i) A weak acid is an acid that only partially dissociates (ionizes) in an aqueous solution, meaning it does…

Question

i) A weak acid is an acid that only partially dissociates (ionizes) in an aqueous solution, meaning it does not release all of its hydrogen ions.

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Here are the solutions to the questions: d) i) A weak acid is an acid that only partially dissociates (ionizes) in an aqueous solution, meaning it does not release all of its hydrogen ions. ii) To calculate the pH of a 0.25 mol dm^−3 solution of ethanoic acid (CH_3COOH) with K_a = 1.8 × 10⁻⁵ mol dm^−3: Step 1: Write the dissociation equilibrium for ethanoic acid. CH₃COOH(aq) CH₃COO^−(aq) + H^+(aq) Step 2: Set up an ICE (Initial, Change, Equilibrium) table for concentrations. Let x be the concentration of H^+ ions at equilibrium. Species — Initial (mol dm^−3) — Change (mol dm^−3) — Equilibrium (mol dm^−3)CH_3COOH — 0.25 — −x — 0.25 − x CH_3COO^− — 0 — +x — x H^+ — 0 — +x — x Step 3: Write the expression for the acid dissociation constant, K_a. K_a = [CH₃COO^−][H^+][CH₃COOH] Step 4: Substitute the equilibrium concentrations into the K_a expression. 1.8 × 10⁻⁵ = ((x)(x))/(0.25 − x) Step 5: Since K_a is small, assume x 0.25, so 0.25 − x ≈ 0.25. 1.8 × 10⁻⁵ = (x²)/(0.25) Step 6: Solve for x. x² = 1.8 × 10⁻⁵ × 0.25; x² = 4.5 × 10⁻⁶; x = √(4.5 × 10⁻⁶); x = 0.002121 mol dm⁻³ This value of x represents [H^+]. Step 7: Calculate the pH. pH = −log[H^+]; pH = −log(0.002121) pH ≈ 2.67 The pH of the solution is 2.67. e) i) A buffer solution is a solution that resists significant changes in pH when small amounts of an acid or a base are added to it. It typically consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. ii) When a small amount of HCl (a strong acid) is added to the buffer solution (a mixture of CH_3COOH and CH_3COONa), the added H^+ ions from HCl react with the conjugate base, ethanoate ions (CH_3COO^−), to form the weak acid, ethanoic acid (CH_3COOH). This consumes the added H^+ ions, preventing a large drop in pH. CH₃COO^−(aq) + H^+(aq) → CH₃COOH(aq)

Species	Initial (mol dm $^{-3}$ )	Change (mol dm $^{-3}$ )	Equilibrium (mol dm $^{-3}$ )
CH $_3$ COOH	0.25	$-x$	$0.25 - x$
CH $_3$ COO $^-$	0	$+x$	$x$
H $^+$	0	$+x$	$x$

I) A weak acid is an acid that only partially dissociates (ionizes) in an aqueous solution, meaning it does not release all of its hydrogen ions.

Answer

Step-by-step solution

Part d(i)

Part d(ii)· 7 steps2.67

Write the dissociation equilibrium for ethanoic acid.

Set up an ICE (Initial, Change, Equilibrium) table for concentrations.

Write the expression for the acid dissociation constant, $K_a$ .

Substitute the equilibrium concentrations into the $K_a$ expression.

Since $K_a$ is small, assume $x \ll 0.25$ , so $0.25 - x \approx 0.25$ .

Solve for $x$ .

Calculate the pH.

Part e

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