chemical calculations for solutions ( ch 12)

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Chemical Calculations for Solutions (Ch 12) Dr. Harris Lecture 12 Suggested HW: Ch 12: 1, 10, 15, 21, 53, 67, 81

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Chemical Calculations for Solutions ( Ch 12). Dr. Harris Lecture 12 Suggested HW: Ch 12: 1, 10, 15, 21, 53, 67, 81. Solutions. As we learned in a previous chapter, solutions are homogenous mixtures, meaning that the components comprising the solution are uniformly dispersed - PowerPoint PPT Presentation

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Page 1: Chemical Calculations for Solutions ( Ch  12)

Chemical Calculations for Solutions (Ch 12)

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

Page 2: Chemical Calculations for Solutions ( Ch  12)

Solutions

• As we learned in a previous chapter, 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)

Page 3: Chemical Calculations for Solutions ( Ch  12)

Solubility

• When NaCl is dissolved, the ions are surrounded by water molecules, and dipole interactions disperse causes the ions to separate and disperse

• As more and more NaCl is added, a point is reached where further dissociation ceases, the salt simply drops to the bottom of the beaker

• The solution is now saturated

• The quantity of NaCl dissolved at this point is its solubility.

Figure Above: Dissolution of NaCl and uniform distribution of solute and solvent.

Page 4: Chemical Calculations for Solutions ( Ch  12)

Electrolytes

• Reminder: Strong electrolytes fully dissociate in water. All strong electrolytes are ionic compounds. These include:• All salts with group 1 cations• All salts with ammonium cations• All salts with nitrate, perchlorate, and acetate anions• Hydroxides of Ca, Sr, and Ba (plus group 1 cations and

ammonium)• All sulfates except Ca and Ba

Page 5: Chemical Calculations for Solutions ( Ch  12)

Strong vs. Weak Electrolytes

CaCl2(s) -----------> Ca2+ (aq) + 2Cl-(aq)

HgCl2(s) ------------->

--------> HgCl2(aq)

--------> HgCl+(aq) + Cl-(aq)

--------> Hg2+(aq) + 2Cl-(aq)

H2O(l)

H2O(l)

100%

99.8%

0.18%

0.02%

Full dissociation of strong electrolyte

Minimal dissociation of weak electrolyte

Page 6: Chemical Calculations for Solutions ( Ch  12)

Strong Acids and Strong Bases Are Strong Electrolytes

• Strong ACIDS• HCl• HBr• HI• HNO3

• HClO4

• H2SO4

• Strong BASES• Hydroxides of group 1

metals• Hydroxides of Ca, Ba,

and Sr

Page 7: Chemical Calculations for Solutions ( Ch  12)

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

Page 8: Chemical Calculations for Solutions ( Ch  12)

• 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)

Page 9: Chemical Calculations for Solutions ( Ch  12)

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

Page 10: Chemical Calculations for Solutions ( Ch  12)

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

Page 11: Chemical Calculations for Solutions ( Ch  12)

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

Page 12: Chemical Calculations for Solutions ( Ch  12)

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

Page 13: Chemical Calculations for Solutions ( Ch  12)

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 of the stock solution, add it to a new container

Page 14: Chemical Calculations for Solutions ( Ch  12)

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

Page 15: Chemical Calculations for Solutions ( Ch  12)

Example

• a.) Explain how would you make a 500 mL stock solution that is 1.0 M NaBr (aq) ? (molar mass NaBr: 102.9 g/mol)

• b.) From this stock solution, you decide to make 100 mL of a 0.10M solution. Explain how you would do this?

Dilution

Page 16: Chemical Calculations for Solutions ( Ch  12)

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)

What volume of an 8.84 M HBr solution is needed to completely react with 3.62 g of MnO2?

Convert MnO2 to moles

Determine the required moles of HBr

Calculate volume

.0416𝑚𝑜𝑙𝑀𝑛𝑂2𝑥4𝑚𝑜𝑙𝐻𝐵𝑟𝑚𝑜𝑙𝑀𝑛𝑂2

=.167𝑚𝑜𝑙𝐻𝐵𝑟

𝑉=𝑛𝑀=

.167𝑚𝑜𝑙8.84𝑚𝑜𝑙 𝐿− 1

=.0188𝐿=18.8𝑚𝐿

.0416𝑚𝑜𝑙𝑀𝑛𝑂2

Page 17: Chemical Calculations for Solutions ( Ch  12)

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)

Page 18: Chemical Calculations for Solutions ( Ch  12)

Titrations

• Knowing that acids and bases neutralize each other, lets imagine that we have an acid/base of unknown concentration.

• How can we find the concentration?

• Perform an acid/base titration

Page 19: Chemical Calculations for Solutions ( Ch  12)

Titrations

• In a titration, an indicator is added to the basic 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 with a known concentration is added to a buret.

• The acid solution is added drop-by-drop until the solution just turns clear (neutralized, pH =7 ).

Page 20: Chemical Calculations for Solutions ( Ch  12)

• 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.100𝐿 =.05𝑀𝑁𝑎𝑂𝐻

.005𝐿𝐻𝐶𝑙 𝑥 1.0𝑚𝑜𝑙𝐻𝐶𝑙𝐿𝐻𝐶𝑙 𝑥 1𝑚𝑜𝑙𝑁𝑎𝑂𝐻

1𝑚𝑜𝑙𝐻𝐶𝑙 = .005𝑚𝑜𝑙𝑁𝑎𝑂𝐻