Chapter 4

Chemical Quantities

and Aqueous Reactions

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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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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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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

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

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

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

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

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

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

Solution Dilution

M1V1 = M2V2

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 = ?

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 = ?

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

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