Calculate the relative concentrations of phenol (Ka = 1.05 x 10-10) and phenolate

ion in a solution with 0.05 M total phenol (pH = 7.5).

Use the Henderson-Hasselbalch equation.

pH = pKa + log (base/acid)
7.5 = pKa + log b/a
Convert Ka to pKa and solve the above equation for (base)/*acid)
Post your work if you get stuck.

The problem asks for relative concentrations and ratio of base/acid or acid/base will give that to you. I don't think you need the 0.05 M total. However, if you wish to calculate the concentration of each and use that for the ratio, take the base/acid ratio as one equation and solve for (acid) and (base) with the second equation of

(acid) + (base) = 0.05 M
After obtaining the ratio of base to acid

To calculate the relative concentrations of phenol and phenolate ion, we will use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where:
- pH is the given pH value (pH = 7.5)
- pKa is the negative logarithm of the acid dissociation constant (pKa = -log(1.05 x 10^-10))
- [A-] is the concentration of the phenolate ion
- [HA] is the concentration of phenol

First, let's calculate the concentration of phenolate ion:

pH = pKa + log([A-]/[HA])

Rearranging the equation, we have:

log([A-]/[HA]) = pH - pKa

Now, substitute the given values:

log([A-]/[HA]) = 7.5 - (-log(1.05 x 10^-10))

log([A-]/[HA]) = 7.5 + 10log(1.05)

log([A-]/[HA]) = 7.5 + (-9.9799)

log([A-]/[HA]) = -2.4799

Next, take the antilog of both sides to obtain the ratio [A-]/[HA]:

[A-]/[HA] = antilog(-2.4799)

[A-]/[HA] ≈ 0.0037

Finally, we can calculate the relative concentrations of phenol and phenolate ion:

Since the total concentration of phenol is 0.05 M, we can write the relation:

[HA] + [A-] = 0.05

Substituting the [A-]/[HA] ratio, we get:

[HA] + 0.0037[HA] ≈ 0.05

1.0037[HA] ≈ 0.05

[HA] ≈ 0.05 / 1.0037

[HA] ≈ 0.0498 M

[A-] ≈ 0.0037 * 0.0498

[A-] ≈ 0.00018426 M

Hence, the relative concentrations of phenol and phenolate ion in the solution are approximately 0.0498 M and 0.00018426 M, respectively.

To calculate the relative concentrations of phenol (C₆H₅OH) and phenolate ion (C₆H₅O⁻) in a solution with a given pH and total phenol concentration, we need to consider the acid-base equilibrium of phenol.

The dissociation of phenol can be represented by the following chemical equation:

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺

The equilibrium constant (Ka) for this reaction is given as 1.05 x 10⁻¹⁰. Ka is the ratio of the concentration of the products (C₆H₅O⁻ and H⁺) to the concentration of the reactant (C₆H₅OH).

At equilibrium, the concentrations of phenol (C₆H₅OH), phenolate ion (C₆H₅O⁻), and hydrogen ion (H⁺) can be expressed as follows:

[C₆H₅OH] = [C₆H₅OH]₀ - [C₆H₅O⁻] (1)
[C₆H₅O⁻] = [H⁺] (2)
[H⁺][C₆H₅O⁻] / [C₆H₅OH] = Ka

In this case, the total phenol concentration ([C₆H₅OH]₀) is given as 0.05 M, and the pH as 7.5. The pH is related to the hydrogen ion concentration ([H⁺]) by the equation:

pH = -log[H⁺]

To find the concentration of hydrogen ion ([H⁺]), we can use the equation:

[H⁺] = 10^(-pH)

Now, let's substitute the values into the equations:

[H⁺] = 10^(-7.5) (from the given pH)
[C₆H₅OH]₀ = 0.05 M (from the given total phenol concentration)
Ka = 1.05 x 10⁻¹⁰ (given equilibrium constant)

Using equation (2), the concentration of phenolate ion ([C₆H₅O⁻]) is equal to [H⁺]:

[C₆H₅O⁻] = [H⁺] = 10^(-7.5)

Using equation (1), the concentration of phenol ([C₆H₅OH]) is:

[C₆H₅OH] = [C₆H₅OH]₀ - [C₆H₅O⁻]
= 0.05 M - 10^(-7.5)

Therefore, the relative concentrations of phenol and phenolate ion in the solution can be calculated as follows:

Relative concentration of phenol = [C₆H₅OH] / [C₆H₅OH]₀
= (0.05 M - 10^(-7.5)) / 0.05 M

Relative concentration of phenolate ion = [C₆H₅O⁻] / [C₆H₅OH]₀
= 10^(-7.5) / 0.05 M

Evaluate these expressions to obtain the numerical values.