Chem 109 - General Chemistry - Spring 2015

Lecture Notes 13: 20 February

Reaction Stoichiometry: Hydrogen "funnel" stoichiometry reaction demo.

Electrolyte Solutions , cont.

Strong Electrolytes

Strong Electrolytes are completely ionized in aqueous solution in concentrations of 1 M or less. Three common classes:

Soluble salts
Strong acids
Strong bases

Strong salts

Virtually all inorganic salts are strong electrolytes in the sense that only ions exist in solution. For example sodium chloride dissociates into hydrated sodium ions (Na⁺_(aq)) and hydrated chloride ions (Cl^-_(aq)). Note that each ion is surrounded by a "shell" of water molecules, countering and diffusing their charges, insulating them from other ions, and keeping them in solution.

Strong acids

Acids which completely dissociate at concentrations of 1 M or less. Thus hydrochloric acid (HCl) dissociates to give H⁺ and Cl^-, nitric acid (HNO₃) gives H⁺ and NO₃^-, and sulfuric acid (H₂SO₄) gives H⁺ and HSO₄^-.

Strong bases

Strong bases dissociate or react completely with water to give hydroxide ions. Thus sodium hydroxide (NaOH) gives Na⁺ and OH^-, and potassium hydroxide gives K⁺ and OH^-.

Weak Electrolytes

Weak electrolytes only partially dissociate in aqueous solution in concentrations of 1 M or less. The most common weak electrolytes are weak acids. For example acetic acid (CH₃OOH) and hydrofluoric acid (HF) both dissociate only slightly (a few percent or less).

Molarity and the Composition of Solutions

The most commonly used concentration term in chemistry is molarity = 1 mole of solute dissolved in enough solvent to give 1 L = M = mol/L. This is the most popular concentration unit at least in part because of the convenience of making molar solutions with volumetric flasks (show flask).

Note that molarity has various meanings in describing solutions. Commonly molarity refers to the amount of substance dissolved in one liter of liquid to give a solution. Thus if 158.52 g (1 mole) of strontium chloride is dissolved in enough water to give one liter we will have a 1 L solution of strontium chloride. Note however, that while the solution is 1M in Sr²⁺, it is 2 M in Cl^- and 3 M in terms of particles!

There are a number of different kinds of problems.

Making molar solutions

Make up a 1.000 L solution of 0.25 M NaCl (note that water is the "default" solvent).

First weigh out 0.25 moles of NaCl

= (0.25 mole)(22.99 g + 35.45 g)/mole = 14.61 g

Next we add water until the total volume is 1.000 L.

Determining Molar Concentration

What is the concentration of a solution made by dissolving 10.00 g of KI in enough water to make 1.000 L

First need to find the number of moles of KI:

(10.00 g) / ({39.10 g+ 126.9 g}/mole) = 6.024 x 10^-2 mole

Thus the concentration will be 6.024 x 10^-2 M

Dilution problems

What is the concentration of a solution resulting from adding 25.0 mL of 0.60 M CaCl₂ to 475 mL of water. (Note - both aqueous, so the volumes are additive.)
First, I like to keep in mind that Moles = Moles, that is we have conservation of mass.

With this in mind, Moles₁ = (volume₁)(Molarity₁) = Moles₂ = (volume₂)(Molarity₂)

(Note the units of volume will cancel, so we don't HAVE to convert to L, though it doesn't hurt.)

First need to determine volumes:

volume₁ = 0.025 L

volume₂ = 0.025 L + 0.475 L = 0.500 L

Can solve as a ratio: (0.025 L) (0.60 M) = (0.500 L) (x)

x = (0.60 M)(0.025 L)/(0.500 L) = 3.00 x 10 ^-2

= 3.0 x 10^-2M

What is the concentration of chloride ion in this solution?

As an inorganic salt, we know it dissociates completely, thus from the formula = CaCl₂

we see that there are (2 Cl^-) / (CaCl₂)

Therefore [Cl^-] = 6.0 x 10 ^-2M

How much 1.000 M MgSO₄ is needed to make 500.0 mL of a 0.25 M solution.

First convert volumes to Liters: 500.0 mL = 0.5000 L

Can solve as a ratio: (0.5000 L) (0.25 M) = (1.0000 M) (x)

x = (0.25M)(0.5000 L)/(1.0000 M) = 0.125 L = 0.13 L

Acid-Base Reactions

Neutralization

When we combine equal numbers of moles of hydrogen ion and hydroxide ion a neutralization occurs. That is, there is no reactive component left, all of the acid has been consumed by all of the base, and water has been synthesized.

Consider the reaction of 50.0 mL of 0.25 M hydrochloric acid with 25.0 mL of 0.50 M sodium hydroxide.

Write a net ionic equation for this reaction:

H⁺ + Cl^- + Na⁺ + OH^- H₂O + Cl^- + Na⁺

Giving: H⁺ + OH^- H₂O

How much acid will be left over? Base? How much water was made (synthesized)?

First find out the moles of each:

acid = (50.0 mL)(1 L/1000 mL)(0.25 mole/L) = 1.25 x 10^-2moles;
base = (25.0 mL)(1 L/1000 mL)(0.50 mole/L) = 1.25 x 10^-2moles.

Since the amounts are identical they will completely neutralize each other. That is they will react completely and both acid and base will be consumed with none left over.
Since the ratio in the equation is 1:1:1 acid:base:water, 1.25 x 10^-2moles of water are produced.

Consider the reaction of a weak acid and strong base: 50.0 mL of 0.25 M acetic acid is reacted with 18.0 mL of 0.50 M sodium hydroxide. Find the number of moles of each of the reactants and products after reaction.

First recall that for the reaction of a weak acid and a strong base the reaction will go until one of the reactants is completely consumed.

Writing the net ionic reaction we get (note that since it is a weak acid, we write it in the undissociated state):

CH₃COOH + OH^- H₂O + CH₃COO^-

First find moles of each reactant:

acid = (50.0 mL)(1 L/1000 mL)(0.25 mole/L) = 1.25 x 10^-2moles

base = (18.0 mL)(1 L/1000 mL)(0.50 mole/L) = 0.900 x 10^-2moles

After reaction all of the base is consumed, so:

base = 0

acetate = 0.900 x 10^-2moles

acid = 1.25 x 10^-2moles - 0.900 x 10^-2moles = 0.35 x 10^-2moles

Water synthesized = 0.900 x 10^-2moles.