Sodium benzoate (C6H5COONa) is a product of sodium hydroxide and benzoic acid (C6H5COOH) (ka = 6.3*10^-5).

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Sodium benzoate (C6H5COONa) is a product of sodium hydroxide and benzoic acid (C6H5COOH) (ka = 6.3*10^-5). What is the pH of 0.12 M Sodium benzoate? Please enter significant figures. (Sample problem 18.12 and follow-up problems) QUESTION 6 Determine whether the following salts are acidic, basic, or neutral: NH4Cl (Basic), Sr(NO3)2 (Neutral), KNO2 (Acidic) QUESTION 7 Determine whether solutions of the following salts are acidic, basic, or neutral at 25°C: Cu(CH3COO)2 (Acidic), NH4F (Neutral), KHC6H6O6 (Basic

The Correct Answer and Explanation is:

Question 6: pH of 0.12 M Sodium Benzoate

Given:

  • Sodium benzoate (C₆H₅COONa) is the salt of benzoic acid (C₆H₅COOH, a weak acid) and sodium hydroxide (NaOH, a strong base).
  • Ka of benzoic acid = 6.3 × 10⁻⁵
  • Molarity of sodium benzoate = 0.12 M

Step 1: Identify the type of salt
Sodium benzoate is a basic salt, because it is formed from the neutralization of a weak acid (benzoic acid) with a strong base (NaOH).

Step 2: Use the Ka of benzoic acid to calculate Kb of benzoate ion Kb=KwKa=1.0×10−146.3×10−5=1.59×10−10K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.3 \times 10^{-5}} = 1.59 \times 10^{-10}Kb​=Ka​Kw​​=6.3×10−51.0×10−14​=1.59×10−10

Step 3: Set up the ICE table for hydrolysis

Let the hydrolysis reaction of the benzoate ion (C₆H₅COO⁻) be: C6H5COO−+H2O⇌C6H5COOH+OH−C_6H_5COO^- + H_2O \rightleftharpoons C_6H_5COOH + OH^-C6​H5​COO−+H2​O⇌C6​H5​COOH+OH−

Initial concentration of C₆H₅COO⁻ = 0.12 M
Change in concentration = –x
Equilibrium [OH⁻] = x Kb=x20.12=1.59×10−10⇒x2=1.91×10−11⇒x=1.91×10−11=4.37×10−6K_b = \frac{x^2}{0.12} = 1.59 \times 10^{-10} \Rightarrow x^2 = 1.91 \times 10^{-11} \Rightarrow x = \sqrt{1.91 \times 10^{-11}} = 4.37 \times 10^{-6}Kb​=0.12×2​=1.59×10−10⇒x2=1.91×10−11⇒x=1.91×10−11​=4.37×10−6

Step 4: Calculate pOH and pH pOH=−log⁡(4.37×10−6)=5.36⇒pH=14−5.36=8.64\text{pOH} = -\log(4.37 \times 10^{-6}) = 5.36 \Rightarrow \text{pH} = 14 – 5.36 = 8.64pOH=−log(4.37×10−6)=5.36⇒pH=14−5.36=8.64

Answer (with correct significant figures):
pH = 8.64

Question 6: Classifying Salts as Acidic, Basic, or Neutral

  1. NH₄Cl: Acidic
    • NH₄⁺ is a weak acid (conjugate acid of weak base NH₃), Cl⁻ is neutral → Acidic
  2. Sr(NO₃)₂: Neutral
    • Sr²⁺ and NO₃⁻ come from strong base and strong acid → Neutral
  3. KNO₂: Basic
    • K⁺ is neutral, NO₂⁻ is the conjugate base of a weak acid (HNO₂) → Basic

Corrected Answers:

  • NH₄Cl: Acidic
  • Sr(NO₃)₂: Neutral
  • KNO₂: Basic

Question 7: Classifying Salts at 25°C

  1. Cu(CH₃COO)₂: Acidic
    • Cu²⁺ is a weak Lewis acid, CH₃COO⁻ is weakly basic → overall acidic
  2. NH₄F: Acidic
    • NH₄⁺ (acidic) and F⁻ (weakly basic), but NH₄⁺ dominates → Acidic
  3. KHC₆H₆O₆ (Potassium hydrogen tartrate): Basic
    • K⁺ is neutral, HC₆H₆O₆⁻ is a weak acid → salt of weak acid → Basic

Corrected Answers:

  • Cu(CH₃COO)₂: Acidic
  • NH₄F: Acidic
  • KHC₆H₆O₆: Basic

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