Phenol C6H5OH partially dissociates in water as shown
C6H5OH + H2O --> H3O+ + C6H5O-
If Ka is 1.6*10^-10 and the concentration of H3O and C6H5O are both 1.0*10^-5 M at equilibrium, what would be the concentration of phenol?
C6H5OH + H2O ==> H3O^+ + C6H5O^-
Ka = (H3O^+)(C6H5O^-)/(C6H5OH)
Plug in values for Ka, (H3O^+), (C6H5O^-), solve for (C6H5OH).
Technically, (C6H5OH) = x-(H3O^+) but (H3O^+) is small in comparison to x that this need not be considered.
Well, let's start by acknowledging that we're talking about a rather high-maintenance molecule here. Phenol always needs to be the center of attention, doesn't it?
Now, since we're dealing with a dissociation reaction, we can use the equation for the dissociation constant (Ka) to solve for the concentration of phenol (C6H5OH).
The equation for Ka is given as:
Ka = [H3O+][C6H5O-] / [C6H5OH]
We know that the concentration of H3O+ and C6H5O- is both 1.0*10^-5 M, and our goal is to find the concentration of C6H5OH. So, we can rearrange the equation:
[C6H5OH] = [H3O+][C6H5O-] / Ka
Substituting the given values, we get:
[C6H5OH] = (1.0*10^-5 M)(1.0*10^-5 M) / 1.6*10^-10
Now, let's do some math:
[C6H5OH] ≈ 6.25 * 10^-1 M
Ah, the concentration of phenol is approximately 0.625 M. That's quite a bold presence for a compound that claims to be partially dissociating. Phenol just can't resist showing off, can it?
Remember, my friend, always keep a safe distance from such attention-seeking molecules. Safety first!
To find the concentration of phenol (C6H5OH) at equilibrium, we can use the expression for the equilibrium constant (Ka) and the given concentrations of H3O+ and C6H5O-.
The equilibrium constant expression for the dissociation of phenol can be written as:
Ka = [H3O+][C6H5O-] / [C6H5OH]
We have the values for [H3O+] = 1.0*10^-5 M and [C6H5O-] = 1.0*10^-5 M.
Now, let's assume the concentration of phenol is 'x'. After partial dissociation, the concentration of H3O+ and C6H5O- is 'x' as well.
Plugging the given values into the equilibrium constant expression:
1.6*10^-10 = (1.0*10^-5)(1.0*10^-5) / x
Simplifying the equation:
1.6*10^-10 = 1.0*10^-10 / x
Cross multiplying:
1.6*10^-10 * x = 1.0*10^-10
Dividing both sides by 1.6*10^-10:
x = (1.0*10^-10) / (1.6*10^-10)
x = 0.625
Therefore, the concentration of phenol (C6H5OH) at equilibrium is approximately 0.625 M.
To find the concentration of phenol (C6H5OH) in the equilibrium mixture, we need to use the dissociation equation and the equilibrium constant (Ka) expression.
The given dissociation equation for phenol in water is:
C6H5OH + H2O ⇌ H3O+ + C6H5O-
The equilibrium constant expression for this reaction is:
Ka = [H3O+][C6H5O-] / [C6H5OH]
We are given that the concentrations of H3O+ and C6H5O- are both 1.0*10^-5 M at equilibrium, and we need to find the concentration of C6H5OH.
Let's use the given values to substitute into the equilibrium constant expression:
1.6*10^-10 = (1.0*10^-5)(1.0*10^-5) / [C6H5OH]
Rearrange the equation to solve for [C6H5OH]:
[C6H5OH] = (1.0*10^-5)(1.0*10^-5) / 1.6*10^-10
Now, we'll perform the calculations:
[C6H5OH] = 1.0*10^-10 / (1.6*10^-10)
[C6H5OH] = 0.625 M
Therefore, the concentration of phenol (C6H5OH) in the equilibrium mixture is 0.625 M.