Sore-throat medications sometimes contain the weak
acid phenol, HC6H5O. A 0.10 M solution of phenol has a pH of 5.43 at 25°C.
What is the acid-dissociation constant Ka for this acid at 25°C?
Hence, what is its degree of ionisation (percent ionisation)?
..................C6H5OH ==> C6H5O^- + H^+
I....................0.1......................0.............0
C....................-x........................x.............x
E....................0.1-x....................x.............x
pH = 5.43 = -log(H^+)
(H^+) = 3.72E-6 but you should confirm that.
Ka = (x)(x)/(0.1-x)
x = 3.72E-6. Substitute and solve for Ka.
% ionization = 100[(x)/0.1]
To find the acid-dissociation constant Ka for phenol at 25°C, we can use the pH of the solution.
First, let's convert the pH to [H⁺] concentration using the formula:
pH = -log[H⁺].
Given that the pH is 5.43, we can calculate the [H⁺] concentration:
[H⁺] = 10^(-pH).
[H⁺] = 10^(-5.43) = 2.74 x 10^(-6) M.
Since phenol is a weak acid, we can assume that the concentration of [H⁺] from the dissociation of phenol is equal to the concentration of phenol that ionizes.
Next, we need to calculate the concentration of phenol [C₆H₅OH]. Since it is a 0.10 M solution of phenol, the concentration is 0.10 M.
Using the equation for the ionization of phenol:
C₆H₅OH ↔ H⁺ + C₆H₅O⁻,
The initial concentration of phenol [C₆H₅OH] is equal to the concentration of undissociated phenol.
Using the ICE method, we can set up the equilibrium expression, where [H⁺]^1 is the concentration of the dissociated H⁺ ions and [C₆H₅O⁻]^1 is the concentration of the dissociated C₆H₅O⁻ ions:
Ka = [H⁺][C₆H₅O⁻] / [C₆H₅OH].
Since we found that [H⁺] concentration is 2.74 x 10^(-6) M and [C₆H₅OH] concentration is 0.10 M, we can substitute these values into the equation to find Ka:
Ka = (2.74 x 10^(-6))(2.74 x 10^(-6)) / (0.10).
Simplifying this expression:
Ka = 7.51 x 10^(-12).
The acid-dissociation constant Ka for phenol at 25°C is 7.51 x 10^(-12).
To calculate the percent ionization (degree of ionization), we can use the formula:
(percent ionization) = ([H⁺] / [C₆H₅OH]) x 100.
Substituting the values we already calculated:
(percent ionization) = (2.74 x 10^(-6) / 0.10) x 100,
(percent ionization) = 0.00274 x 100,
(percent ionization) = 0.274%.
Therefore, the degree of ionization (percent ionization) for phenol in this solution is 0.274%.
To find the acid-dissociation constant (Ka) and the degree of ionization (percent ionization) for phenol (HC6H5O), we can use the given pH of the solution.
Step 1: Determine the concentration of hydrogen ions (H+) in the solution.
Using the pH value, we can calculate the concentration of H+ ions using the equation:
pH = -log[H+]
Rearranging the equation gives:
[H+] = 10^(-pH)
[H+] = 10^(-5.43) = 2.8 x 10^(-6) M
Step 2: Find the initial concentration of the acid (phenol).
Since the solution is 0.10 M, the initial concentration of phenol ([C6H5OH]) is 0.10 M.
Step 3: Calculate the concentration of the dissociated acid ([C6H5O-]).
Since phenol is a weak acid, it partially dissociates in water.
Let x be the concentration of [C6H5O-].
Then, [H+] = [C6H5O-] = x (approximately)
Step 4: Write the chemical equation for the dissociation of phenol.
HC6H5O ⇌ H+ + C6H5O-
Step 5: Set up the expression for the acid-dissociation constant (Ka).
Ka = ([H+][C6H5O-]) / [HC6H5O]
Substituting the values:
Ka = (x)(x) / (0.10 - x)
Step 6: Calculate the value of Ka.
Since the solution has a small concentration of phenol, we can assume that x is negligible compared to 0.10.
Therefore, we can approximate 0.10 - x as approximately 0.10.
Ka ≈ (x)(x) / 0.10
Now, we know that [H+] is equal to x, which is 2.8 x 10^(-6) M.
Substitute this value into the equation:
Ka ≈ (2.8 x 10^(-6))(2.8 x 10^(-6)) / 0.10
Ka ≈ 7.84 x 10^(-12) / 0.10
Ka ≈ 7.84 x 10^(-11)
Therefore, the acid-dissociation constant (Ka) for phenol at 25°C is approximately 7.84 x 10^(-11).
To find the degree of ionization (percent ionization):
Degree of ionization = ([C6H5O-] / [HC6H5O]) * 100
Substituting the values:
Degree of ionization = (x / 0.10) * 100
Now, substitute the value of x:
Degree of ionization = (2.8 x 10^(-6) / 0.10) * 100
Degree of ionization = 2.8 x 10^(-5) * 100
Degree of ionization ≈ 0.0028%
Therefore, the degree of ionization (percent ionization) for phenol at 25°C is approximately 0.0028%.