Chemistry Exam Notes

pH and pOH Calculations

pH = 4.37 is given.
The relationship between pH and pOH is pH + pOH = 14.
Therefore, pOH = 14 - 4.37 = 9.63.
The formula to find hydrogen ion concentration is pH = -\log[H^+].
So, 4.37 = -\log[H^+]. To find the hydrogen ion concentration, use [H^+] = 10^{-4.37}.
Similarly, pOH = -\log[OH^-], so 9.63 = -\log[OH^-]. To find the hydroxide ion concentration, use [OH^-] = 10^{-9.63} = 2.3 \times 10^{-10}.
Alternatively, use the relationship [H^+][OH^-] = 1 \times 10^{-14}. This can be used to find either [H^+] or [OH^-] if one is known.

Moles of NaOH Added

Given 15 milliliters of 0.15 molar NaOH.
Convert milliliters to liters: 15 \text{ mL} = 0.015 \text{ L}.
Molarity is defined as moles per liter, so 0.15 \text{ M} = 0.15 \frac{\text{moles}}{\text{L}}.
Calculate moles of NaOH: 0.015 \text{ L} \times 0.15 \frac{\text{moles}}{\text{L}} = 0.00225 \text{ moles}.

Equivalence Point in Titration

Benzoic acid is a weak acid.
NaOH is a strong base.
When a weak acid is titrated with a strong base, the equivalence point is basic (pH > 7).
This is because at the equivalence point, all that remains is the weak conjugate base.
Weak acid + strong base: equivalence pH > 7 (basic).
Weak base + strong acid: equivalence pH < 7 (acidic).
Strong acid + strong base: equivalence pH = 7 (neutral).
Explanation: A weak acid reacting with a strong base creates a weak conjugate base. At the equivalence point, the acid is neutralized, leaving only the weak base, which makes the solution basic.

Mixture of Benzoic Acid and NaCl Titration

A mixture of benzoic acid and NaCl is titrated with 0.15 M NaOH.
24.78 mL of NaOH is required to reach the equivalence point.
Convert mL to L: 24.78 \text{ mL} = 0.02478 \text{ L}.
Moles of NaOH used: 0.02478 \text{ L} \times 0.15 \frac{\text{moles}}{\text{L}} = 0.003717 \text{ moles}.
The mole ratio of benzoic acid to NaOH is 1:1.
Therefore, moles of benzoic acid = moles of NaOH = 0.003717 moles.
Mass of benzoic acid = moles × molar mass. (Molar mass of benzoic acid ≈ 122.12 g/mol).
Mass of benzoic acid = 0.003717 \text{ moles} \times 122.12 \frac{\text{g}}{\text{mol}} = 0.454 \text{ grams}.
Percent of benzoic acid in the mixture: \frac{0.454}{0.7529} \times 100\% = 60.3\%.

Indicator Choice for Titration

The best indicator changes color near the equivalence point.
In this case, the equivalence point is around pH 8.5 to 9.
Phenol red (pH range 6.8 - 8.2) is the best choice because it changes color within this range.
Thymolphthalein is an option but may change too late (at a higher pH).

Buffer Solutions

A buffer is best when the moles of acid and conjugate base are equal.
When the moles of acid and conjugate base are equal, then the pH equals equals the pKa.
pH = pKa + \log(\frac{[base]}{[acid]}). If [base] = [acid], then \log(\frac{[base]}{[acid]}) = \log(1) = 0, so pH = pKa.
pKa = -\log(Ka). The pKa closest to 7.5 corresponds to the best buffer at that pH.
Buffers resist change in pH.

Buffer Example: Acetic Acid and Sodium Acetate

Mixing acetic acid and sodium acetate creates a buffer.
The pH is 4.73. Adding a strong acid (HCl) does not change the pH significantly because the buffer neutralizes it.
The base (acetate) reacts with the added acid: H^+ + C2H3O2^- \rightarrow HC2H3O2.
The reaction shifts to maintain equilibrium. Buffers contain both a weak acid and a conjugate base to neutralize both acids and bases, maintaining a stable pH (e.g., blood).

Mass Calculation Example

Calculate the mass of CH3NH3Cl given 0.0025 moles.
Mass = moles × molar mass. (Molar mass of CH3NH3Cl ≈ 67.52 g/mol).
0.0025 \text{ moles} \times 67.52 \frac{\text{g}}{\text{mol}} = 0.1688 \text{ grams}.

Burette Usage

To prepare the burette for titration, rinse it first with deionized water, then with the solution that will be used in the titration (e.g., 0.1 M CH3NH2).
This ensures that any residual water in the burette does not dilute the solution, which would change its concentration.

Buffer Calculations with pKa

Given pK_b = 10.64, and equal amounts (0.0025 moles) of base and conjugate acid are added.
Since pH = pKa + \log(\frac{[base]}{[acid]}) and the amounts are equal, pH = pKa.

Lewis Structures and Formal Charges

Formal charge = (Valence electrons) - (Non-bonding electrons) - ($\frac{1}{2}$ Bonding electrons).
In CO_2, diagram z is better because all formal charges are zero.
Carbon: 4 - 0 - ($\frac{1}{2}$ \times 8) = 0.
Oxygen: 6 - 4 - ($\frac{1}{2}$ \times 4) = 0.
Hybridization of carbon in CO_2 is sp (linear geometry).

Sublimation and Enthalpy

Sublimation is the phase transition from solid to gas.
CO2(s) \rightarrow CO2(g)
The enthalpy change for sublimation is positive (endothermic) because heat must be added to convert a solid to a gas.

Ideal Gas Law Calculation

To calculate the volume of the balloon with CO_2 gas, use the ideal gas law: PV = nRT.
List of known values:
- n (moles) = 0.1931 moles
- P (pressure) = 0.998 atm
- T (temperature) = 21°C = 294.15 K
- R (ideal gas constant) = 0.0821 L atm / (mol K) [include this, it's needed for PV=nRT]

Acid-Base Identification

If HSO4^− is acting as an acid, its conjugate base is SO4^{2-}.
Acid-base pairs differ by a proton (H^+).

Equilibrium and Reaction Quotient

When CO2 increases, H2CO_3 formation increases.
Q = \frac{[Products]}{[Reactants]}. If you increase the products then Q will increase.
If Q is less than K, the equilibrium will shift forward to increase product formation until Q = K again.

Titration & Hydroxide Calculations

Calculate hydroxide concentration from pH: pH + pOH = 14.
pOH = 14 - 9.68 = 4.32.
[OH^-] = 10^{-4.32} = 4.8 \times 10^{-5} M.

Kb Expression and Calculation

For the reaction HCO3^- + H2O \rightleftharpoons H2CO3 + OH^-, the Kb expression is Kb = \frac{[H2CO3][OH^-]}{[HCO_3^-]}.
Use an ICE table to find equilibrium concentrations.
- Initial: [HCO3^-] = 0.1 M, [H2CO_3] = 0, [OH^-] = 0.
- Change: [HCO3^-] = -x, [H2CO_3] = +x, [OH^-] = +x.
- Equilibrium: [HCO3^-] = 0.1 - x, [H2CO_3] = x, [OH^-] = x.
Since [OH^-] = 4.8 \times 10^{-5} M, x = 4.8 \times 10^{-5}. Assume x is small (it usually is) so that 0.1 –x is about 0.1.
Kb = \frac{(4.8 \times 10^{-5})^2}{(0.1 - 4.8 \times 10^{-5})} \approx \frac{(4.8 \times 10^{-5})^2}{0.1} = 2.3 \times 10^{-8}.

Equilibrium Constant & Temperature

The equilibrium constant (Kb) is constant at a given temperature; only temperature changes it.

Identifying Acids

Strong acid vs. weak acid: Strength doesn't affect the amount of base needed for neutralization.
Mole ratio is key and since the problem states 1:1 that simplifies the math.
Neutralization requires the same number of moles regardless of acid strength.