Write out the chemical reaction for the dissociation of acetic acid, CH3COOH, in water and expression of its equilibrium constant. b/ What is the pH of a 5M acetic acid solution given its Ka= 1.7510-5at 25oC. c/ Given that the Kaof acetic is 1.7010-5at 5oC.Calculate the Kaof the acetic acid at 95oC.
Write out the chemical reaction for the dissociation of acetic acid, CH3COOH, in water and expression of its equilibrium constant.
b/ What is the pH of a 5M acetic acid solution given its Ka= 1.75*10-5at 25oC.
c/ Given that the Kaof acetic is 1.70*10-5at 5oC.Calculate the Kaof the acetic acid at 95oC.
The Correct Answer and Explanation is :
Part a: Dissociation of Acetic Acid and Expression of Ka
The dissociation of acetic acid ((\text{CH}_3\text{COOH})) in water is represented as:
[
\text{CH}_3\text{COOH} \, + \, \text{H}_2\text{O} \, \rightleftharpoons \, \text{CH}_3\text{COO}^- \, + \, \text{H}_3\text{O}^+
]
The equilibrium constant for this reaction ((K_a)) is expressed as:
[
K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]}
]
Where:
- ([\text{CH}_3\text{COOH}]) = concentration of undissociated acetic acid
- ([\text{CH}_3\text{COO}^-]) = concentration of acetate ion
- ([\text{H}_3\text{O}^+]) = concentration of hydronium ion
Part b: pH of a 5 M Acetic Acid Solution at 25°C
Given:
[
K_a = 1.75 \times 10^{-5}, \, [\text{CH}_3\text{COOH}] = 5 \, \text{M}
]
Using the relationship:
[
K_a = \frac{x^2}{[CH_3COOH]_0 – x}
]
Where (x) represents ([\text{H}_3\text{O}^+]), and assuming (x \ll 5), simplify to:
[
K_a \approx \frac{x^2}{5}
]
Solve for (x):
[
x = \sqrt{K_a \cdot 5} = \sqrt{1.75 \times 10^{-5} \cdot 5} = \sqrt{8.75 \times 10^{-5}} \approx 9.35 \times 10^{-3} \, \text{M}
]
[
\text{pH} = -\log_{10} [\text{H}3\text{O}^+] = -\log{10} (9.35 \times 10^{-3}) \approx 2.03
]
Thus, the pH is approximately 2.03.
Part c: Calculate (K_a) at 95°C Using Van’t Hoff Equation
The Van’t Hoff equation relates (K_a) to temperature:
[
\ln\left(\frac{K_{a2}}{K_{a1}}\right) = \frac{\Delta H^\circ}{R} \left(\frac{1}{T_1} – \frac{1}{T_2}\right)
]
Assume (\Delta H^\circ \approx 30 \, \text{kJ/mol}) for acetic acid dissociation. Convert temperatures to Kelvin:
[
T_1 = 5^\circ \text{C} = 278 \, \text{K}, \, T_2 = 95^\circ \text{C} = 368 \, \text{K}
]
Plug in:
[
\ln\left(\frac{K_{a2}}{1.70 \times 10^{-5}}\right) = \frac{30,000}{8.314} \left(\frac{1}{278} – \frac{1}{368}\right)
]
Calculate the right-hand side:
[
\ln\left(\frac{K_{a2}}{1.70 \times 10^{-5}}\right) \approx 2.293
]
[
\frac{K_{a2}}{1.70 \times 10^{-5}} \approx e^{2.293} \approx 9.91
]
[
K_{a2} \approx 1.70 \times 10^{-5} \cdot 9.91 \approx 1.69 \times 10^{-4}
]
Thus, (K_a) at 95°C is approximately (1.69 \times 10^{-4}).
Explanation
Acetic acid dissociates partially, with its equilibrium constant ((K_a)) describing this balance. To calculate pH, we solve for ([\text{H}_3\text{O}^+]) using the simplified quadratic equation. The higher (K_a) at 95°C (compared to 5°C) reflects increased dissociation due to the endothermic nature of the process, consistent with Le Chatelier’s principle.