UNIT 3 - SCALARS/POWER SETS/EXAM QUESTIONS

Predict the pH when each salt below is dissolved in water KCN (basic) LiCl (acidic) (NH4)SCN (basic)

- If a salt comes from a strong base, and a weak acid the solution will be basic - If a salt comes from a strong acid and a weak base, the solution will be acidic - If both are strong/weak, the solution will be neutral KOH and HCN LiOH and HCl

What makes solutions ideal

- similar IMFs - relatively small concentrations

At 37 °C, Kw = 3.98 × 10-14. Calculate the pH of water at 37 °C.

1. [H3O+] = [OH-] 2. Kw = [H3O+] = [OH-] = sqrt (3.98 x 10^-14) = 1.99 x 10^-7 3. pH = -log[H+] -> pH = -log[1.99 x 10^-7] 4. pH = 6.7

Greater than or less than and why? Ka of HClO3 Ka of H2SO3 pH of HBrO2 pH of HIO2 % ionization of 0.1MHOI % ionization of 0.1M HIO2

> ? (not enough data, we don't have concentration) <

Comparing pH of Various 0.1 M Acid Solutions HA(aq) + H2O(l) = A-(aq) + H3O+ (aq) What is the correct order of Ka for the hydrolysis reaction of each acid: A) Ka (HNO3 ) > Ka (HF) > Ka (CH3COOH) > Ka (HCN) B) Ka (HNO3 ) < Ka (HF) < Ka (CH3COOH) < Ka (HCN) C) Not enough data is shown to answer this question D) I don't know

A) Ka (HNO3 ) > Ka (HF) > Ka (CH3COOH) > Ka (HCN)

For the following, predict the products and write balanced reactions: A. HF (aq) dissolved in water: B. Ca(OH)2 (aq) and HCl (aq) are mixed:

A. HF (aq) dissolved in water: HF(aq) + H2O(l) ⇌ H3O+(aq) + F-(aq) - No NIE B. Ca(OH)2 (aq) and HCl (aq) are mixed: Ca(OH)2 (aq) + HCl(aq) ⇌ 2H2O + CaCl2(aq) Ionic equation: Ca2+ + 2OH- + H+ + Cl- 2H+ -> 2H2O + Ca2+ + 2Cl- - NIE: 2OH- + 2H -> 2H2O

For each of the following reactions, fill in the missing compound, and label the acid, base, conjugate acid, and conjugate base: NH3 + HCl ⇌ HNO3 + H2O ⇌ OH- + H2 O ⇌ H2 O + H2 O ⇌

Cl- + NH4+ NO3- + H3O+ H2O + OH- OH- + H3O+

Morpholine (C4H9NO) pKb = 5.68 at 25 °C. What is its pH at 0.01 M?

Given that pKb = 5.68, we can calculate Kb as follows: Kb = 10^(-pKb) = 10^(-5.68) ≈ 1.71 * 10^(-6) Now, let's set up the rice table: Reaction C4H9NO + H2O ⇌ C4H9NH+ + OH- Initial. 0.01 0 0 0 Change -x -x x x Equilibrium 0.01-x - x x x Since Kb = [C4H9NH+][OH-] / [C4H9NO], we can write the following: Kb = x^2 / (0.01 - x) Now, we can assume that x is very small compared to 0.01 (since it's a weak base), so we can neglect it in the denominator: Kb ≈ x^2 / 0.01 Now, let's solve for x (the concentration of OH-): 1.71 * 10^(-6) ≈ x^2 / 0.01 x^2 ≈ 1.71 * 10^(-8) x ≈ √(1.71 * 10^(-8)) ≈ 1.31 * 10^(-4) Now that we have the concentration of OH-, we can find the pOH: pOH = -log[OH-] = -log(1.31 * 10^(-4)) ≈ 3.88 pH = 14 - pOH = 14 - 3.88 ≈ 10.12

Acid Hydrolysis and % Ionization For the hydrolysis of a weak acid, use Le Chatelier's principle to discuss how diluting the acid will affect the equilibrium. As a weak acid is diluted, the pH , and the % ionization . A) Increases; increases B) Increases; decreases C) Decreases; increases D) Decreases; decreases

In summary, when we dilute a weak acid by adding more water, the equilibrium of the hydrolysis reaction shifts to the left, making the solution less acidic and increasing the concentration of the weak acid and water. This process helps maintain a balance in the solution and is guided by Le Chatelier's principle. A) Increases; increases

Using Ka to calculate pH Determine the pH of 0.1 M HOCl (Ka = 3.5×10-8 at 25 °C). Rice Table

R: HOCl(aq) + H2O(l) = H3O+(aq) + OCl-(aq) - write out a balanced chemical reaction equation I: 0.1 M. N/A 0 0 - write out the initial concentration (before water is added) C: -x N/A +x +x E: (0.1-x) N/A +x +x Ka = products/reactants Ka = [H3O+][OCl-]/[HOCl] 3.5 x 10^-8 = (x)(x)/(0.1-x) - you can assume 0.1-x is 0.1 because it is so small - Therefore, 3.5 x 10^-8 = (x^2)/(0.1) x = 5.9 x 10^-5 pH = -log [H3O+ or H+] -log[5.9 x 10^-5] = 4.2

How does our model of an acid as an H+ donor account for each observation? - The addition of acid lowers the pH - Acid "neutralizes" base - Acid is corrosive - Acid solutions are conductive

The addition of acid lowers the pH - An acid can donate a hydrogen ion to water, creating H3O+, thus increasing the concentration of H3O+, and by definition of pH, this lowers the pH Acid "neutralizes" base - When an acid is added to a basic solution, the hydrogen ions are donated to the hydroxide ions creating the already fully ionized base Acid is corrosive - Hydrogen ions can extract electrons from another atom or molecule, thus forming H2 gas and oxidizing the other material. This oxidation can thus break down or corrode the other material. Acid solutions are conductive - Acid solutions contain hydrogen ions and conjugate base anions. These charged particles can serve to conduct electricity by moving through the solution.What is an acid?

The pH of a 0.1 M solution of NH3 in water is 11.12 at 25 °C. Write the Kb expression for NH3 and the chemical reaction it corresponds to. What is the value of Kb? (Make a Rice table) - If we know the pH, we can solve for pOH in basic conditions

Write the Kb expression for NH3 and the chemical reaction it corresponds to. If we have pH we can find concentrations, if we have concentrations we can get to pH Kb = [NH4+][OH-]/[NH3] pOH = 14 - pH (11.12) = 2.9 [OH-] = 10^(pOH) [OH-] = 10^(-2.9) x = 1.3 x 10^-3 M Kb = x^2/(0.1 - x [basically 0.1]) Kb = x^2/0.1 = 1.8 × 10^−5 R: NH3 + H2O(l) = NH4+ + OH- - write out a balanced chemical reaction equation I: 0.1 M. N/A 0 0 - write out the initial concentration (before water is added) C: -x N/A +x +x E: (0.1-x) N/A +x +x