You can now complete Lab 19

A soluble salt is an ionic compound that dissolves in water.

An insoluble salt is an ionic compound that does not dissolve in water.

A soluble salt dissolves in water. ◦Breaks apart into ions

Insoluble salts do not dissolve in water.◦Attraction between ions is to strong for

water to break apart

The solubility rules predict whether a salt is soluble or insoluble in water. Table 8.8

When solutions of salts are mixed, a solid forms when ions of an insoluble salt combine.

Precipitate- an insoluble compound that forms as a result a reaction

The precipitate forms when ions in solution recombine in a way that makes an insoluble compound.

Pb(NO3)2(aq) + 2 KI(aq) --->

2 KNO3(aq) + PbI2(s)

Pb(NO3)2 (aq) + NaBr (aq) ? (s) + ? (aq)

Approach: Consider all possible ion combinations and determine which is insoluble (ie, which is the solid) based on solubility rules.

You should be able to identify when a precipitate will form based on the solubility of the products.

When iron (III) chloride reacts with sodium hydroxide in water, what are the products? Write a balanced equation and indicate if there are any precipitates.

You can now complete Worksheet 1

The concentration of a solution is the amount of solute dissolved in a specific amount of solution (solutions = solute + solvent).

amount of soluteamount of solution

The percent concentration describes the amount of solute that is dissolved in 100 parts of solution.

amount of solute100 parts solution

The mass percent (%m/m)- concentration of the solution that describes the mass of solute in every 100 g of solution.

Concentration is the percent by mass of solute in a solution.mass percent = g of solute x 100% g of solution

grams of solute + grams of solvent

50.0 g KCl solution

Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution).

g of KCl = 8.0 gg of solvent (water) = 42.0 gg of KCl solution = 50.0 g

8.0 g KCl (solute) x 100 = 16% (m/m) KCl

50.0 g KCl solution

Mass Percent: Is the g of solute in 100 g of solution.

mass percent = g of solute 100 g of

solution10% (m/m) = 10 g of solute in 100g of solution

15% (m/m) = 15 g of solute in 100 g of solution.

The mass/volume percent (%m/v) Concentration is the ratio of the mass

in grams (g) of solute in a volume (mL) of solution.mass/volume % = g of solute x 100% mL of solution

A percent mass/volume solution is prepared by weighing out the grams of solute (g) and adding water to give the final volume of the solution.

Mass/volume percent (%m/v) is calculated from the grams of solute (g KCl) and the volume of solution (mL KCl solution).g of KI = 5.0 g KImL of KI solution = 250.0 mL

5.0 g KI (solute) x 100 = 2.0%(m/v) KI

250.0 mL KI solution

Is the g of solute in 100 mL of solution.mass/volume % = g of solute 100 mL of solution

10% (m/v) = 10 g of solute in 100 mL of solution

15% (m/v) = 15 g of solute in 100 mL of solution

The volume percent (%v/v) Concentration is the percent volume

(mL) of solute (liquid) to volume (mL) of solution.volume % (v/v) = mL of solute x100% mL of solution

Is the mL of solute in 100 mL of solution.volume % (v/v) = mL of solute 100 mL of solution

10% (v/v) = 10 mL of solute in 100 mL of solution

15% (v/v) = 15 mL of solute in 100 mL of solution

Two conversion factors can be written for any type of % value. Table 8.9

How many grams of NaCl are needed to prepare

250 g of a 10.0% (m/m) NaCl solution?

1. Write the 10.0 %(m/m) as conversion factors.10.0 g NaCl and 100 g solution

100 g solution 10.0 g NaCl

2. Use the factor that cancels given (g solution).250 g solution x 10.0 g NaCl = 25 g NaCl

100 g solution

How much (in g) of KI do you need to make a 2% (m/v) if the final volume is 250 mL?

Molarity is a concentration unit for the moles of solute in the liters (L) of solution.

Molarity (M) = moles of solute = moles liter of solution L

Examples:2.0 M HCl = 2.0 moles HCl

1 L

6.0 M HCl= 6.0 moles HCl 1 L

A 1.0 M NaCl solution is prepared by weighing out 58.5 g NaCl ( 1.0 mole) and adding water to make 1.0 liter of solution.

What is the molarity of a NaOH solution prepared by adding 4.0 g of solid NaOH to water to make 0.50 L of solution ?

1. Determine the moles of solute. 4.0 g NaOH x 1 mole NaOH = 0.10 mole 40.0 g NaOH

2. Calculate molarity. 0.10 mole = 0.20 mole = 0.20 M

NaOH 0.50 L 1 L

What is the molarity of a solution made by dissolving 12.5 g NaCl in water to make 500. mL of solution?

What is the molarity of a solution made by dissolving 12.5 g KCl in water to make 500. mL of solution?

The units in molarity can be used to write conversion factors. Table 8.10

How many liters of a 2.00M NaCl solution do you need to provide 67.3 g of NaCl

1. Determine the moles of solute desired.

67.3 g NaCl x 1 mole NaCl = 1.15 mole 58.5 g NaCl

2. Calculate volume. 1.15 mole x 1L = 0.575 L

NaCl 2.00 mole

How many grams of KCl is needed to prepare 4.0L of of a 0.20 M KCl solution?

Diluting a solution Is the addition of water. Decreases concentration.

Concentrated DilutedSolution Solution

In a dilution, the amount (moles or grams) of solute does not change.Amount solute (initial) =Amount solute (dilute)

A dilution increases the volume of the solution.

Amount solute Amount solute

Volume (initial) Volume (increased)

As a result, the concentration of the solution decreases.

Add water to the 3.0 M solution to lower its concentration to 0.50 M

Dilute the solution!

3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

But how much water But how much water do we add?do we add?

How much water is added?

The important point is that --->

moles of NaOH in ORIGINAL solution = moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solutionmoles of NaOH in FINAL solution

PROBLEM: You have 50.0 mL of PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 3.0 M NaOH and you want 0.50

M NaOH. What do you do?M NaOH. What do you do?

Amount of NaOH in original solution =

M • V =

(3.0 mol/L)(0.050 L) = 0.15 mol NaOH

Amount of NaOH in final solution must also = 0.15 mol NaOH

Volume of final solution =

(0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L

Conclusion:add 250 mL of

water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH. 3.0 M NaOH 0.50 M NaOH

H2O

Concentrated Dilute

What is the molarity of an HCl solution prepared by

adding 0.50 L of water to 0.10 L of a 12 M HCl.

1. Calculate the moles of HCl. 0.10 L x 12 moles HCl = 1.2

moles HCl 1 L

What is the molarity of an HCl solution prepared by

adding 0.50 L of water to 0.10 L of a 12 M HCl.

1. Calculate the moles of HCl. 0.10 L x 12 moles HCl = 1.2

moles HCl 1 L

2. Determine the volume after dilution. 0.10 L + 0.50 L = 0.60 L solution

What is the molarity of an HCl solution prepared by

adding 0.50 L of water to 0.10 L of a 12 M HCl.1. Calculate the moles of HCl. 0.10 L x 12 moles HCl = 1.2 moles

HCl 1 L

2. Determine the volume after dilution. 0.10 L + 0.50 L = 0.60 L solution3. Calculate the molarity of the diluted HCl.

1.2 moles HCl = 2.0 M HCl 0.60 L

The dilution calculation can be expressed as an equation.

C1V1 = C2V2 C = concentration For molar concentration, the equation is

M1V1 = M2V2

(moles) = (moles after dilution )

For percent concentration, %1V1 = %2V2

(grams) = (grams after dilution )

A shortcut

Cinitial • Vinitial = Cfinal • Vfinal

How many milliliters of 6.0 M NaOH solution are needed to prepare 1.0 L of a 0.15 M NaOH solution?M1 = 6.0 M M2 = .15 MV1 = ??? V2 = 1.0 LRearrange the dilution equation for V1

How many milliliters of 6.0 M NaOH solution are needed to prepare 1.0 L of a 0.15 M NaOH solution?M1 = 6.0 M M2 = .15 MV1 = ??? V2 = 1.0 LRearrange the dilution equation for V1

V1 = M2V2 = 0.15 M x 1.0 L

M1 6.0 M= 0.025 L x 1000 mL = 25 mL

1 L

Describe how to make a 0.12 M solution CuCl from 10mL of a 0.29 M solution of CuCl.