒𝐫(𝐎𝐇)𝟐 = 𝟏. 𝟔𝟐 𝐠 𝐒𝐫(𝐎𝐇)𝟐 ×
𝟏 𝐦𝐨𝐥 𝐍𝐚𝐎𝐇
𝟒𝟎.𝟎𝟎 𝐠 𝐍𝐚𝐎𝐇
𝟏 𝐦𝐨𝐥 𝐒𝐫(𝐎𝐇)𝟐
𝟏𝟐𝟏.𝟔𝟒 𝐠 𝐒𝐫(𝐎𝐇)𝟐
= 𝟎. 𝟎𝟏𝟔𝟒 𝐦𝐨𝐥 𝐍𝐚𝐎𝐇
= 𝟎. 𝟎𝟏𝟑𝟑 𝐦𝐨𝐥 𝐒𝐫(𝐎𝐇)𝟐
mols Na+ = mols NaOH = 0.0164 mol
𝑪=
𝐧
𝐕
=
𝟎.𝟎𝟏𝟔𝟒 𝐦𝐨𝐥
𝟎.𝟏𝟎𝟎 𝐋
= 𝟎. 𝟏𝟔𝟒 𝐌 𝐍𝐚+
(𝐚𝐪)
mols Sr2+ = mols 𝐒𝐫(𝐎𝐇)𝟐 = 0.0133 mol
𝑪=
𝐧
𝐕
=
𝟎.𝟎𝟏𝟑𝟑 𝐦𝐨𝐥
𝟎.𝟏𝟎𝟎 𝐋
𝟐+
= 𝟎. 𝟏𝟑𝟑 𝐌 𝐒𝐫(𝐚𝐪)
mols OH– = (1 × mols NaOH) + (2 × mols Sr(OH)2)
= 0.0164 mol + (2 x 0.0133) mol = 0.0430 mol OH–
𝑪=
𝐧
𝐕
=
𝟎.𝟎𝟒𝟑𝟎 𝐦𝐨𝐥
𝟎.𝟏𝟎𝟎 𝐋
−
= 𝟎. 𝟒𝟑𝟎 𝐌 𝐎𝐇(𝐚𝐪)
Page 6 of 10
BCLN CHEMISTRY 11 - Rev. July/2014
Electrolytes:
1. Why are covalent solutions unable to conduct electricity?
They do not form ions
2. Name 3 common non-electrolytes
Sugar, ethanol and ethylene glycol. Other answers are possible
3. Fill in the following table
Ion Charge
Ion Name
Positive
Cation
Negative
Anion
4. Why does not much current flow through a weak electrolyte solution?
Because most of the compound stays in molecular form meaning there are not many ions
5. How can you create an electrolyte other than dissolving a compound?
Melt an ionic compound
Page 7 of 10
BCLN CHEMISTRY 11 - Rev. July/2014
Chemical Reactions and Solutions:
1. Use the reaction below to answer the following questions. Remember that M = mol/L
K2CO3(s) + 2HCl(aq)  2KCl(aq) + H2O(l) + CO2(g)
a) What volume in mL of 3.00 M hydrochloric acid is required to completely react with 20.0 g of
potassium carbonate?
Molar Mass K2CO3 = 2 K + 1 C + 3 O = 2(39.10) + 12.01 + 3(16.00) = 138.21 g/mol
𝟐𝟎. 𝟎 𝐠 𝐊 𝟐 𝐂𝐎𝟑 ×
𝟏 𝐦𝐨𝐥 𝐊 𝟐 𝐂𝐎𝟑
𝟏𝟑𝟖.𝟐𝟏 𝐠 𝐊 𝟐 𝐂𝐎𝟑
×
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
𝟏 𝐦𝐨𝐥 𝐊 𝟐 𝐂𝐎𝟑
×
𝟏 𝐋 𝐇𝐂𝐥
𝟑.𝟎𝟎 𝐦𝐨𝐥 𝐇𝐂𝐥
= 0.0965 L
= 96.5 mL HCl
b) Determine the number of litres of carbon dioxide gas that can be created at STP when 300. mL
of 3.00 M hydrochloric acid reacts completely.
𝟎. 𝟑𝟎𝟎 𝐋 𝐇𝐂𝐥 ×
𝟑.𝟎𝟎 𝐦𝐨𝐥 𝐇𝐂𝐥
𝟏 𝐋 𝐇𝐂𝐥
×
𝟏 𝐦𝐨𝐥 𝐂𝐎𝟐
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
×
𝟐𝟐.𝟒 𝐋 𝐂𝐎𝟐
𝟏 𝐦𝐨𝐥 𝐂𝐎𝟐
= 𝟏𝟎. 𝟏 𝐋 𝐂𝐎𝟐
c) What volume in mL of 0.250 M hydrochloric acid is required to produce 8.00 g of water?
Molar Mass H2O = 2 H + 1 O = 2(1.01) + 16.00 = 18.02 g/mol
𝟖. 𝟎𝟎 𝐠 𝐇𝟐 𝐎 ×
𝟏 𝐦𝐨𝐥 𝐇𝟐 𝐎
𝟏𝟖.𝟎𝟐 𝐠 𝐇𝟐 𝐎
×
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
𝟏 𝐦𝐨𝐥 𝐇𝟐 𝐎
×
𝟏 𝐋 𝐇𝐂𝐥
𝟎.𝟐𝟓𝟎 𝐦𝐨𝐥 𝐇𝐂𝐥
= 3.55 L HCl
= 3550 mL HCl
d) What mass of potassium carbonate is needed to completely react with 375 mL of 1.65 M
hydrochloric acid?
𝟎. 𝟑𝟕𝟓 𝐋 𝐇𝐂𝐥 ×
𝟏.𝟔𝟓 𝐦𝐨𝐥 𝐇𝐂𝐥
𝟏 𝐋 𝐇𝐂𝐥
×
𝟏 𝐦𝐨𝐥 𝐊 𝟐 𝐂𝐎𝟑
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
Page 8 of 10
×
𝟏𝟑𝟖.𝟐𝟏 𝐠 𝐊 𝟐 𝐂𝐎𝟑
𝟏 𝐦𝐨𝐥 𝐊 𝟐 𝐂𝐎𝟑
= 𝟒𝟐. 𝟖 𝐠 𝐊 𝟐 𝐂𝐎𝟑
BCLN CHEMISTRY 11 - Rev. July/2014
2. What is the molar concentration of each ion if 153.7g of tantalum (V) phosphate are dissolved in
350. mL of solution?
Molar Mass Ta3(PO4)5 = 3 Ta + 5 P + 20 O = 3(180.95) + 5(30.97) + 20(16.00) = 1017.70 g/mol
𝟏𝟓𝟑. 𝟕𝐠 𝐓𝐚𝟑 (𝐏𝐎𝟒 )𝟓 𝐱
𝑪=
𝐧
𝐕
=
𝟎.𝟏𝟓𝟏 𝐦𝐨𝐥
𝟎.𝟑𝟓𝟎 𝐋
𝟏 𝐦𝐨𝐥 𝐓𝐚𝟑 (𝐏𝐎𝟒 )𝟓
𝟏𝟎𝟏𝟕.𝟕𝟎 𝐠
𝐓𝐚𝟑 (𝐏𝐎𝟒 )𝟓
= 𝟎. 𝟏𝟓𝟏 𝐦𝐨𝐥 𝐓𝐚𝟑 (𝐏𝐎𝟒 )𝟓
= 𝟎. 𝟒𝟑𝟏 𝐌 𝐓𝐚𝟑 (𝐏𝐎𝟒 )𝟓
Dissociation equation: Ta3(PO4)5(aq)  3 Ta5+(aq) + 5 PO43-(aq)
Concentration:
0.431M
1.29 M
2.16 M
3. What is the concentration of each ion in solution when 97.7 grams of Be3P2 and 121 grams of
(NH4)3P are dissolved to make 500. mL of solution?
Molar Mass Be3P2 = 3 Be + 2 P = 3(9.01) + 2(30.97) = 88.97 g/mol
Molar Mass (NH4)3P = 3 N + 12 H + 1 P = 3(14.01) + 12(1.01) + 1(30.97) = 85.12 g/mol
𝐦𝐨𝐥𝐬 𝐁𝐞𝟑 𝐏𝟐 = 𝟗𝟕. 𝟕𝐠 𝐁𝐞𝟑 𝐏𝟐 𝐱
𝟏 𝐦𝐨𝐥 𝐁𝐞𝟐 𝐏𝟑
𝟖𝟖.𝟗𝟕𝐠 𝐁𝐞𝟐 𝐏𝟑
𝐦𝐨𝐥𝐬 (𝐍𝐇𝟒 )𝟑 𝐏 = 𝟏𝟐𝟏𝐠 (𝐍𝐇𝟒 )𝟑 𝐏 𝐱
= 𝟏. 𝟏𝟎 𝐦𝐨𝐥 𝐁𝐞𝟑 𝐏𝟐
𝟏 𝐦𝐨𝐥 (𝐍𝐇𝟒 )𝟑 𝐏
𝟖𝟓.𝟏𝟐𝐠 (𝐍𝐇𝟒 )𝟑 𝐏
= 𝟏. 𝟒𝟐 𝐦𝐨𝐥 (𝐍𝐇𝟒 )𝟑 𝐏
mols Be2+ = 3 x mols 𝐁𝐞𝟑 𝐏𝟐 = 3(1.10) = 3.30 mol
𝑪=
𝐧
𝐕
=
𝟑.𝟑𝟎 𝐦𝐨𝐥
𝟎.𝟓𝟎𝟎 𝐋
= 𝟔. 𝟔𝟎 𝐌 𝐁𝐞𝟐+
(𝐚𝐪)
mols NH4+ = 3 x mols (NH4)3P = 3(1.42) = 4.26 mol
𝑪=
𝐧
𝐕
=
𝟒.𝟐𝟔 𝐦𝐨𝐥
𝟎.𝟓𝟎𝟎 𝐋
+
= 𝟖. 𝟓𝟐 𝐌 𝐍𝐇𝟒(𝐚𝐪)
mols P3- = 2 x mols Be3P2 + mols (NH4)3P = 2(1.10) + 1.42
= 3.62 mol P3𝑪=
𝐧
𝐕
=
𝟑.𝟔𝟐 𝐦𝐨𝐥
𝟎.𝟓𝟎𝟎 𝐋
𝟑−
= 𝟕. 𝟐𝟒 𝐌 𝐏(𝐚𝐪)
Page 9 of 10
BCLN CHEMISTRY 11 - Rev. July/2014
4. What is the molar concentration of each ion when 250. mL of 0.500 M NH4F (aq) are added to
750. mL of 0.600 M (NH4)2CrO4 (aq) ?
Dilutions
[NH4 F ] = 𝟎. 𝟓𝟎𝟎 𝐌 𝐱
𝟐𝟓𝟎 𝐦𝐥
𝟏𝟎𝟎𝟎 𝐦𝐋
[(NH4 )2 CrO4 ] = 𝟎. 𝟔𝟎𝟎 𝐌 𝐱
= 𝟎. 𝟏𝟐𝟓 𝐌
𝟕𝟓𝟎 𝐦𝐥
𝟏𝟎𝟎𝟎 𝐦𝐋
= 𝟎. 𝟒𝟓𝟎 𝐌
Dissociation equation: NH4F(aq)  NH4+(aq) + F-(aq)
Concentration:
0.125 M
0.125 M
0.125 M
Dissociation equation: (NH4)2CrO4(aq) 2NH4+(aq) + CrO42-(aq)
Concentration:
0.450 M
0.900 M
0.450 M
[NH4+] = 0.125 M + 0.900 M = 1.025 M
[F-] = 0.125 M
[CrO42-] = 0.450 M
Answers:
Molar Concentration:
2. 2.1 M
3. 0.592 M MgSO4
6. 4.80 L
7. 432 mL
4. 1.58 mol H2O2
5. 4.91g nitre
Dissociation and Ionic Equations:
3. 11.0 M Al3+(aq) and 16.5 M SO42-(aq)
5. 0.164 M Na+(aq),
4. 0.548 M Ni3+(aq) and 0.822 M SO42-(aq)
0.133 M Sr2+(aq) and
0.430 M OH–(aq)
Chemical Reactions and Solutions:
1. a) 96.5 mL HCl
b) 10.1 L CO2 c) 3550 mL HCl
2. 1.29 M Ta5+(aq) and 2.16 M PO43-(aq)
3. 6.60 M Be2+(aq),
8.52 M NH4+(aq) and
7.24 M P3-(aq)
4. 1.025 M NH4+(aq), 0.125 M F-(aq), and 0.450 M CrO42-(aq)
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d) 42.8 g K2CO3