What would be the pH of a 0.1 M aqueous solution of the phenolate ion, C6H5O–? (Ka for phenol, C6H5OH, is 1.3 x 10–10) I got the answer 5.44 but I think it's wrong. Do I have to subtract it from 14?

My answer was slightly different for pOH than yours (5.11) but yes you need to subtract from 14 since I suspect you solved for OH^-.

Call phenolate ion P^-, then
.......P^- + HOH ==> HP + OH^-
I....0.1..............0....0
C.....-x..............x.....x
E....0.1-x...........x.......x

Kb for phenolate = (Kw/Ka for phenol) = (x)(x)/(0.1-x) and solve for x = (OH^-) then convert to pOH and pH.

I reworked the problem and realized I didn't find Kb at first and plugged in Ka instead. Thanks!

To find the pH of a solution of the phenolate ion (C6H5O–), we can use the equation for the pOH:

pOH = -log10[OH-]

Since the concentration of OH- is equivalent to the concentration of the phenolate ion, we can use the given concentration of 0.1 M.

C6H5O– + H2O ⇌ C6H5OH + OH–

We can assume that the initial concentration of OH- is 0 and that the reaction proceeds nearly to completion, so the concentration of OH- at equilibrium is 0.1 M.

Plugging this value into the pOH equation:

pOH = -log10(0.1) = 1

To find the pH, we subtract the pOH from 14:

pH = 14 - pOH = 14 - 1 = 13

Therefore, the pH of a 0.1 M aqueous solution of the phenolate ion (C6H5O–) is estimated to be 13.

To determine the pH of the solution of the phenolate ion, you need to calculate the concentration of OH- ions in the solution, and then convert it to the pH scale.

Here's how you can solve it step by step:

1. First, write the chemical equation for the dissociation of phenol (C6H5OH) into the phenolate ion (C6H5O-):

C6H5OH ⇌ C6H5O- + H+

2. Determine the concentration of the phenolate ion (C6H5O-) in the solution. In this case, it is given as 0.1 M.

3. Since phenolate (C6H5O-) is a base, it will react with water to produce OH- ions. The reaction can be written as:

C6H5O- + H2O ⇌ C6H5OH + OH-

4. Write the equilibrium expression for this reaction:

Kb = [C6H5OH][OH-] / [C6H5O-]

Since phenol (C6H5OH) is a more popular compound, the base dissociation constant (Kb) is usually not provided. However, it can be related to the acid dissociation constant (Ka) for the reaction of phenol with water:

Kb = Kw / Ka

Kw is the ion product of water, which is 1.0 x 10^-14 at 25°C.

Therefore, Kb = 1.0 x 10^-14 / 1.3 x 10^-10 = 7.69 x 10^-5

5. Use the equilibrium expression to determine the concentration of OH- ions in the solution.

Kb = [C6H5OH][OH-] / [C6H5O-]

Since [C6H5OH] is not given, we can assume that its concentration will be much lower in comparison to [OH-]. This happens because phenol is a weak acid and will not dissociate completely, whereas the reaction goes towards OH- production.

Therefore, we can approximate that [C6H5OH] ≈ 0 M.

[OH-] = (Kb * [C6H5O-]) / [C6H5OH]

[OH-] = (7.69 x 10^-5 * 0.1 M) / 0 M = Infinity

The concentration of OH- ions in the solution is infinite.

6. Since the concentration of OH- is extremely high, the resulting pH will be above 14, which is beyond the pH scale.

Hence, you do not subtract it from 14. The solution is more accurately described as being basic.

Therefore, the pH of a 0.1 M aqueous solution of the phenolate ion (C6H5O-) is not defined on the pH scale as it is a concentrated basic solution.