Solutions
Calculate the molality of a solution made by dissolving 8.5 g of sucrose (C₁₂H₂₂O₁₁, Molar Mass = 342 g/mol) in 225 g of water.
0.111 m
Calculate [H⁺] and pH for a 0.30 M solution of HNO₃ (strong acid).
[H⁺] = 0.30 M (strong acids fully dissociate)
pH = -log(0.30) = 0.52
For the reaction 2N₂O(g) ⇌ 2N₂(g) + O₂(g), write the Kp expression.
Kp = [P(N₂)]² × P(O₂) / [P(N₂O)]²
What does Q represent and how does it relate to Kc?
Q (reaction quotient) has the same mathematical form as Kc but uses current concentrations instead of equilibrium concentrations. It tells you which direction the reaction will shift:
If Q < Kc → reaction shifts forward
If Q > Kc → reaction shifts reverse
If Q = Kc → system is at equilibrium
Explain why adding a non-volatile solute to a liquid decreases its vapor pressure.
Non-volatile solute particles occupy space at the liquid surface, reducing the number of solvent molecules at the surface that can escape. This lowers the rate of evaporation, decreasing the vapor pressure below that of pure solvent.
For formic acid, Ka = 1.8 × 10⁻⁴. What is pKa?
pKa = -log(Ka) = -log(1.8 × 10⁻⁴) = 3.74
If you decrease the volume of a system at equilibrium for 2A(g) ⇌ B(g), which direction does it shift? Why?
Shifts right (toward products).
At 25°C, a solution has [H⁺] = 2.5 × 10⁻¹¹ M. Calculate [OH⁻] using Kw and determine if the solution is acidic, basic, or neutral.
[OH⁻] (4.0 × 10⁻⁴) >> [H⁺] (2.5 × 10⁻¹¹), the solution is basic.
Calculate the mole fraction of ethanol in a solution containing 4.6 g of ethanol (molar mass = 46 g/mol) dissolved in 36 g of water.
χ(ethanol) = 0.048
A weak base solution has a pH of 10.5 at 25°C. Calculate the [OH⁻] concentration.
[OH⁻] = 3.16 × 10⁻⁴ M
For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = -198 kJ. If temperature is increased, does Kc increase or decrease?
Kc decreases.
The reaction is exothermic (ΔH < 0). Increasing temperature shifts equilibrium reverse (toward reactants) by Le Châtelier's Principle. This means fewer products at equilibrium, so Kc must decrease.
What is the conjugate acid of NH₃? What is the conjugate base of HCO₃⁻?
Conjugate acid of NH₃ is NH₄⁺ (ammonia accepts a proton)
Conjugate base of HCO₃⁻ is CO₃²⁻ (bicarbonate loses a proton)
A solution has a boiling point elevation of 1.5°C. If Kb for water is 0.512°C/m and the solute is a non-electrolyte, what is the molality?
2.9 m
Rank by increasing acid strength and explain: HF, H₂O, acetic acid (CH₃COOH), HCl. Hint: Consider bond strength, electronegativity, and stability of conjugate base.
H₂O < CH₃COOH < HF< HCl
At 500 K, Kc = 1.44 for H₂(g) + I₂(g) ⇌ 2HI(g). If [H₂] = 0.050 M, [I₂] = 0.050 M, and [HI] = 0.32 M, calculate Q and determine reaction direction.
Since Q (40.96) >> Kc (1.44), the reaction shifts left (reverse) toward reactants.
Explain why adding excess salt to pure water can change the pH of the water.
If the salt contains an ion that undergoes hydrolysis (like NH₄⁺ or acetate), the solution pH can change. NH₄⁺ is acidic: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, lowering pH.
Acetate is basic: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, raising pH.
A neutral salt like NaCl has no effect.
Two solutions have identical solute particles but different solvents. Solution A uses water (Kb = 0.512°C/m) and Solution B uses benzene (Kb = 2.53°C/m). Which has a larger boiling point elevation? Why?
Solution B has the larger boiling point elevation. Since both have the same molality, and ΔTb = Kb × m, the solution with the larger Kb (benzene) will have the larger elevation. Benzene has weaker intermolecular forces than water, so solute particles disrupt the solution more dramatically.
Explain why HClO₄ is a much stronger acid than HClO. Consider the structure and electron-withdrawing effects.
HClO₄: Cl in +7 oxidation state with 4 electron-withdrawing O atoms. These O atoms pull electron density through inductive effects, highly polarizing the O-H bond and making H⁺ easy to lose. The conjugate base ClO₄⁻ distributes negative charge across 4 oxygens.
HClO: Cl in +1 oxidation state with only 1 O atom. Much less electron withdrawal, weaker O-H polarization, and less stable conjugate base. Result: HClO is 10⁸ times weaker.
The decomposition of N₂O₄ has Kc = 0.60 at 298 K for N₂O₄(g) ⇌ 2NO₂(g). Starting with 0.80 mol of N₂O₄ in a 2.0 L vessel, find all equilibrium concentrations.
[N₂O₄] = 0.40 - 0.185 = 0.22 M
[NO₂] = 2(0.185) = 0.37 M
Calculate Kp from Kc for the reaction CO(g) + Cl₂(g) ⇌ COCl₂(g) at 400 K if Kc = 49.5. (R = 0.0821 L·atm/(mol·K))
Kp = Kc(RT)^Δn
Kp = 1.51