1. MOLARITY

2. Aqueous Solutions
Much of the chemistry that affects us occurs among substances dissolved in water (proteins, salts, carbohydrates, etc.)
The most common type of solution is a solid dissolved in a liquid. The dissolved solid is the solute, the liquid is the solvent.
Solutes and solvents do not react, merely co-exist, as is the case with an aqueous solution like salt water
NaCl (s) > NaCl(aq)
H2O (L)

3. Concentration (Molarity)
Chemical reactions often take place when two solutions are mixed. To perform stoichiometric calculations in such cases, we must know two things:
The overall balanced reaction
The amount of solute present in each solution
The concentration or MOLARITY of a solute describes the number of solute ions/molecules in a certain volume of solvent
Molarity, represented by the letter M, is defined as the moles of solute per liter of solution.
𝐌= 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏(𝑳)

4. Preparing an Aqueous Solution (Ex. 250mL of 1.43M Ammonium Dichromate)
Typically, a volumetric flask is used to prepare solutions. Volumetric flasks come in a wide array of sizes, and are marked to indicate a specific volume of solution.
For example, 250 mL of a 1.43M (NH4)2Cr2O7 (aq) is prepared by adding the appropriate mass of the salt to a 250 mL volumetric flask and filling up to the mark.

5. Examples
0.5g of Cobalt (II) chloride are dissolve in enough H2O to produce 10 mL of solution. What is the concentration of CoCl2 ?
𝐂𝐨𝐧𝐯𝐞𝐫𝐭 𝐦𝐚𝐬𝐬 𝐭𝐨 𝐦𝐨𝐥𝐞𝐬: g CoC l 2 x 1 mol CoC l g CoC l 2 = mol
𝐂𝐨𝐧𝐯𝐞𝐫𝐭 𝐯𝐨𝐥𝐮𝐦𝐞 𝐭𝐨 𝐋𝐢𝐭𝐞𝐫𝐬: 10 mL solution x 10 −3 L mL =0.010 L
𝐌𝐨𝐥𝐚𝐫𝐢𝐭𝐲 𝐨𝐟 𝐂𝐨𝐂𝐥𝟐(𝐚𝐪)= 𝑚𝑜𝑙 𝐿 𝐻 2 𝑂 =𝟎.𝟑𝟖 𝐌
How many moles of CoCl2 would be present in 6.3 mL of this solution?
L solution x 0.38 mol CoC l 2 1 L solution = mol CoC l 2

6. Al(NO3)3 ------> Al3+(aq) + 3NO3-(aq)
Example
15 g of Aluminum nitrate, Al(NO3)3, is dissolved in enough water to produce 200 mL of solution. What is the molarity of nitrate in the solution?
Aluminum nitrate will dissociate into aluminum and nitrate ions, as according to the chemical formula:
Al(NO3) > Al3+(aq) + 3NO3-(aq)
Therefore, every mole of aluminum nitrate yields 3 moles of nitrate
H2O(L)
15 g Al(N O 3 ) 3 x mol Al(N O 3 ) g Al(N O 3 ) 3 x 𝟑 𝐦𝐨𝐥 𝐍𝐎 𝟑 − 𝟏 𝐦𝐨𝐥 𝐀𝐥(𝐍 𝐎 𝟑 ) 𝟑 = .211 mol N O 3 −
Nitrate concentration= mol N O 3 − L =𝟏.𝟎𝟓 𝐌

7. Dilution
In many instances (especially in lab), you may need to prepare a solution of some desired concentration from a pre-existing stock solution.
For example, consider a concentrated detergent like Tide®. If wanted to wash a shirt, you wouldn’t just dump Tide® on it.
Instead, you add a cap-full or so (aliquot) to a large volume of water to attain a manageable solution.
This action of “watering down” the detergent to a useable state is called dilution. The bottle of Tide® is the stock solution.

8. Dilution V1 V2
Keep in mind that dilution does not change the total moles of solute, only the molarity.
We know that the moles (n) of solute in V liters of a solution with molarity M is:
n = MV
The moles of solute present before addition of water (n1) must be same as the moles of solute present after (n2)
Therefore:
𝐧 𝟏 = 𝐧 𝟐
𝑴 𝟏 𝑽 𝟏 = 𝑴 𝟐 𝑽 𝟐

9. How to perform a Dilution
Take an aliquot (V1) of the stock solution, add it to a new container
High concentration
stock solution of concentration M1
After mixing, we have a dilute solution with volume V2 and concentration M2
Aliquot of stock solution with volume V1 and concentration M1.
Dilute with solvent to desired volume, V2
𝐧= 𝐌 𝟏 𝐕 𝟏
𝐧= 𝐌 𝟐 𝐕 𝟐

10. Example
You need to perform an experiment using NaOH (aq). At your disposal is 1L of a concentrated stock solution of 19.1 M NaOH (aq). This is much too concentrated for your intended purpose. You would instead prefer to have 1L of a 1.0M solution. How would you perform this dilution?
We are given:
initial concentration of the NaOH stock (M1 = 19.1 M),
the desired diluted concentration of NaOH (M2 = 1.0 M),
and the final volume of the diluted solution (V2 = 1 L).
We need to find the volume of the aliquot (V1)
𝐌 𝟏 𝐕 𝟏 = 𝐌 𝟐 𝐕 𝟐
19.1 𝑀 𝑉 1 = 1.0 𝑀 (1𝐿)
V 1 =0.052 L =52 mL
Add 52 mL of the stock solution to 948 mL of water and mix to obtain 1L of a 1.0 M NaOH (aq) solution.

11. Group Examples
a.) Explain how would you make a 500 mL stock solution that is 0.1 M Sodium Selenide.
b.) From this stock solution, you decide to perform a dilution to prepare 100 mL of a M solution. Explain how you would do this.
Dilution

12. Applying Molarity to Stoichiometry
For reactions of solutions, we can use molarity to calculate product yields
Example:
MnO2(s) + 4HBr(aq) -----> MnBr2(aq) + Br2(L) + 2H2O(L)
3.62 g of MnO2 is added to 25 mL of a 0.85M HBr(aq) solution. Determine the mass of Br (L) formed.
𝑀𝑛 𝑂 2 : 3.62𝑔 𝑀𝑛 𝑂 2 𝑥 𝑚𝑜𝑙 𝑀𝑛 𝑂 2 87𝑔 𝑀𝑛 𝑂 2 = 𝑚𝑜𝑙 𝑀𝑛 𝑂 2
𝐻𝐵𝑟: 𝑚𝑜𝑙 𝐻𝐵𝑟 1 𝐿 𝑥 𝐿=0.021 𝑚𝑜𝑙 𝐻𝐵𝑟
Limiting Reactant !
0.021 𝑚𝑜𝑙 𝐻𝐵𝑟 𝑥 1 𝑚𝑜𝑙 𝐵 𝑟 2 4 𝑚𝑜𝑙 𝐻𝐵𝑟 𝑥 𝑔 𝐵 𝑟 2 1 𝑚𝑜𝑙 𝐵𝑟 2 =0.84 𝑔 𝐵 𝑟 2

13. Single and Double Replacement Reactions
Before proceeding, let’s define two important types of ionic reactions that occur in solution:
Single Replacement reactions (an element reacts with an electrolyte; the element becomes a cation and the cation becomes and element)
Double Replacement reactions (reaction between two salts where the cations and anions swap positions)

14. Group Example
A beaker contains 500mL of 0.34M nitric acid, HNO3(aq). A 1.07g chunk of silver metal is added to the beaker. A single replacement reaction proceeds. Calculate the mass of hydrogen produced, and the molarity of silver (I) nitrate (aq) if the reaction has a 90% yield. You may assume that the change in the volume of the solution following the addition of Ag is negligible.

15. Metal Displacement
Single replacement reactions occur with metals when one metal is less stable (more reactive) in its solid state than the other.
In the reaction below, Zn displaces Ag because Zn is more reactive:
Cu(s) + 2AgNO3 (aq) Cu(NO3)2(aq) + 2Ag(s)
More reactive metals prefer to exist as aqueous ions instead of neutral atoms. Therefore, a more reactive metal will displace a less active metal. The opposite will NOT occur.
An activity series is provided the powerpoint.

17. Group Work
As shown in the table, elemental Li has the highest activity, so it is the most likely to react. It will displace any other metal. Potassium will displace any metal except Li. And so on.
Predict the products.
Li (s) + Ca(ClO4)2 (aq)
Na (s) + ZnSO4(aq)
Al (s) + KNO3 (aq)

18. Metal Replacement Involving Acids
When a metal reacts with a binary acid (HX), the metal replaces the hydrogen atom to yield a salt and elemental hydrogen.
Zn(s) + 2HCl (aq) ZnCl2(aq) + H2(g)

19. Redox
Single replacement reactions are examples of red-ox (reduction-oxidation) reactions
A reduction reaction occurs when the oxidation state (charge) of an element/ion becomes more negative during the course of a reaction (i.e. the species gains electrons)
In an oxidation process, the oxidation state of an element/ion becomes more positive during a reaction (species loses electrons)

20. Redox Consider the following single replacement reaction:
Zn(s) + Cu SO4 (aq) Zn SO4 (aq) + Cu (s)
On the reactant side, we have elemental Zn. The charge on any pure element is 0
On the product side, we have a Zn2+ ion. Since the charge of Zn has gone from 0 to 2+, Zn has undergone an oxidation. Zn loses 2 electrons. Where did they go???
On the product side, we have elemental Cu, so Cu has undergone a reduction from 2+ to 0 by taking electrons from Zn.
On the reactant side, we have a Cu2+ ion.

21. Zn(s) + Cu SO4 (aq) Zn SO4 (aq) + Cu (s)
Redox
Zn(s) + Cu SO4 (aq) Zn SO4 (aq) + Cu (s)
We have identified the reduction and oxidation processes in the reaction above
O: Zn0  Zn2+ + 2e-
R: Cu2+ + 2e-  Cu0
RED-OX REACTIONS
Because Cu2+ gets reduced, it is the oxidizing agent. Cu2+ takes electrons away from more the more reactive Zn (oxidizes Zn).
Because Zn gets oxidized, it is the reducing agent. In other words, Zn electrons are used for the reduction of Cu2+

22. Zn(s) + Cu SO4 (aq) Zn SO4 (aq) + Cu (s)

23. Ex. Rusting
Reduced
4Fe(s) + 3O2(g) 2Fe2O3(s)
Oxidized