Solutions
Define "solution" and give one example at the molecular level.
A solution is a homogeneous mixture of solute dissolved in solvent; e.g., NaCl (ions) dispersed in H₂O
State the definition of "solubility" and what a solubility curve shows.
Solubility = max amount of solute that dissolves in given solvent at a certain temperature; curve plots solubility vs temperature.
Identify whether HBr is an acid or a base and state the ion it produces in water.
HBr is an acid; produces Br⁻ and H₃O⁺ (or H⁺) in water.
State two factors that affect reaction rate and briefly explain how they change collision frequency or effectiveness.
Temperature and concentration (also catalysts, surface area, pressure for gases).
What is the molarity of 0.202 mol KCl in 7.98 L? (Show formula and final value.)
M = n / V = 0.202 mol / 7.98 L = 0.0253 M (answer: 0.0253 M).
What is the difference between a suspension and a colloid? Give one real-world example of each.
Suspension: particles visible and settle (e.g., muddy water). Colloid: intermediate particle size, do not settle (e.g., milk).
Given that solubility of KNO₃ at 50°C is about 80 g per 100 g H₂O, classify a solution that contains 60 g KNO₃ in 100 g H₂O at 50°C (unsaturated, saturated, or supersaturated) and explain why.
60 g < 80 g → unsaturated (there is capacity to dissolve more).
Complete and balance the ionization of sulfuric acid in water and name the ions formed.
H₂SO₄ + 2 H₂O → 2 H₃O⁺ + SO₄²⁻ (balanced; produces two hydronium per molecule).
Explain, using collision theory, why increasing temperature generally increases reaction rate.
Higher T → molecules move faster → more frequent and higher-energy collisions → greater fraction exceed activation energy.
How many grams of HCl are present in 0.70 L of a 0.33 M HCl solution? (Show steps.)
0.33 mol/L × 0.70 L = 0.231 mol; mass = 0.231 mol × 36.46 g/mol = 8.42 g (≈ 8.4 g).
Explain why dissolving sodium chloride in water is considered a physical change even though ionic bonds separate.
Dissolving NaCl separates ions (Na⁺, Cl⁻) by hydration; no new substances formed—physical change.
Describe three ways to increase the rate at which a solid solute dissolves in a liquid and explain the molecular rationale for each.
Increase T (more kinetic energy → more effective collisions), stir (brings fresh solvent to solute surface), grind (increases surface area).
Explain why strong acids are strong and weak acids are weak in terms of molecular dissociation and equilibrium.
Strong acids fully dissociate (equilibrium far right), weak acids partially dissociate (equilibrium lies left); explained by acid dissociation constant, Ka.
Define dynamic equilibrium for a reversible chemical reaction and state what is constant at equilibrium.
At dynamic equilibrium forward and reverse rates equal; concentrations remain constant.
A sample contains 0.044 g Hg in 200.0 g water. Calculate concentration in ppb. (Show work.)
(mass solute / mass solution) × 10⁹ = (0.044 g / 200 g) × 10⁹ = 220,000 ppb.
Describe how polarity determines whether two substances will form a solution. Use ethanol + water and oil + water as examples.
Polar solutes dissolve in polar solvents (ethanol + water mix via H-bonding). Oil is nonpolar so it does not form solution with polar water.
A student dissolves 55 g NH₃ in 100 g water at 20°C. If the solubility of NH₃ at 20°C is 30 g per 100 g H₂O, classify the sample and explain how it could have been prepared.
55 g > 30 g → supersaturated or unstable—likely prepared by dissolving at higher T then cooling without crystallization.
Describe what happens to blue litmus paper in an acidic solution and explain the chemical basis for the color change.
Blue litmus turns red in acid due to protonation of dye molecules changing molecular structure and light absorption.
For the reversible reaction A + B ⇌ C + D, explain qualitatively how adding more A shifts the equilibrium and why (Le Chatelier’s principle).
Adding A shifts equilibrium right to consume added A, producing more C and D until new equilibrium.
Calculate the freezing point depression for a 1.50 m solution of glucose in water. (Use appropriate constant and show steps.)
i·Kf·m; for glucose i = 1, Kf (water) = 1.86 °C·kg/mol, m = 1.50 → ΔTf = 1×1.86×1.50 = 2.79 °C; freezing point = 0.00 − 2.79 = −2.79 °C.
Explain, in molecular terms, what happens to solute and solvent particles when a solute is added until the solution is saturated.
At saturation, dissolving and crystallization rates equal; adding solute produces no net dissolution.
Explain concentration units: molarity, molality, % (m/m), ppm/ppb. For each give the formula and one scenario where that unit is more appropriate than molarity.
Molarity = mol solute / L solution. Molality = mol solute / kg solvent. % (m/m) = mass solute / mass solution ×100. ppm = mg solute / kg solution; ppb = µg solute / kg solution. Use molality for colligative properties (depends on moles per kg solvent), ppm/ppb for trace contaminants.
Write the neutralization reaction between Be(OH)₂ and HCl. Balance it and specify the salt produced.
Be(OH)₂ + 2 HCl → BeCl₂ + 2 H₂O (balanced). Salt: BeCl₂.
Describe one experimental method chemists can use to increase the rate of a desired reaction while minimizing undesirable side reactions; explain trade-offs.
Use a catalyst to lower activation energy (increases rate) while controlling temperature to limit side reactions; trade-off: catalysts may also lower selectivity or require separation.
Volume dilution: What volume (mL) of 12.00 M H₂SO₄ is needed to prepare 1.00 L of 0.500 M H₂SO₄? (Show calculation and answer.)
M1V1 = M2V2 → V1 = (M2V2)/M1 = (0.500×1000 mL)/12.00 = 41.67 mL (≈ 41.7 mL).