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A solution contains 0.09 M HCl, 0.09 M CHClCOOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10^-5, calculate Ka for CHCl2COOH.?
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A solution contains 0.09 M HCl, 0.09 M CHClCOOH and 0.1 M CH3COOH. The...
Given Information:
- Concentration of HCl in the solution: 0.09 M
- Concentration of CHClCOOH in the solution: 0.09 M
- Concentration of CH3COOH in the solution: 0.1 M
- pH of the solution: 1
- Ka for acetic acid (CH3COOH): 10^-5
Calculating the concentration of H3O+
The pH of a solution is a measure of the concentration of H3O+ ions. In this case, we are given that the pH is 1. The pH scale is logarithmic, so a pH of 1 corresponds to a concentration of H3O+ of 0.1 M.
Calculating the concentration of CH3COO-
Since acetic acid (CH3COOH) is a weak acid, it will partially dissociate in water to form its conjugate base, acetate ion (CH3COO-). We can assume that the concentration of CH3COOH is approximately equal to the concentration of CH3COO- because the dissociation is small.
Writing the equilibrium expression for acetic acid
The equilibrium expression for the dissociation of acetic acid is:
CH3COOH ⇌ CH3COO- + H+
The equilibrium constant (Ka) for this reaction is given as 10^-5.
Calculating the concentration of H+ and CH3COO-
Since we know the equilibrium constant (Ka) for the reaction and the concentration of CH3COOH, we can use the equation for Ka to calculate the concentrations of H+ and CH3COO-.
Ka = [CH3COO-][H+]/[CH3COOH]
Let x be the concentration of CH3COO- and H+ ions formed from the dissociation of CH3COOH.
Since the concentration of CH3COOH is 0.1 M and the dissociation is small, we can assume that the concentration of CH3COO- and H+ ions formed is also x.
Using the equilibrium constant expression, we can write the equation as:
10^-5 = x * x / (0.1 - x)
Calculating x
Simplifying the equation, we get:
10^-5 = x^2 / (0.1 - x)
Rearranging the equation:
x^2 = 10^-5 * (0.1 - x)
Expanding and rearranging further:
x^2 + 10^-5 * x - 10^-6 = 0
We can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 1, b = 10^-5, and c = -10^-6.
Solving the equation using the quadratic formula, we get two possible values for x.
Calculating Ka for CHCl2COOH
Now that we have the concentration of CH3COO-, we can use the equation for Ka to calculate the Ka for CHCl2COOH.
Ka = [CHCl2COO-][H+]/[CHCl2COOH]
Since the concentration of CH3COO- is
- Concentration of HCl in the solution: 0.09 M
- Concentration of CHClCOOH in the solution: 0.09 M
- Concentration of CH3COOH in the solution: 0.1 M
- pH of the solution: 1
- Ka for acetic acid (CH3COOH): 10^-5
Calculating the concentration of H3O+
The pH of a solution is a measure of the concentration of H3O+ ions. In this case, we are given that the pH is 1. The pH scale is logarithmic, so a pH of 1 corresponds to a concentration of H3O+ of 0.1 M.
Calculating the concentration of CH3COO-
Since acetic acid (CH3COOH) is a weak acid, it will partially dissociate in water to form its conjugate base, acetate ion (CH3COO-). We can assume that the concentration of CH3COOH is approximately equal to the concentration of CH3COO- because the dissociation is small.
Writing the equilibrium expression for acetic acid
The equilibrium expression for the dissociation of acetic acid is:
CH3COOH ⇌ CH3COO- + H+
The equilibrium constant (Ka) for this reaction is given as 10^-5.
Calculating the concentration of H+ and CH3COO-
Since we know the equilibrium constant (Ka) for the reaction and the concentration of CH3COOH, we can use the equation for Ka to calculate the concentrations of H+ and CH3COO-.
Ka = [CH3COO-][H+]/[CH3COOH]
Let x be the concentration of CH3COO- and H+ ions formed from the dissociation of CH3COOH.
Since the concentration of CH3COOH is 0.1 M and the dissociation is small, we can assume that the concentration of CH3COO- and H+ ions formed is also x.
Using the equilibrium constant expression, we can write the equation as:
10^-5 = x * x / (0.1 - x)
Calculating x
Simplifying the equation, we get:
10^-5 = x^2 / (0.1 - x)
Rearranging the equation:
x^2 = 10^-5 * (0.1 - x)
Expanding and rearranging further:
x^2 + 10^-5 * x - 10^-6 = 0
We can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 1, b = 10^-5, and c = -10^-6.
Solving the equation using the quadratic formula, we get two possible values for x.
Calculating Ka for CHCl2COOH
Now that we have the concentration of CH3COO-, we can use the equation for Ka to calculate the Ka for CHCl2COOH.
Ka = [CHCl2COO-][H+]/[CHCl2COOH]
Since the concentration of CH3COO- is
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A solution contains 0.09 M HCl, 0.09 M CHClCOOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10^-5, calculate Ka for CHCl2COOH.?
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A solution contains 0.09 M HCl, 0.09 M CHClCOOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10^-5, calculate Ka for CHCl2COOH.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A solution contains 0.09 M HCl, 0.09 M CHClCOOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10^-5, calculate Ka for CHCl2COOH.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution contains 0.09 M HCl, 0.09 M CHClCOOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10^-5, calculate Ka for CHCl2COOH.?.
A solution contains 0.09 M HCl, 0.09 M CHClCOOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10^-5, calculate Ka for CHCl2COOH.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A solution contains 0.09 M HCl, 0.09 M CHClCOOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10^-5, calculate Ka for CHCl2COOH.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution contains 0.09 M HCl, 0.09 M CHClCOOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10^-5, calculate Ka for CHCl2COOH.?.
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