Notes
Solutions
- "Like dissolves like." Solids usually dissolve better when hot; gases dissolve better when cold and under pressure.
- Molarity: $\displaystyle M = \frac{\text{mol solute}}{\text{L solution}}$. Dilution: $M_1V_1 = M_2V_2$.
- Colligative properties depend on particle count: $\Delta T_f = iK_fm$, $\Delta T_b = iK_bm$ (water: $K_f = 1.86$, $K_b = 0.512$ °C/m).
Acids and bases
| Acid | Base | |
|---|---|---|
| Arrhenius | makes $\text{H}^+$ in water | makes $\text{OH}^-$ in water |
| Brønsted-Lowry | proton donor | proton acceptor |
| pH | below 7 | above 7 |
| Litmus | turns blue litmus red | turns red litmus blue |
- $\text{pH} = -\log[\text{H}^+]$, $\text{pH} + \text{pOH} = 14$, and each pH unit is a factor of 10.
- Neutralization: acid + base $\rightarrow$ salt + water. In a titration, use moles and the mole ratio; for 1 : 1, $M_aV_a = M_bV_b$.
Rates and equilibrium
- Rate goes up with temperature, concentration, surface area and catalysts.
- At equilibrium, forward and reverse rates are equal. $K_{eq}$ = products over reactants, raised to coefficients.
- Le Chatelier: the system shifts to relieve a stress.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle \frac{0.250}{0.5000} = 0.500$ M.
$\displaystyle M_2 = \frac{(2.00)(25.0)}{100.0} = 0.500$ M.
Nitric acid is strong, so $[\text{H}^+] = 1.0 \times 10^{-3}$ M and $\text{pH} = 3.00$.
$\text{NH}_3$ gains one $\text{H}^+$ to become $\text{NH}_4^+$.
$\displaystyle M_a = \frac{(0.150)(20.0)}{25.0} = 0.120$ M.
Adding a reactant is a stress the system relieves by using it up, shifting toward products.
Free response
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Free-response question
A student finds the concentration of acetic acid in vinegar. She titrates 10.00 mL of vinegar with 0.500 M NaOH, using phenolphthalein. The solution turns light pink after 16.70 mL of NaOH is added. The reaction is $\text{HC}_2\text{H}_3\text{O}_2 + \text{NaOH} \rightarrow \text{NaC}_2\text{H}_3\text{O}_2 + \text{H}_2\text{O}$.
(a) Calculate the moles of NaOH used.
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$\text{mol} = M \times V = 0.500 \times 0.01670 = 0.00835$ mol NaOH.
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(b) Calculate the molarity of acetic acid in the vinegar.
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The mole ratio is 1 : 1, so there are 0.00835 mol of acetic acid. $\displaystyle M = \frac{0.00835}{0.01000} = 0.835$ M.
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(c) Acetic acid has a molar mass of 60.06 g/mol. Find the grams of acetic acid in 1.00 L of vinegar, and the percent by mass if 1.00 L of vinegar has a mass of 1000 g.
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$0.835 \text{ mol} \times 60.06 \text{ g/mol} = 50.2$ g per liter. Percent by mass $\displaystyle = \frac{50.2}{1000} \times 100\% = 5.0\%$, which matches typical vinegar.
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