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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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Chemical Calculations for Solutions (Ch 12)
Dr. HarrisLecture 12Suggested HW: Ch 12: 1, 10, 15, 21, 53, 67, 81
Solutions
โข As we learned in a previous chapter, solutions are homogenous mixtures, meaning that the components comprising the solution are uniformly dispersed
โข The most common type of solution is a solid dissolved in a liquid. The dissolved solid is the solute, the liquid is the solvent.
โข 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)
Solubility
โข When NaCl is dissolved, the ions are surrounded by water molecules, and dipole interactions disperse causes the ions to separate and disperse
โข 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
โข The solution is now saturated
โข The quantity of NaCl dissolved at this point is its solubility.
Figure Above: Dissolution of NaCl and uniform distribution of solute and solvent.
Electrolytes
โข Reminder: Strong electrolytes fully dissociate in water. All strong electrolytes are ionic compounds. These include:โข All salts with group 1 cationsโข All salts with ammonium cationsโข All salts with nitrate, perchlorate, and acetate anionsโข Hydroxides of Ca, Sr, and Ba (plus group 1 cations and
ammonium)โข All sulfates except Ca and Ba
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
Strong Acids and Strong Bases Are Strong Electrolytes
โข Strong ACIDSโข HClโข HBrโข HIโข HNO3
โข HClO4
โข H2SO4
โข Strong BASESโข Hydroxides of group 1
metalsโข Hydroxides of Ca, Ba,
and Sr
pH Scale
โข 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
โข The concentration of a solute describes the number of solute ions/molecules in a certain volume of solvent
โข 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
โข 30 g of NaCl are dissolved in 450 mL of H2O. What is the concentration of NaCl?
๐ฃ๐๐๐ข๐๐ :450๐๐ฟ๐ป2๐ x10โ 3 ๐ฟ๐๐ฟ =0.450 ๐ฟ๐ป2๐
๐๐๐๐๐ : 30๐๐๐๐ถ๐ ๐ฅ ๐๐๐๐๐๐ถ๐58.45๐๐๐๐ถ๐=.513๐๐๐๐๐๐ถ๐
๐ถ๐๐๐๐๐๐ก๐๐๐ก๐๐๐๐๐ ๐๐๐ถ๐= .513๐๐๐๐๐๐ถ๐.450 ๐ฟ๐ป2๐
=1.14๐
โข How many moles of NaCl are there in 500 mL of this solution?
.500 ๐ฟ๐ ๐๐๐ข๐ก๐๐๐๐ฅ 1.14๐๐๐๐๐๐ถ๐1๐ฟ๐ ๐๐๐ข๐ก๐๐๐ =0.57๐๐๐
Example
โข 15 g of Aluminum nitrate, Al(NO3)3, is dissolved in 200 mL of H2O. What is the concentration of nitrate in the solution?
โข Aluminum nitrate will dissociate into aluminum and nitrate ions, as according to the chemical formula:
Al(NO3)3 ------> Al3+(aq) + 3NO3-(aq)
โข Therefore, every mole of aluminum nitrate yields 3 moles of nitrate
H2O(L)
15๐ ๐ด๐ ยฟ๐๐๐ก๐๐๐ก๐ ๐๐๐๐๐๐๐ก๐๐๐ก๐๐๐=
.542๐๐๐.200๐ฟ =๐ .๐๐๐ด
Dilution
โข In many instances (especially in lab), you may need to prepare a solution of some desired concentration from a pre-existing stock solution.
โข An example of this would be a water enhancer, like Mio. You wouldnโt drink the Mio directly because it is extremely sweet.
โข Instead, you add a small amount (aliquot) to your water, until youโve attained the desired level of taste and sweetness.
โข This is dilution. The flavored water is our diluted aqueous solution, and the bottle of Mio is the stock solution
Dilution
โข Keep in mind that dilution does not change the total moles of solute, only the molarity.
โข We know that the moles (n) of solute in V liters of a solution with molarity M is:
n = MV
โข Therefore, we know the concentration of a solution before and after dilution:
๐ด๐๐ฝ๐=๐ด๐๐ฝ ๐
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 of the stock solution, add it to a new container
Example
โข 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: โข initial concentration of the NaOH stock (M1 = 19.1 M), โข the desired final concentration of NaOH (M2 = 3.0 M), โข and the final volume of the solution (V2 = 0.5 L). We need to find the volume of the aliquot (V1)
๐ด๐๐ฝ๐=๐ด๐๐ฝ ๐ (19.1๐ ) (๐ 1 )=(3.0๐ )(0.500๐ฟ)
(๐ 1 )=(3.0๐ )(0.500๐ฟ)
(19.1๐ )=.0785๐ฟ=78.5๐๐ฟ
โข 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
โข 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)
โข b.) From this stock solution, you decide to make 100 mL of a 0.10M solution. Explain how you would do this?
Dilution
Applying Molarity to Stoichiometric Calculations
โข For reactions of solutions, we can use molarity to calculate product yields
โข 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
.0416๐๐๐๐๐๐2๐ฅ4๐๐๐๐ป๐ต๐๐๐๐๐๐๐2
=.167๐๐๐๐ป๐ต๐
๐=๐๐=
.167๐๐๐8.84๐๐๐ ๐ฟโ 1
=.0188๐ฟ=18.8๐๐ฟ
.0416๐๐๐๐๐๐2
The Reaction of Strong Acids and Strong Bases is A Double-Replacement Reaction Known as a Neutralization Reaction
โข When acids and bases react, they neutralize each other, and the product is salt and water
HCl (aq) + NaOH(aq) H2O(L) + NaCl(aq)
โข 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
โข Knowing that acids and bases neutralize each other, lets imagine that we have an acid/base of unknown concentration.
โข How can we find the concentration?
โข Perform an acid/base titration
Titrations
โข In a titration, an indicator is added to the basic solution.
โข In the example to the right, as long as the pH is above 7 (basic) the indicator will make the solution pink.
โข An exact volume of an acid solution with a known concentration is added to a buret.
โข The acid solution is added drop-by-drop until the solution just turns clear (neutralized, pH =7 ).
โข Say we have 100 mL of a basic NaOH solution of an unknown concentration.
โข 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)
โข 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 = .005๐๐๐.100๐ฟ =.05๐๐๐๐๐ป
.005๐ฟ๐ป๐ถ๐ ๐ฅ 1.0๐๐๐๐ป๐ถ๐๐ฟ๐ป๐ถ๐ ๐ฅ 1๐๐๐๐๐๐๐ป
1๐๐๐๐ป๐ถ๐ = .005๐๐๐๐๐๐๐ป