Chemical Calculations for Solutions (Ch 12)

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

Solutions

• 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 NaCl(aq)

NaCl (s) -----> NaCl(aq) H2O (L)

• The concentration of a solute describes the number of solute ions/molecules in a certain volume of solvent

• Concentration is most commonly expressed using MOLARITY, represented by the letter M. Molarity is defined as the moles of solute per liter of solution.

Concentration (Molarity)

Examples

• 30 g of NaCl are dissolved in 450 mL of H2O. What is the concentration of NaCl?

𝑣𝑜𝑙𝑢𝑚𝑒 :450𝑚𝐿𝐻2𝑂 x10− 3 𝐿𝑚𝐿 =0.450 𝐿𝐻2𝑂

𝑚𝑜𝑙𝑒𝑠 : 30𝑔𝑁𝑎𝐶𝑙 𝑥 𝑚𝑜𝑙𝑁𝑎𝐶𝑙58.45𝑔𝑁𝑎𝐶𝑙=.513𝑚𝑜𝑙𝑁𝑎𝐶𝑙

𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑜𝑓 𝑁𝑎𝐶𝑙= .513𝑚𝑜𝑙𝑁𝑎𝐶𝑙.450 𝐿𝐻2𝑂

=1.14𝐌

• How many moles of NaCl are there in 500 mL of this solution?

.500 𝐿𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑥 1.14𝑚𝑜𝑙𝑁𝑎𝐶𝑙1𝐿𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 =0.57𝑚𝑜𝑙

Example

• 15 g of Aluminum nitrate, Al(NO3)3, is dissolved in 200 mL of H2O. What is the concentration of nitrate in the solution?

• Aluminum nitrate will dissociate into aluminum and nitrate ions, as according to the chemical formula:

Al(NO3)3 ------> Al3+(aq) + 3NO3-(aq)

• Therefore, every mole of aluminum nitrate yields 3 moles of nitrate

H2O(L)

15𝑔 𝐴𝑙 ¿𝑁𝑖𝑡𝑟𝑎𝑡𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛=

.542𝑚𝑜𝑙.200𝐿 =𝟐 .𝟕𝟏𝑴

Dilution

• In many instances (especially in lab), you may need to prepare a solution of some desired concentration from a pre-existing stock solution.

• An example of this would be a water enhancer, like Mio. You wouldn’t drink the Mio directly because it is extremely sweet.

• Instead, you add a small amount (aliquot) to your water, until you’ve attained the desired level of taste and sweetness.

• This is dilution. The flavored water is our diluted aqueous solution, and the bottle of Mio is the stock solution

Dilution

• 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

• Therefore, we know the concentration of a solution before and after dilution:

𝑴𝟏𝑽𝟏=𝑴𝟐𝑽 𝟐

V1 V2

How to perform a Series Dilution

High concentrationstock solution of concentration M1

Aliquot of stock solution with volume V1 and concentration M1.

Dilute with solvent to desired volume, V2

After complete mixing, we have a dilute solution with volume V2 and concentration M2

Take an aliquot (V1) of the stock solution, add it to a new container

Example

• A concentrated stock solution of NaOH is 19.1 M. How would you prepare 500 mL of a 3.0 M solution?

We are given: • initial concentration of the NaOH stock (M1 = 19.1 M), • the desired final concentration of NaOH (M2 = 3.0 M), • and the final volume of the solution (V2 = 0.5 L). We need to find the volume of the aliquot (V1)

𝑴𝟏𝑽𝟏=𝑴𝟐𝑽 𝟐 (19.1𝑀 ) (𝑉 1 )=(3.0𝑀 )(0.500𝐿)

(𝑉 1 )=(3.0𝑀 )(0.500𝐿)

(19.1𝑀 )=.0785𝐿=78.5𝑚𝐿

• A 78.5 mL aliquot of the stock solution is added to 421.5 mL of water to make a 3.0 M solution

Example

• a.) Explain how would you make a 500 mL stock solution that is 0.1 M NaBr (aq) ? (molar mass NaBr: 102.9 g/mol)

• b.) From this stock solution, you decide to make 100 mL of a 0.01 M solution. Explain how you would do this?

Dilution

Applying Molarity to Stoichiometric Calculations

• 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.

𝐻𝐵𝑟 : 0.85𝑚𝑜𝑙𝐻𝐵𝑟1𝐿 𝑥 0.025𝐿=0.021𝑚𝑜𝑙𝐻𝐵𝑟

𝑀𝑛𝑂2 :3.62𝑔𝑀𝑛𝑂2𝑥𝑚𝑜𝑙𝑀𝑛𝑂2

87𝑔𝑀𝑛𝑂2=0.041𝑚𝑜𝑙𝑀𝑛𝑂2

Limiting Reactant !0.021𝑚𝑜𝑙𝐻𝐵𝑟 𝑥

1𝑚𝑜𝑙𝐵𝑟24𝑚𝑜𝑙𝐻𝐵𝑟 𝑥

159.8𝑔 𝐵𝑟 21𝑚𝑜𝑙 𝐵𝑟 2

=0.84𝑔 𝐵𝑟 2

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

HCl (aq) + NaOH(aq) H2O(L) + NaCl(aq)

• This is a double replacement reaction. The net ionic equation is:

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) H2O(L) + Na+(aq) + Cl-(aq)

H+(aq) + OH-(aq) H2O(L)

Titrations

• 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

pH Scale

• 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.

BasesAcids

WATER

Titrations

• 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.

• We titrate with 5 mL of 1.0 M HCl, and the solution just turns clear.

TitrationsBefore titration

After titration

NaOH(aq) + HCl(aq) H2O(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

Concentration of base solution = .005𝑚𝑜𝑙.100 𝐿 =.05𝑀𝑁𝑎𝑂𝐻

.005𝐿𝐻𝐶𝑙 𝑥 1.0𝑚𝑜𝑙𝐻𝐶𝑙𝐿𝐻𝐶𝑙 𝑥 1𝑚𝑜𝑙𝑁𝑎𝑂𝐻

1𝑚𝑜𝑙𝐻𝐶𝑙 = .005𝑚𝑜𝑙𝑁𝑎𝑂𝐻