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Chapter 4

Solutions:Concentrations,

Dilution,Stoichiometry

(Sections 4.5,4.6)

Lecture Presentation

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Solutions

• Solutions are defined as homogeneous mixtures of two or more pure substances.

• The solvent is present in greatest abundance.

• All other substances are solutes.

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Dissociation

• When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them.

• This process is called dissociation.

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Dissociation

• An electrolyte is a substances that dissociates into ions when dissolved in water.

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Solutions

• An electrolyte is a substance that dissociates into ions when dissolved in water.

• A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

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Electrolytes and Nonelectrolytes

Soluble ionic compounds tend to be electrolytes.

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Electrolytes and Nonelectrolytes

Molecular compounds tend to be nonelectrolytes, except for acids and bases.

Mass Percent

Ways of Expressing ConcentrationWays of Expressing Concentration

100solution of mass total

solution incomponent of masscomponent of % mass

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Molarity• Two solutions can contain the same

compounds but be quite different because the proportions of those compounds are different.

• Molarity is one way to measure the concentration of a solution:

moles of solutevolume of solution in liters

Molarity (M) =

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Mixing a Solution

• To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute.

• The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

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Dilution• One can also dilute a more concentrated

solution by– Using a pipet to deliver a volume of the solution to a

new volumetric flask, and– Adding solvent to the line on the neck of the new flask.

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DilutionThe molarity of the new solution can be determined from the equation

Mc Vc = Md Vd,

where Mc and Md are the molarity of the concentrated and dilute solutions, respectively, and Vc and Vd are the volumes of the two solutions.

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Using Molarities inStoichiometric Calculations

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Titration

Titration is an analytical technique in which one can calculate the concentration of a solute in a solution.

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Practice Problems

1. What is the percent by mass of a solution made by dissolving 6.83 g of ethanol to produce 100. mL of a solution with a density 0.889 g/mL?

 2. A. How many grams of sugar would you need to make 900.0 g of a sugar solution that is 18.6 % by mass sugar?  B. How many grams of water do you need to make 900.0 g of a sugar

solution that is 18.6 % by mass sugar? 

3. What is the molarity of a solution made by dissolving 22.5 g of Li2SO4 in enough water to make 650 mL of solution?

4. 250.0 mL of 0.100 M AgNO3 solution contains _____ grams of silver nitrate.

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Practice Problems5. How many grams of NaOH (MW = 40.0) are there in 500.0 mL of a

0.225 M NaOH solution?

6. The total concentration of ions in a 0.625 M solution of HCl is ________.

7. How many grams of CH3OH must be added to water to prepare 150 mL of a solution that is 1.0 M CH3OH?

8. What is the concentration (M) of KCl in a solution made by mixing 30.0 mL of 0.100 M KCl with 50.0 mL of 0.100 M KCl?

9. What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 230.

mL of solution?

.

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Practice Problems10. How many grams of H3PO4 are in 265 mL of a 1.50 M solution of H3PO4?

11. A stock solution of HNO3 is prepared and found to contain 13.5 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of

0.500 L, the concentration of the diluted solution is __________ M.

12. How much water must be added to a 5.00 M sulfuric acid solution to obtain 250. mL of 1.00 M sulfuric acid solution?

13.To make 250.0 mL of 0.500 M KI solution, _____ mL of 6.00 M KI must be used.

14. A solution is prepared by adding 1.60 g of solid NaCl to 50.0 mL of 0.100 M CaCl2. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

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Practice Problems15. A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form

200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.

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Practice Problems – Solution Stoichiometry1. What mass (g) of AgBr is formed when 35.5 mL of 0.184 M AgNO3 is treated with an excess of aqueous hydrobromic acid?

2.A 13.8 mL solution of 0.176 M H3PO4 (aq) is to be titrated with 0.110 M NaOH (aq). What volume (mL) of base will it take to reach the equivalence point?

3. Oxalic acid is a diprotic acid. Calculate the percent of oxalic acid in a solid given that a 0.7984-g sample of that solid required 37.98 mL of 0.2283 M NaOH for neutralization.

4. Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide. Lead(II) iodide is virtually insoluble in water so that the reaction appears to go to completion. How many milliliters of 3.550 M

HI(aq) must be added to a solution containing 0.700 mol of Pb(NO)3)2 (aq) to completely precipitate the lead?

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Practice Problems – Solution Stoichiometry5. A 186 g sample of sodium oxide was reacted with excess water to form

aqueous sodium hydroxide. This sodium hydroxide was reacted with phosphoric acid to produce sodium phosphate. How many milliliters of 6.00 M phosphoric acid solution is required to react with all the sodium hydroxide produced?

6. The mass percentage of chloride ion in a sample of sea water can be determined by titrating the sample with silver nitrate, which produces a precipitate. In one test, a 25.0 mL sample of sea water required 39.47 mL

of a 0.3029M silver nitrate solution to precipitate out all of the chloride ions.a. Write the balanced chemical equation for the reaction between silver nitrate and chloride ion.b. Calculate the mass percentage of chloride ion in the sea water if the density of sea water is 1.03 g/mL.

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Practice Problems – Solution Stoichiometry7. A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

8. How many liters of a 0.75 M solution of Ca(NO3)2 will be required to react with 148 g of Na2CO3?