Two types to discuss this semester:
| Solutions | Aqeueous | General |
|---|---|---|
| solvent | water | Major component of the mixture |
| solute | Substance dissolved in the water (primarily ionic compounds/salts) | Substance dissolved in the solvent |
| Solution Types | Reactions | Equations |
|---|---|---|
| Strong electrolytes | Acid/Base | Molecular |
| Non-electrolytes | Precipitation | Total Ionic |
| Weak electrolytes | Reduction-Oxidation or Oxidation-Reduction (redox) | ? |
Strong electrolytic (complete dissolution) solution
\[ NaCl (s) \rightarrow Na^{+} (aq) + Cl^{-} (aq) \]
Non-electrolytic solution
Weak electrolytic solution
Arrhenius definition of acids and bases
| Strong Acids | Strong Bases |
|---|---|
| \(HCl\) hydrochloric | \(LiOH\) |
| \(HBr\) Hydrobromic | \(NaOH\) |
| \(HI\) Hydroiodic | \(KOH\) |
| \(HNO_{3}\) Nitric | \(RbOH\) |
| \(HClO_{4}\) Perchloric | \(CsOH\) |
| \(HClO_{3}\) Chloric acid | \(Ca(OH)_{2}\) |
| \(H_{2}SO_{4}\) Sulfuric | \(Sr(OH)_{2}\) |
| \(Ba(OH)_{2}\) |
Neutral elements and monatomic ions:
Polyatomic species:
Reduction-Oxidation and Oxidation-Reduction rxns are known as “redox” rxns.
Combusion Rxn:
\[ CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O \]
| Elements | Reactant Side | Product Side | Difference | Charge |
|---|---|---|---|---|
| C | -4 | +4 | +8 | 8 |
| H | +1 | +1 | 0 | 0 |
| O | 0 | -2 | -2 | -2 |
Associated idea:
Consider the rxn:
\[ 2Al (s) + 3CuSO_{4} (aq) \rightarrow Al_{2}(SO_{4})_{3} (aq) + 3 Cu (s) \]
Total ionic (aqeueous separate):
\[ 2Al (s) + 3Cu^{2+} (aq) + 3SO_{4}^{2-} (aq) \rightarrow 2Al^{3+} + 3SO_{4}^{2-} (aq) + 3 Cu (s) \]
Notice the \(3SO_{4}^{2-}\) is a spectator ion, so we can write the net ionic (excluding the spectator ion):
\[ 2Al (s) + 3Cu^{2+} (aq) \rightarrow 2Al^{3+} + 3 Cu (s) \]
| Elements | Initial Ox. Num | Final Ox. Num | Diff | Change |
|---|---|---|---|---|
| Al | 0 | +3 | +3 | Oxidation |
| Cu | +2 | 0 | -2 | Reduction |
Half rxn is a balanced chemical equation that includes electrons to maintain charge balance.
Consider aluminum from the previous rxn:
\[ Al \rightarrow Al^{+3} \]
Charge is imbalanced, must add electrons to the product side
\[ Al \rightarrow Al^{+3} + 3e^{-} \]
Now, the charge is balanced.
Notice, that aluminum loses 3 electrons, so oxidation is loss of electrons
Consider copper from the previous rxn:
\[ Cu^{2+} + 2e^{-} \rightarrow Cu \]
Reactants need to gain 2 electrons, so reduction is gain of electrons
Acronym: OIL RIG (oxidation is loss of electrons) (reduction is gain of electrons)
\[ Al \rightarrow Al^{3+} + 3e^{-} \]
\[ Cu^{2+} + 2e^{-} \rightarrow Cu \]
Let’s double the first and triple the second
\[ 2Al \rightarrow 2Al^{3+} + 6e^{-} \]
\[ 3Cu^{2+} + 6e^{-} \rightarrow 3Cu \]
Electrons cancel, so we are left with:
\[ 2Al + 3Cu^{2+} \rightarrow 2Al^{3+} + 3Cu \]
\[ 2 H_{2}O_{2} \rightarrow + 2H_{2}O + O_{2} \]
| Element | Init | Final | Change |
|---|---|---|---|
| H | +1 | +1 | 0 |
| O | -1 | -2/0 | -1/+1 |
A disproportionation rxn means one element is both reduced and oxidized.
We use molarity to figure out how much solute is in a solution
\[ \text{conc} = \frac{n}{v} = \frac{0.715\text{ mol}}{0.495\text{ L}} = 1.44\text{ mol/L} = 1.44\text{ M} \]
\[ \text{conc} = \frac{n}{v} \]
\[ n = \text{conc} * V = 0.50\text{ M} * 0.085\text{ L} = 0.042\text{ mol KI} \]
\[ \frac{0.575}{1.23} = 0.467\text{ L} \]
\[ \frac{135}{} * \frac{1}{342.3} = 0.394\text{ mol} \]
\[ V = \frac{\text{mol}}{M} = \frac{0.394}{3.30} = 0.1195\text{ L} = 120. \text{ mL} \]
Adding 127.4 g of \(NaOH\) (40.00 g/mol) to make 500.0 mL of solution.
\[ \frac{127.4}{} * \frac{1}{40.00} = 3.185\text{ mol NaOH} \]
\[ \frac{n}{v} = \frac{3.185}{.5} = 6.370\text{ M} \]
Using the 6.370 M stock solution, how to get 0.100 M of NaOH 250.0 mL volumetric flask; how much of the stock solution needs to be diluted?
\[ M_{1}V_{1} = M_{2}V_{2} \]
\[ V_{1} = V_{2} \frac{M_{2}}{M_{1}} = 250.0 * \frac{0.1000}{6.370} = 3.925\text{ mL} \]
Can also solve this by finding number of moles
What is in a solution when we say, “A solution of nominal concentration of 0.01M”?
Suppose sodium chloride dissolve to Na and Cl: the nominal concentration of the individual ions are the same as the salt itself, since the quantities are 1:1.