Solutions Chapter 14

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

solute•The substance that is dissolved

•examples: sugar, salt

solvent•Substance that does the dissolving–example: water, ethanol

•Aqueous solutions-use water as solvent

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.

Calculating the strength of a solution

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

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.

Volume Percent

Volume of solute present in a total volume of solution.

Volume Percent (v/v) = volume of solute

/ volume of solution x 100%

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 solution’s volume percent?

% (v/v) ethanol = 36 ml ethanol / 150 ml total x 100%

Volume Percent ethanol = 24 %

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

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

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.

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.

Solubility•The extent to which a

solute will dissolve–expressed in grams of solute per 100g of solvent

– ‘likes dissolve likes’

•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

Solubility & liquids•Miscible- two liquids that

dissolve in each other completely

•immiscible- liquids that are insoluble in one another–oil & vinegar

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

•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?

saturated solution

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

unsaturated solution

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

Supersaturated solution

•Solution contains more solute than it can ‘hold’–too much

•Dilute solution- contains a small amount of solute

•Concentrated solution- contains large amount of solute

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

Solubility Curve

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

saturated

unsaturated

supersaturated

temperature

Solute (g) per 100 g H2O

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)

• 92 g of NaNO3 are added to 100ml of water at 25°C and mixed. What type of solution is it?

• 80 g of NaNO3 are added to 100ml of water at 25°C and mixed. What type of solution is it?

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

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

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)

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

Molarity• Number of moles of solute per volume

of solution in liters

moles of solute• molarity (M) = liters of solution mol L

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

Molarity• Convert mass of solute to moles (using

molar mass of NaOH). Then we can divide by volume

• molar mass of solute = 40.0 g11.5 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

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

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

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

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

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/mol1.00 g ethanol / 46.08 g/mol= 0.0217 mol

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

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

• 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

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

Finding moles to calculate gramsHow 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

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

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

Diluting solutionsDiluting solutions

MM11VV11 = M = M22VV22

• M1 = molarity of stock solution (initial)

• V1 = volume of stock solution (initial)

• M2 = molarity of dilute solution

• V2 = volume of dilute solution

MM11VV11 = M = M22VV22

• 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 = ?

MM11VV11 = M = M22VV22

• Solve for V1

• V1 = M2 x V2

M1

0.400M x 100.00 mL

2.00M

= 20.0 mL

MM11VV11 = M = M22VV22

• How many milliliters of a solution of 4.00M KI are needed to prepare 0.250 L of 0.760M KI?

• Mı=4.00M Vı=?

• M2=0.760M

• V2=0.250 L

V1=M2xV2/M1

• V1=(0.760M)(0.250 L)/4.00M

• V1=0.0475 L

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