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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.