Suggested HW: Ch 12: 1, 10, 15, 21, 53, 67, 81

• Much of the chemistry that affects us occurs among substances dissolved in water (proteins, salts, carbohydrates, etc.)
• Solutions are homogenous mixtures, meaning that the components comprising the solution are uniformly dispersed
• 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)
H
2
O (L)

• Chemical reactions often take place when two solutions are mixed.
To perform stoichiometric calculations in such cases, we must know two things:
1.
The overall balanced reaction
2.
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. 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆
𝐌 = 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏(𝑳)

• 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 (NH
4
)
2
Cr
2
O
7
(aq) is prepared by adding the appropriate mass of the salt to a 250 mL volumetric flask and filling up to the mark.

• 0.5g of Cobalt (II) chloride are dissolve in enough H mL of solution. What is the concentration of CoCl
2
2
O to produce 10
?
𝐂𝐨𝐧𝐯𝐞𝐫𝐭 𝐦𝐚𝐬𝐬 𝐭𝐨 𝐦𝐨𝐥𝐞𝐬: 0.5 g CoCl
2 x
1 mol CoCl
2
130 g CoCl
2
= 0.0038 mol
𝐂𝐨𝐧𝐯𝐞𝐫𝐭 𝐯𝐨𝐥𝐮𝐦𝐞 𝐭𝐨 𝐋𝐢𝐭𝐞𝐫𝐬: 10 mL solution x
10
−3 mL
L
= 0.010 L
𝐌𝐨𝐥𝐚𝐫𝐢𝐭𝐲 𝐨𝐟 𝐂𝐨𝐂𝐥
𝟐
(𝐚𝐪) =
0.0038 𝑚𝑜𝑙
0.010 𝐿 𝐻
2
𝑂
= 𝟎. 𝟑𝟖 𝐌
• How many moles of CoCl
2 would be present in 6.3 mL of this solution?
0.0063 L solution x
0.38 mol CoCl
2
1 L solution
= 0.0014 mol CoCl
2

• 15 g of Aluminum nitrate, Al(NO
3
)
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(NO
3
)
3
------> Al 3+ (aq) + 3NO
3
(aq)
H
2
O(L)
• Therefore, every mole of aluminum nitrate yields 3 moles of nitrate
15 g Al(NO
3
)
3 x mol Al(NO
3
)
3
213 g Al(NO
3
)
3 x
𝟑 𝐦𝐨𝐥 𝐍𝐎
−
𝟑
𝟏 𝐦𝐨𝐥 𝐀𝐥(𝐍𝐎
𝟑
)
𝟑
= .211
mol NO
−
3
Nitrate concentration =
.211 mol NO −
3
.200 L
= 𝟏. 𝟎𝟓 𝐌

• 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.

• 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 (n
1
) must be same as the moles of solute present after (n
2
) 𝐧
𝟏
= 𝐧
𝟐
• Therefore:
𝑴
𝟏
𝑽
𝟏
= 𝑴
𝟐
𝑽
𝟐
V
1
V
2

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

• 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 (M
1
= 19.1 M),
• the desired diluted concentration of NaOH (M
2
= 1.0 M),
• and the final volume of the diluted solution (V
2
= 1 L).
 We need to find the volume of the aliquot (V
1
)
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.

• 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 0.035 M solution. Explain how you would do this.
Dilution

• For reactions of solutions, we can use molarity to calculate product yields
• Example:
MnO
2
(s) + 4HBr(aq) -----> MnBr
2
(aq) + Br
2
(L) + 2H
2
O(L)
3.62 g of MnO
2 is added to 25 mL of a 0.85M HBr(aq) solution. Determine the mass of Br (L) formed.
𝑀𝑛𝑂
2
: 3.62𝑔 𝑀𝑛𝑂
2 𝑥 𝑚𝑜𝑙 𝑀𝑛𝑂
2
87𝑔 𝑀𝑛𝑂
2
= 0.041 𝑚𝑜𝑙 𝑀𝑛𝑂
2
𝐻𝐵𝑟:
0.85 𝑚𝑜𝑙 𝐻𝐵𝑟 𝑥 0.025 𝐿 = 0.021 𝑚𝑜𝑙 𝐻𝐵𝑟
1 𝐿
Limiting Reactant !
0.021 𝑚𝑜𝑙 𝐻𝐵𝑟 𝑥
1 𝑚𝑜𝑙 𝐵𝑟
2
4 𝑚𝑜𝑙 𝐻𝐵𝑟 𝑥
159.8 𝑔 𝐵𝑟
2
1 𝑚𝑜𝑙 𝐵𝑟
2
= 0.84 𝑔 𝐵𝑟
2

• A beaker contains 500mL of 0.34M nitric acid, HNO
3
(aq). A
1.07g chunk of silver metal is added to the beaker, and a single replacement reaction proceeds, yielding silver (I) nitrate
(aq) and hydrogen gas.
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.

• Recall from chapter 10 that strong electrolytes are ionic compounds that fully dissociate in water. Acids and bases that are strong electrolytes are known as strong acids/bases . Know these!

• At this point, we will not go into full detail of pH. However, it is important to know how acids and bases are distinguished. The pH scale allows us to do this.
Acids Bases
WATER

The Reaction of Strong Acids and Strong Bases is A Double-
Replacement Reaction Known as a Neutralization Reaction
• When acids and bases react, they neutralize each other, and the product is salt and water
H Cl (aq) + Na OH (aq) H
2
O(L) + Na Cl (aq)
• This is a double replacement reaction. The net ionic equation is:
H + (aq) + Cl (aq) + Na + (aq) + OH (aq) H
2
O(L) + Na + (aq) + Cl (aq)
H + (aq) + OH (aq) H
2
O(L)

• Knowing that acids and bases neutralize each other, lets imagine that we have an acid or base of unknown concentration.
• How can we find the concentration?
• Perform a titration

• In a titration, an indicator is added to the base solution.
• In the example to the right, as long as the pH is above 7 (basic) the indicator will make the solution pink.
• An exact volume of an acid solution is added to a buret.
• The acid solution is added drop-by-drop until the solution just turns clear (neutralized, pH =7 ).
• At this point, you have a stoichiometric equivalent of acid and base. Therefore, if you know the concentration of one, you can immediately determine the concentration of the other.

• Say we have 100 mL of a basic NaOH solution of an unknown concentration.
Before titration
After titration
• We titrate with 5 mL of 1.0 M HCl, and the solution just turns clear.
NaOH(aq) + HCl(aq)  H
2
O(L) + NaCl(aq)
• We know that the acid and base are completely neutralized , and none is left in solution.
Moles of acid added = Stoichiometric equivalent of base
.005 𝐿 𝐻𝐶𝑙 𝑥
1.0 𝑚𝑜𝑙 𝐻𝐶𝑙
𝐿 𝐻𝐶𝑙 𝑥
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= .005 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐻𝐶𝑙
Concentration of base solution =
.005 𝑚𝑜𝑙
.100 𝐿
= .05 𝑀 𝑁𝑎𝑂𝐻

• It is found that 24.6 mL of 0.30M H
2
SO
4 is required to neutralize 20.0 mL of NaOH (aq). Determine the molarity of NaOH (aq), and the molarity of the salt product .
• To test the purity of NaOH tablets, a stock solution is made by 0.40g sample is dissolved into 100 mL of water. This solution is then titrated with a separate 0.100M HCl (aq) solution. It is found that 25 mL of the NaOH (aq) solution is neutralized by 23.2 g of the HCl (aq) solution. Determine the % of the tablet that is NaOH .