Ethanamine, CH3CH2NH2, is a weak base. pKa(CH3CH2NH3+) = 10.6 (a) (i) Write an equation to show the reaction of ethanamine CH3CH2NH2 with water. CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OH- (ii) Explain, with reference to the equation and the species in solution, what is meant by the terms weak and base. In the equation, ethanamine (CH3CH2NH2) reacts with water to form ethanammonium ion (CH3CH2NH3+) and hydroxide ion (OH-). A weak base is a substance that partially dissociates in water, meaning it does not completely ionize. In this case, ethanamine only partially reacts with water to form the ethanammonium ion and hydroxide ion. (b) (i) Determine the pKb for ethanamine. pKb = 14 – pKa pKb = 14 – 10.6 pKb = 3.4 (ii) Calculate the pH of a 0.109 mol L-1 solution of ethanamine. To calculate the pH, we need to find the pOH first. pOH = -log10[OH-] pOH = -log10(0.109) pOH = 0.962 pH + pOH = 14 pH = 14 – pOH pH = 14 – 0.962 pH = 13.038
The Correct Answer and Explanation is:
Corrected Answer:
(a)
(i)
Equation:
CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃⁺ + OH⁻
(ii)
Ethanamine (CH₃CH₂NH₂) acts as a weak base, meaning it only partially reacts with water to produce its conjugate acid (ethanammonium ion, CH₃CH₂NH₃⁺) and hydroxide ions (OH⁻). A base is a substance that accepts protons (H⁺), and in this reaction, ethanamine accepts a proton from water. Because the reaction does not go to completion, and a significant amount of unreacted CH₃CH₂NH₂ remains in solution, ethanamine is classified as a weak base.
(b)
(i)
pKa(CH₃CH₂NH₃⁺) = 10.6
pKb = 14 − pKa = 14 − 10.6 = 3.4
(ii) Calculate the pH of a 0.109 mol L⁻¹ ethanamine solution
We start by calculating the concentration of OH⁻ using the base dissociation constant (Kb):
Step 1: Calculate Kb from pKb pKb=3.4⇒Kb=10−3.4≈3.98×10−4\text{pKb} = 3.4 \Rightarrow K_b = 10^{-3.4} \approx 3.98 \times 10^{-4}
Step 2: Set up ICE table for CH₃CH₂NH₂ ⇌ CH₃CH₂NH₃⁺ + OH⁻
Let initial [CH₃CH₂NH₂] = 0.109 mol L⁻¹
At equilibrium:
[OH⁻] = x
CH₃CH₂NH₂ ⇌ CH₃CH₂NH₃⁺ + OH⁻
Initial: 0.109 0 0
Change: -x +x +x
Equilibrium: 0.109−x x x Kb=x20.109−x≈x20.109K_b = \frac{x^2}{0.109 – x} \approx \frac{x^2}{0.109}
Assuming x is small, 3.98×10−4=x20.109⇒x2=(3.98×10−4)(0.109)⇒x2=4.34×10−5⇒x=4.34×10−5≈6.59×10−33.98 \times 10^{-4} = \frac{x^2}{0.109} \Rightarrow x^2 = (3.98 \times 10^{-4})(0.109) \Rightarrow x^2 = 4.34 \times 10^{-5} \Rightarrow x = \sqrt{4.34 \times 10^{-5}} \approx 6.59 \times 10^{-3}
So [OH⁻] ≈ 6.59 × 10⁻³ mol L⁻¹
Step 3: Calculate pOH and then pH pOH=−log(6.59×10−3)≈2.18⇒pH=14−2.18=11.82\text{pOH} = -\log(6.59 \times 10^{-3}) \approx 2.18 \Rightarrow \text{pH} = 14 – 2.18 = \boxed{11.82}
Summary
Ethanamine (CH₃CH₂NH₂) is a weak base that reacts with water by accepting a proton to form the ethanammonium ion (CH₃CH₂NH₃⁺) and hydroxide ion (OH⁻). The equilibrium lies significantly to the left, indicating only partial ionization, a hallmark of weak bases. The extent of this ionization is quantified by the base dissociation constant (Kb), which is derived from the pKa of its conjugate acid. Since pKa = 10.6, the corresponding pKb = 14 − 10.6 = 3.4, and Kb = 10⁻³.⁴ ≈ 3.98 × 10⁻⁴.
To determine the pH of a 0.109 mol L⁻¹ ethanamine solution, we use the Kb expression and solve for [OH⁻], assuming the change in concentration is small compared to the initial base concentration. The resulting hydroxide concentration is found to be approximately 6.59 × 10⁻³ mol L⁻¹. Using this, the pOH is calculated as −log(6.59 × 10⁻³) ≈ 2.18, leading to a final pH of 14 − 2.18 = 11.82.
The original approach incorrectly used the full base concentration (0.109 mol L⁻¹) as the [OH⁻] concentration, which is only valid for strong bases that fully dissociate. Since ethanamine is a weak base, this assumption leads to a significant overestimation of pH. The correct method involves using the equilibrium expression for weak base dissociation and solving for the hydroxide ion concentration using Kb.
