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Chapter 4b Chemistry Slides
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Transcript of Chapter 4b Chemistry Slides
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Chapter 4
Chemical Quantities
and Aqueous Reactions
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2014 Pearson Education, Inc.
Solutions
A homogeneous mixture of two or more substances
Solvent
major component
Solute
minor component
An aqueous solution has water as the solvent
Example: water + salt
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2014 Pearson Education, Inc.
Solutions
molarity (M) = concentration of a solution
amount of solute (in mol) volume of solution (in L)
molarity (M) =
mol L
M =
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2014 Pearson Education, Inc.
Solution Preparation 1. Add solute to a flask
2. Add water to reach desired volume
Example:
Make 1 L of a 1 M NaCl solution
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Example
25.5 g KBr x = 0.214286 mol KBr 119.00 g KBr
1 mol KBr
What is the molarity of a solution that contains 25.5 g KBr in a 1.75 L solution? Molar mass of KBr = 119.00 g/mol
amount of solute (in mol) volume of solution (in L)
molarity (M) =
molarity (M) = 1.75 L solution
0.214286 mol KBr = 0.122449 M = 0.123 M
0.123 mol/L
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Example
45.4 g NaNO3 x = 0.5341177 mol NaNO3 85.00 g NaNO3
1 mol NaNO3
Calculate the molarity of a solution made by adding 45.4 g NaNO3 to a flask and dissolving it in water to a volume of 2.50 L. Molar mass of NaNO3 = 85.00 g/mol
amount of solute (in mol) volume of solution (in L)
molarity (M) =
molarity (M) = 2.50 L solution
0.5341177 mol NaNO3 = 0.213647 M = 0.214 M
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Example
1.50 M KBr = 0.250 L solution
mol KBr
What mass of KBr (g) is needed to make 250.0 mL of a 1.50 M KBr solution? Molar mass of KBr = 119.00 g/mol
amount of solute (in mol) volume of solution (in L)
molarity (M) =
1 mol KBr
119.00 g KBr = 44.6 g KBr
0.375 mol KBr
0.375 mol KBr x
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Molarity Conversion Factors
1 L solution
0.500 mol NaCl 0.500 M NaCl or 1 L solution
0.500 mol NaCl
Example
How many liters of a 0.125 M NaOH solution contain 0.255 mol of NaOH?
0.255 mol NaOH x = 2.04 L solution 0.125 mol NaOH
1 L solution
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Example
0.758 M C12H22O11 = 1.55 L solution
mol C12H22O11
How many grams of sucrose (C12H22O11) are in 1.55 L of a 0.758 M sucrose solution? (molar mass of sucrose = 342.30 g/mol)
amount of solute (in mol) volume of solution (in L)
molarity (M) =
1 mol C12H22O11
342.30 g C12H22O11 = 402 g C12H22O11
1.1749 mol C12H22O11
1.1749 mol C12H22O11 x
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Solution Dilution
Concentrated form of a solution is referred to as a stock solution
We can dilute a stock solution using the following formula:
M1V1 = M2V2
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Solution Dilution
M1V1 = M2V2
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Example
To what volume should you dilute 0.200 L of a 15.0 M NaOH solution to obtain a 3.00 M NaOH solution?
M1V1 = M2V2 M1 = 15.0 M V1 = 0.200 L M2 = 3.00 M
Solve for V2 : V2 = M2
M1V1
V2 = 3.00 mol/L
15.0 mol/L x 0.200 L = 1.00 L
V2 = ?
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Example
To what volume should you dilute 100.0 mL of a 5.00 M CaCl2 solution to obtain a 0.750 M CaCl2 solution?
M1V1 = M2V2 M1 = 5.00 M V1 = 100.0 mL M2 = 0.750 M
Solve for V2 : V2 = M2
M1V1
V2 = 0.750 mol/L
5.00 mol/L x 0.1000 L = 0.667 L
= 0.1000 L
V2 = ?
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Example
What volume of a 6.00 M NaNO3 solution should you use to make 0.525 L of a 1.20 M NaNO3 solution?
Solve for V1 : V1 = M1
M2V2
V1 = 6.00 mol/L
1.20 mol/L x 0.525 L = 0.105 L
M1V1 = M2V2
M1 = 6.00 M V1 = ? M2 = 1.20 M V2 = 0.525 L
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Example
What volume (L) of a 0.150 M KCl solution will completely react with 0.150 L of a 0.175 M Pb(NO3)2 solution?
0.150 L Pb(NO3)2 x 1 L Pb(NO3)2 solution
0.175 mol Pb(NO3)2
= 0.350 L KCl solution
2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)
0.150 M 0.175 M V = ? 0.150 L 1 mol Pb(NO3)2
2 mol KCl
x 1 mol Pb(NO3)2
2 mol KCl
x 0.150 mol KCl
1 L KCl solution