answer:To determine the effect of increasing the temperature on the ionization constant (Ka) of acetic acid, we need to consider the Van't Hoff equation, which relates the temperature dependence of the equilibrium constant (Ka) to the enthalpy change (ΔH) of the reaction: ln(K2/K1) = (ΔH/R) * (1/T1 - 1/T2) Where: K1 and K2 are the ionization constants at temperatures T1 and T2, respectively ΔH is the enthalpy change of the reaction R is the gas constant (8.314 J/mol·K) T1 and T2 are the initial and final temperatures in Kelvin, respectively First, we need to convert the given temperatures from Celsius to Kelvin: T1 = 25°C + 273.15 = 298.15 K T2 = 45°C + 273.15 = 318.15 K The enthalpy change (ΔH) for the ionization of acetic acid is approximately -55 kJ/mol. Since we need the value in J/mol for the Van't Hoff equation, we convert it: ΔH = -55,000 J/mol Now, we can plug the values into the Van't Hoff equation: ln(K2/K1) = (-55,000 J/mol) / (8.314 J/mol·K) * (1/298.15 K - 1/318.15 K) To solve for K2, we first need to find the value of ln(K2/K1): ln(K2/K1) ≈ 3.03 Now, we can solve for K2: K2 = K1 * e^(3.03) However, we need the initial ionization constant (K1) of acetic acid at 25°C. The Ka value for acetic acid at 25°C is approximately 1.8 x 10^(-5). K2 = (1.8 x 10^(-5)) * e^(3.03) K2 ≈ 1.5 x 10^(-4) So, the ionization constant (Ka) of acetic acid increases from 1.8 x 10^(-5) at 25°C to approximately 1.5 x 10^(-4) at 45°C. This increase in Ka indicates that the ionization of acetic acid is more favorable at higher temperatures.

answer:To find the percent ionization of the weak acid HA at 25°C, we can use the ionization constant (Ka) and the initial concentration of the acid. The ionization of the weak acid can be represented by the following equilibrium: HA ⇌ H+ + A- Ka = [H+][A-] / [HA] Since the initial concentration of HA is 0.1 M, we can assume that the change in concentration due to ionization is small compared to the initial concentration. Let x be the concentration of H+ and A- ions formed due to ionization. Then, the equilibrium concentrations can be represented as: [H+] = [A-] = x [HA] = 0.1 - x Now, we can substitute these values into the Ka expression: 2.5 x 10^-5 = x^2 / (0.1 - x) Since x is small compared to 0.1, we can approximate (0.1 - x) as 0.1: 2.5 x 10^-5 = x^2 / 0.1 Now, solve for x: x^2 = 2.5 x 10^-5 * 0.1 x^2 = 2.5 x 10^-6 x = sqrt(2.5 x 10^-6) x ≈ 1.58 x 10^-3 Now, we can calculate the percent ionization: Percent ionization = (x / 0.1) * 100 Percent ionization ≈ (1.58 x 10^-3 / 0.1) * 100 Percent ionization ≈ 1.58% At 25°C, the percent ionization of the weak acid HA is approximately 1.58%. When the temperature is increased to 45°C, the ionization constant Ka will also increase due to the increased temperature. However, without knowing the exact relationship between temperature and Ka or the new value of Ka at 45°C, we cannot determine the new percent ionization of the weak acid. In general, though, we can expect the percent ionization to increase as the temperature increases.

answer:To determine the effect of temperature on the ionization constant (Ka) of acetic acid, we can use the van 't Hoff equation: ln(Ka2/Ka1) = -∆H/R * (1/T2 - 1/T1) where Ka1 and Ka2 are the ionization constants at temperatures T1 and T2, respectively, ∆H is the enthalpy change for the ionization reaction, and R is the gas constant (8.314 J/mol·K). First, we need to convert the temperatures from Celsius to Kelvin: T1 = 25°C + 273.15 = 298.15 K T2 = 35°C + 273.15 = 308.15 K Next, we need to find the enthalpy change (∆H) for the ionization reaction of acetic acid. The literature value for the enthalpy change of acetic acid ionization is approximately -55.2 kJ/mol. Now we can plug the values into the van 't Hoff equation: ln(Ka2/(1.76 x 10^-5)) = -(-55,200 J/mol) / (8.314 J/mol·K) * (1/308.15 K - 1/298.15 K) Solving for Ka2: Ka2 = (1.76 x 10^-5) * exp[-(-55,200 J/mol) / (8.314 J/mol·K) * (1/308.15 K - 1/298.15 K)] Ka2 ≈ 2.08 x 10^-5 So, the ionization constant (Ka) of acetic acid at 35°C is approximately 2.08 x 10^-5. As the temperature increases, the ionization constant also increases, which means that acetic acid will ionize more at higher temperatures.

answer:To solve this problem, we will use the Van't Hoff equation, which relates the temperature dependence of the equilibrium constant (K) to the standard enthalpy change (ΔH) of the reaction: ln(K₂/K₁) = -ΔH/R * (1/T₂ - 1/T₁) Where: - K₁ and K₂ are the ionization constants at temperatures T₁ and T₂, respectively - ΔH is the standard enthalpy change of the reaction - R is the gas constant (8.314 J/mol·K) - T₁ and T₂ are the initial and final temperatures in Kelvin, respectively First, we need to convert the temperatures from Celsius to Kelvin: T₁ = 25°C + 273.15 = 298.15 K T₂ = 40°C + 273.15 = 313.15 K Now, we can plug the values into the Van't Hoff equation: ln(K₂/(1.8 x 10⁻⁵)) = -(-25,000 J/mol) / (8.314 J/mol·K) * (1/313.15 K - 1/298.15 K) Solve for K₂: ln(K₂/(1.8 x 10⁻⁵)) = 25,000 / 8.314 * (0.003196 - 0.003356) ln(K₂/(1.8 x 10⁻⁵)) ≈ 3.006 Now, we can solve for K₂: K₂ = (1.8 x 10⁻⁵) * e^(3.006) K₂ ≈ 1.49 x 10⁻² So, the new value of the ionization constant for the weak acid at 40°C is approximately 1.49 x 10⁻².