Chem Acid Base Key

Review of Acids and Bases Directions: Use your two Acid-Base note handouts and previous practice problems as tools to complete the following questions. You will need a scientific calculator. 1. Identify each of the following as an acid (A), a base (B), neither (N), or either (E). [H+] > [OH-] A pH < 7 A KI N KOH B Feels slippery B H₂SO₄ A pH > 7 B CaBr₂ N [H+] = [OH-] N CH₄ N Ba(OH)₂ B HC₂H₃O₂ A Tastes bitter B [H+] < [OH-] B H₂O E 2. Write the balanced chemical equation for the dissociation of the following acids and bases. a. HNO₃(aq) → H⁺ + NO₃⁻ b. Ba(OH)₂(aq) → Ba²⁺ + 2OH⁻ c. H₃PO₄(aq) ⇌ 3H⁺ + PO₄³⁻ d. KOH(aq) → K⁺ + OH⁻ 3. Name each of the following compounds. HCl hydrochloric acid HF hydrofluoric acid HNO₃ nitric acid Sr(OH)₂ strontium hydroxide HBr hydrobromic acid H₂SO₄ sulfuric acid Al(OH)₃ aluminum hydroxide NaOH sodium hydroxide 4. Complete the following acid-base neutralization reactions. a. LiOH + HCl → H₂O + LiCl b. H₂SO₄ + Ba(OH)₂ → 2H₂O + BaSO₄ c. H₃PO₄ + 3KOH → 3H₂O + K₃PO₄ d. Ca(OH)₂ + 2HNO₃ → 2H₂O + Ca(NO₃)₂ 5. Circle the ionizable hydrogen atoms in the acids below. H₃C-COOH H-Cl 6. Identify the acid (A), base (B), conjugate acid (CA) and conjugate base (CB) in the following reactions. NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ B A CA CB HCN + H₂O ⇌ CN⁻ + H₃O⁺ A B CB CA pH = -log[H⁺] pOH = -log[OH⁻] pH + pOH = 14 [H⁺] = antilog(-pH) [OH⁻] = antilog(-pOH) 7. A solution has [H⁺] = 9.5 x 10⁻⁹ M. Find the [OH⁻] then circle if the solution is acidic, basic, or neutral. (9.5 x 10⁻⁹)(x) = 1 x 10⁻¹⁴ [OH⁻] = 1.1 x 10⁻⁵ M [OH⁻] > [H⁺] Is the solution: acidic basic neutral 8. A solution has [OH⁻] = 2.5 x 10⁻⁶ M. Find the [H⁺] then circle if the solution is acidic, basic, or neutral. (x)(2.5 x 10⁻⁶) = 1.0 x 10⁻¹⁴ [H⁺] = 4.0 x 10⁻⁹ M [OH⁻] > [H⁺] Is the solution: acidic basic neutral 9. A solution has [H⁺] = 3.7 x 10⁻¹⁰ M. Find the [OH⁻] then circle if the solution is acidic, basic, or neutral. (3.7 x 10⁻¹⁰)(x) = 1 x 10⁻¹⁴ [OH⁻] = 2.7 x 10⁻⁵ M [OH⁻] > [H⁺] Is the solution: acidic basic neutral 10. Calculate the [H+] if pH = 3.9. [H+] = antilog(-pH) = 10^(-3.9) ≈ 1.26 x 10^(-4) M 11. Calculate the [OH-] if pOH = 9. [OH-] = antilog(-pOH) = 10^(-9) = 1.0 x 10^(-9) M 12. Calculate the [H+] if pOH = 11.28. pH + pOH = 14 14 - 11.28 = pH = 2.72 [H+] = antilog(-2.72) = 10^(-2.72) ≈ 0.0019 M 13. Complete the following table: | [H⁺] | [OH⁻] | pH | pOH | |---|---|---|---| | 1.0 x 10⁻⁸ | 1.0 x 10⁻⁶ | 8 | 6 | | 1.0 x 10⁻³ | 1.0 x 10⁻¹¹ | 3 | 11 | | 4.57 x 10⁻¹² | 2.17 x 10⁻³ | 11.34 | 2.66 | | 4.3 x 10⁻⁴ | 2.34 x 10⁻¹¹ | 3.37 | 10.63 14. Find the pH of a 5.3 x 10⁻⁴ M HNO₃ solution. HNO₃ → H⁺ + NO₃⁻ [H⁺] = 5.3 x 10⁻⁴ M pH = -log[H⁺] = -log(5.3 x 10⁻⁴) ≈ 3.28 15. Find the pH of a 5.3 x 10⁻⁶ M HCl solution. HCl → H⁺ + Cl⁻ [H⁺] = 5.3 x 10⁻⁶ M pH = -log[H⁺] = -log(5.3 x 10⁻⁶) ≈ 5.28 16. Find the pH of a 5.3 x 10⁻⁶ M H₂SO₄ solution. H₂SO₄ → 2H⁺ + SO₄²⁻ [H⁺] = 1.06 x 10⁻⁵ M pH = -log[H⁺] = -log(1.06 x 10⁻⁵) ≈ 4.97 17. Find the pOH of a 2 x 10⁻³ M NaOH solution. NaOH → Na⁺ + OH⁻ [OH⁻] = 2 x 10⁻³ M pOH = -log[OH⁻] = -log(2 x 10⁻³) ≈ 2.70 Explain why many compounds that contain hydrogen atoms (like benzene) are not classified as acids. Acids must have ionizable 1 H+ ions (polar bonds) that can be donated to a base. Benzene contains nonpolar C-H bonds.   What does the value of Ka or Kb indicate about acids and bases? K = products/reactants Large Ka or Kb = Strong acid or base