answer:To solve this problem, we can use the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]) where pH is the pH of the solution, pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the undissociated acid. In this case, we are given the pKa (4.2) and the initial concentration of benzoic acid ([HA] = 0.1 M). Since benzoic acid is a weak acid, only a small amount of it will dissociate into its conjugate base (benzoate ion) and a proton (H+). We can represent this dissociation as follows: C6H5COOH (benzoic acid) ↔ C6H5COO- (benzoate ion) + H+ (proton) Let x be the concentration of benzoic acid that dissociates. Then, the concentration of the benzoate ion [A-] will also be x, and the concentration of the undissociated benzoic acid [HA] will be (0.1 - x). Now we can plug these values into the Henderson-Hasselbalch equation: pH = 4.2 + log (x / (0.1 - x)) To solve for x, we can first simplify the equation by removing the logarithm. We can do this by taking the antilog (10^x) of both sides of the equation: 10^(pH - 4.2) = x / (0.1 - x) Since the pH is what we want to find, we can rearrange the equation to isolate the pH: pH = 4.2 + log (x / (0.1 - x)) Now, we can use the fact that the pKa is equal to the pH when the concentrations of the acid and its conjugate base are equal ([A-] = [HA]). In this case, x = 0.1 - x, so 2x = 0.1, and x = 0.05. Plugging this value back into the Henderson-Hasselbalch equation, we get: pH = 4.2 + log (0.05 / (0.1 - 0.05)) pH = 4.2 + log (0.05 / 0.05) pH = 4.2 + log (1) Since the log (1) = 0, the pH of the solution is: pH = 4.2 So, the pH of a 0.1 M solution of benzoic acid with a pKa of 4.2 at 25°C is 4.2.

answer:To find the pH of a 0.1 M solution of acetic acid, we can use the formula for the ionization constant (Ka) and the definition of pH. First, let's set up an equilibrium expression for the ionization of acetic acid: CH3COOH (aq) <=> CH3COO- (aq) + H+ (aq) At equilibrium, let x be the concentration of CH3COO- and H+ ions. Since the initial concentration of acetic acid is 0.1 M, the equilibrium concentration of acetic acid will be (0.1 - x) M. Now, we can write the expression for Ka: Ka = [CH3COO-][H+] / [CH3COOH] Given that Ka = 1.8 x 10^-5, we can plug in the equilibrium concentrations: 1.8 x 10^-5 = (x)(x) / (0.1 - x) Since Ka is very small, we can assume that x is also very small compared to 0.1, so we can simplify the equation: 1.8 x 10^-5 ≈ x^2 / 0.1 Now, we can solve for x: x^2 = 1.8 x 10^-5 * 0.1 x^2 = 1.8 x 10^-6 x = √(1.8 x 10^-6) x ≈ 1.34 x 10^-3 Now that we have the concentration of H+ ions (x), we can find the pH using the definition of pH: pH = -log[H+] pH = -log(1.34 x 10^-3) pH ≈ 2.87 So, the pH of a 0.1 M solution of acetic acid is approximately 2.87.

answer:To solve this problem, we need to first determine the amount of ethanoic acid and sodium hydroxide that will react with each other. Since sodium hydroxide is a strong base, it will react completely with the ethanoic acid, which is a weak acid. First, we need to find the limiting reactant. The reaction between ethanoic acid and sodium hydroxide can be represented as: CH₃COOH + NaOH → CH₃COONa + H₂O The initial moles of each reactant can be calculated as follows: moles of CH₃COOH = 0.1 M × volume (L) moles of NaOH = 0.2 M × volume (L) Since the stoichiometry of the reaction is 1:1, we can determine the limiting reactant by comparing the initial moles of each reactant. Let's assume the volume of the solution is 1 L for simplicity: moles of CH₃COOH = 0.1 mol moles of NaOH = 0.2 mol Since there are fewer moles of CH₃COOH, it is the limiting reactant. The moles of CH₃COOH and NaOH that will react are: moles of CH₃COOH reacted = 0.1 mol moles of NaOH reacted = 0.1 mol After the reaction, the moles of each species in the solution are: moles of CH₃COOH remaining = 0 mol moles of NaOH remaining = 0.2 mol - 0.1 mol = 0.1 mol moles of CH₃COONa formed = 0.1 mol Since CH₃COONa is the conjugate base of CH₃COOH, it will hydrolyze in water to produce OH⁻ ions. The hydrolysis reaction can be represented as: CH₃COONa + H₂O → CH₃COOH + OH⁻ Now we can use the pKa of ethanoic acid to find the equilibrium constant (Kb) for the hydrolysis reaction: pKa + pKb = 14 pKb = 14 - pKa = 14 - 4.76 = 9.24 Kb = 10^(-pKb) = 10^(-9.24) ≈ 5.75 × 10⁻¹⁰ Next, we can set up an ICE table for the hydrolysis reaction: Initial: [CH₃COO⁻] = 0.1 M [OH⁻] = 0 M [CH₃COOH] = 0 M Change: [CH₃COO⁻] = -x [OH⁻] = +x [CH₃COOH] = +x Equilibrium: [CH₃COO⁻] = 0.1 - x [OH⁻] = x [CH₃COOH] = x Now we can write the expression for Kb and solve for x: Kb = [OH⁻][CH₃COOH] / [CH₃COO⁻] 5.75 × 10⁻¹⁰ = (x)(x) / (0.1 - x) Assuming x is much smaller than 0.1, we can simplify the equation: 5.75 × 10⁻¹⁰ ≈ x² / 0.1 x² ≈ 5.75 × 10⁻¹¹ x ≈ √(5.75 × 10⁻¹¹) ≈ 7.58 × 10⁻⁶ Since x represents the concentration of OH⁻ ions at equilibrium, we can now calculate the pOH and pH of the solution: pOH = -log[OH⁻] = -log(7.58 × 10⁻⁶) ≈ 5.12 pH = 14 - pOH = 14 - 5.12 ≈ 8.88 Therefore, the pH of the solution that results from the reaction of 0.1 M ethanoic acid and 0.2 M sodium hydroxide at 25°C is approximately 8.88.

answer:The pKa value of acetic acid is 4.76. To calculate the pH of a 0.1 M solution of acetic acid, we can use the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]) In this case, [HA] is the concentration of acetic acid (0.1 M), and [A-] is the concentration of acetate ions. Since we don't know the concentration of acetate ions, we can use the ionization constant (Ka) of acetic acid to find it. The Ka of acetic acid is 10^(-pKa) = 10^(-4.76) ≈ 1.74 x 10^(-5). Let x be the concentration of acetate ions ([A-]) and acetic acid that ionizes. Then, the ionization equation is: Ka = (x)(x) / (0.1 - x) 1.74 x 10^(-5) = x^2 / (0.1 - x) Solving for x, we get x ≈ 1.33 x 10^(-3) M. This is the concentration of acetate ions and the amount of acetic acid that ionizes. Now we can use the Henderson-Hasselbalch equation: pH = 4.76 + log (1.33 x 10^(-3) / (0.1 - 1.33 x 10^(-3))) pH ≈ 2.87 The pH of the 0.1 M solution of acetic acid is approximately 2.87. To find the percent ionization of acetic acid in this solution, we can use the following formula: Percent ionization = (concentration of ionized acetic acid / initial concentration of acetic acid) x 100 Percent ionization = (1.33 x 10^(-3) / 0.1) x 100 ≈ 1.33% The percent ionization of acetic acid in this 0.1 M solution is approximately 1.33%.