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.