Solutions Chapter 14

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• Solutions Chapter 14

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solution

• Homogeneous mixture of 2 or more substances in a
  single physical state
• particles in a solution are very small
• particles in a solution are evenly distributed
• particles in a solution will not separate

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solute

• The substance that is dissolved
• examples sugar, salt

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solvent

• Substance that does the dissolving
• example water, ethanol
• Aqueous solutions-use water as solvent

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Like dissolves like

• A solute will dissolve best in a solvent with
  similar intermolecular forces.
• If the intermolecular forces are too different
  the solute will not dissolve in that solvent.

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Calculating the strength of a solution

• Often the strength of a solution can be expressed
  in terms of percent.

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Percent Solutions can be calculated 2 ways.

• by volume
• This compares the volume of solute to the total
  volume of solution.

• by mass
• This compares the mass of solute to the total
  mass of solution.

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Volume Percent

• Volume of solute present in a total volume of
  solution.
• Volume Percent (v/v) volume of solute
• / volume of solution x 100

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Calculating volume percent

• A solution is prepared by dissolving 36 ml of
  ethanol in water to a final volume of
• 150 ml what is the solutions volume percent?
• (v/v) ethanol 36 ml ethanol / 150 ml total x
  100
• Volume Percent ethanol 24

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Volume percent

• If 15.0ml of acetone is diluted to 500ml with
  water what is the (v/v) of the prepared
  solution?
• (v/v) 15.0ml / 500ml x 100
• v/v 3.0 acetone

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Mass percent

• Way to describe solutions composition
• mass of solute present in given mass of solution
• mass percent mass of solute
• mass of solution
• grams of solute
• grams of solute grams of solvent

X 100
X 100
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Mass percent

• A solution is prepared by dissolving 1.0g of
  sodium chloride in 48 g of water. The solution
  has a mass of 49 g, and there is 1.0g of solute
  (NaCl) present. Find the mass percent of solute.

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Mass Percent

• A solution is prepared by mixing 1.00g of
  ethanol, C2H5OH, with 100.0 g of water.
  Calculate the mass percent of ethanol in this
  solution.

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Solubility

• The extent to which a solute will dissolve
• expressed in grams of solute per 100g of solvent
• likes dissolve likes

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• Not every substance dissolves in every other
  substance
• soluble- capable of being dissolved
• salt
• insoluble- does not dissolve in another
• oil does not dissolve in water

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Solubility liquids

• Miscible- two liquids that dissolve in each other
  completely
• immiscible- liquids that are insoluble in one
  another
• oil vinegar

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• The compositions of the solvent and solute will
  determine if the substance will dissolve
• stirring
• temperature
• surface area of the dissolving particles

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• A solution is prepared by mixing 2.8 g of sodium
  chloride with 100 g of water. What is the mass
  percent of NaCl?
• What is the volume percent alcohol when you add
  sufficient water to 700mL of isopropyl alcohol to
  obtain 1000mL of solution?

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saturated solution

• contains the maximum amount of solute for a given
  quantity of solvent
• no more solute will dissolve

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unsaturated solution

• contains less solute than a saturated solution
• could use more

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Supersaturated solution

• Solution contains more solute than it can hold
• too much

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• Dilute solution- contains a small amount of
  solute
• Concentrated solution- contains large amount of
  solute

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Solubility

• Table salt at room temperature, 37.7 g can be
  dissolved in 100 ml of H2O

• Sugar at room temperature, 200 g can be
  dissolved in 100 ml of H2O

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Solubility Curve
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Solubility curve

• Determines solubility of substances at specific
  temperatures
• with raising temperature solids increase in
  solubility
• with increase in temperature gases decrease in
  solubility
• ex fish die

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• on the line- saturated (can not hold
  anymore)
• above the line- supersaturated (holding more than
  it can)
• below the line- unsaturated (can hold more solute)

supersaturated
Solute (g) per 100 g H2O
saturated
unsaturated
temperature
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• 92 g of NaNO3 are added to 100ml of water at 25C
  and mixed. What type of solution is it?
• 80 g of NaNO3 are added to 100ml of water at 25C
  and mixed. What type of solution is it?

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• What is the solubility of NaNO3 in 100g of H2O at
  20C?
• What is the solubility of NaNO3 in 200g of H2O at
  20C?

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Concentration of solutions

• Concentration of a solution is the amount of
  solute in a given amount of solvent
• most common measurements of concentration are
• molarity
• (mole fraction)

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Concentration of solutions

• Concentration of a solution is the amount of
  solute in a given amount of solvent
• most common measurements of concentration are
• molarity
• (mole fraction) not discussed in this class

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Molarity

• Number of moles of solute per volume of solution
  in liters
• moles of solute
• molarity (M)
• liters of solution
• mol
• L

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Molarity

• Calculate the molarity of a solution prepared by
  dissolving 11.5g of solid NaOH in enough water to
  make 1.50 L of solution.
• Given
• mass of solute 11.5 g NaOH
• vol of solution 1.50 L
• molarity is moles of solute per liters of
  solution

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Molarity

• Convert mass of solute to moles (using molar mass
  of NaOH). Then we can divide by volume
• molar mass of solute 40.0 g
• 11.5 g NaOH x 1 mol NaOH
• 40.0 g NaOH
• 0.288 mol NaOH
• 1.50 L solution

0.288 mol NaOH
0.192 M NaOH
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Molarity

• Calculate the molarity of a solution prepared by
  dissolving 1.56 g of gaseous HCl into enough
  water to make 26.8 mL of solution.
• Given mass of solute (HCl) 1.56 g
• volume of solution 26.8 mL

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Molarity

• Molarity is moles per liters
• we have to change 1.56 g HCl to moles HCl and
  then change 26.8 mL to liters
• molar mass of HCl 36.5 g
• 1.56 g HCl x 1 mol HCl
• 36.5 g HCl

0.0427 mol HCl 4.27 x 10-2 mol HCl
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Molarity

• Change the volume from mL to liters
• 1 L 1000 mL
• 26.8 mL x 1 L
• 1000 mL

0.0268 L 2.68 x 10-2 L
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molarity

• Finally, divide the moles of solute by the liters
  of solution
• molarity 4.27 x 10-2 mol HCl
• 2.68 x 10-2 L
• 1.59 M HCl

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molarity

• Calculate the molarity of a solution prepared by
  dissolving 1.00 g of ethanol, C2H5OH, in enough
  water to give a final volume of 101mL.
• Molarity moles of solute/ L of solution
• Moles of ethanol MM ethanol 46.08 g/mol
• 1.00 g ethanol / 46.08 g/mol 0.0217 mol

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Solution volume 101 ml convert to liters

• 101 ml / 1000ml per liter 0.101 L
• Molarity (M) moles / L
• M 0.0217 moles / 0.101 L
• Molarity 0.215 M ethanol

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molarity

• One saline solution contains 0.90 g NaCl in
  exactly 1.0 L of solution. What is the molarity
  of the solution?
• Calculate the moles of NaCl
• MM NaCl 58.44 g / mol

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• 0.90 g NaCl / 58.44 g / mol 0.015 moles
• Volume 1.0 L
• Molarity 0.015 moles / 1.0 L
• Molarity 0.015 M NaCl

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molarity

• A solution has a volume of 250 mL and contains
  7.0 x 10?¹ mol NaCl. What is its molarity?
• Convert volume to liters
• 250 ml / 1000 ml per L 0.25 L
• M 7.0 x 10?¹ / 0.25 L
• Molarity of NaCl 2.8 M

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Finding moles to calculate grams

• How many grams of solute is needed to prepare
  300. ml of 3.2 M KCl solution?
• Use the molarity relationship the find the number
  of mol.
• moles L x M
• Convert volume to L
• 300. ml x (1 L/1000ml) 0.300 L
• Calculate mol
• moles 0.300 L x 3.2M 0.96 mol
• Convert mol to grams
• 0.96 mol KCl x (74.5g/mol) 72 g

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Finding volume

• How many liters of 0.442 M MgS can be made with
  27.3 g of MgS? MM of MgS 56 g/mol.
• Use the relationship
• L mol / M
• Convert g to mol
• 27.3g x (1 mol/56 g) 0.488 mol
• Calculate liters
• L 0.488 mol/ 0.442 M 1.10 L

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dilution

• Diluting a solution
• reduces the number of moles of solute per unit
  volume
• the total number of moles of solute in solution
  does not change

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Diluting solutions

• M1V1 M2V2
• M1 molarity of stock solution (initial)
• V1 volume of stock solution (initial)
• M2 molarity of dilute solution
• V2 volume of dilute solution

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M1V1 M2V2

• How many milliliters of aqueous 2.00M MgSO4
  solution must be diluted with water to prepare
  100.00 mL of aqueous 0.400M MgSO4?
• M1 2.00M MgSO4
• M2 0.400M MgSO4
• V2 100.00 mL MgSO4
• V1 ?

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M1V1 M2V2

• Solve for V1
• V1 M2 x V2
• M1
• 0.400M x 100.00 mL
• 2.00M
• 20.0 mL

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M1V1 M2V2

• How many milliliters of a solution of 4.00M KI
  are needed to prepare 0.250 L of 0.760M KI?
• Mi4.00M Vi?
• M20.760M
• V20.250 L

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V1M2xV2/M1

• V1(0.760M)(0.250 L)/4.00M
• V10.0475 L

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If 0.250 L of a 5.00 M HBr solution is diluted to
2.00 L with water what will the resulting
concentration be?

• M1 5.00 M V1 0.250 L
• M2 ? V2 2.00 L
• M2 M1 x V1 / V2
• M2 (5.00M)(0.250L) / (2.00 L)
• M2 0.625M