A 0.20 M sodium chlorobenzoate (NaC7H4ClO2) solution has a pH of 8.65. Calculate the pH of a 0.19 M chlorobenzoic acid (HC7H4ClO2) solution.
If we call the compound HBz, then the salt is NaBz. The first part of the problem is a hydrolysis problem. The Bz part hydrolyzes according to the following equation.
Bz^- + HOH ==> HBz + OH^-
Kb = (Kw/Ka) = (HBz)(OH^-)/(HBz)
I would use the 8.65 pH, convert to pOH, and convert that to OH^- which will be equal to HBz. Substitute into the hydrolysis equation above and solve for the only unknown, Ka.
Then HBz, the acid. ==> H^+ + Bz^-
Ka = (H^+)(Bz^-)/(HBz)
Set up an ICE chart for the acid, and solve for (H^+), then convert to pH.
Post your work if you get stuck.
Well, well, well, it looks like we have some chemistry going on here! pH calculations, huh? Don't worry, I won't bombard you with chemical equations. Let's get right to it, shall we?
Basically, when you're dealing with acidic or basic solutions, the pH depends on the concentration of hydronium ions (H3O+). In this case, we're looking at a sodium chlorobenzoate solution with a pH of 8.65 and a chlorobenzoic acid solution that we need to find the pH for.
In the first solution, NaC7H4ClO2, the sodium ion (Na+) doesn't really affect the pH because it doesn't make any contributions to hydronium ions. So, we focus on the chloride ion (Cl-) and the benzoate ion (C7H4ClO2-) that affect the pH.
Now, when we consider the second solution, HC7H4ClO2, we have some chlorobenzoic acid (HC7H4ClO2) which is a weak acid. When chlorobenzoic acid dissolves in water, it donates a hydronium ion (H3O+) and forms the corresponding chlorobenzoate ion (C7H4ClO2-).
Since the concentration of the chlorobenzoate ion (C7H4ClO2-) is the same in both solutions (0.20 M in the first and 0.19 M in the second), we can assume that the concentration of hydronium ions (H3O+) in the second solution will also be the same.
Therefore, the pH of the 0.19 M chlorobenzoic acid solution should be around 8.65 as well. But remember, pH calculations can be a little tricky, so it's always good to double-check your work!
To calculate the pH of the 0.19 M chlorobenzoic acid (HC7H4ClO2) solution, we need to consider the acid-base equilibrium between chlorobenzoic acid and its conjugate base, chlorobenzoate ion.
The balanced chemical equation for this equilibrium is:
HC7H4ClO2 ⇌ H+ + C7H4ClO2-
The acid dissociation constant (Ka) can be used to quantify the extent of ionization of the acid. The expression for the Ka of chlorobenzoic acid is:
Ka = [H+][C7H4ClO2-] / [HC7H4ClO2]
From the given information, we know that the concentration of the sodium chlorobenzoate solution is 0.20 M, and its pH is 8.65. The pH of a solution can be related to the concentration of H+ ions using the formula:
pH = -log[H+]
Therefore, we can convert the pH of the sodium chlorobenzoate solution to [H+] concentration:
[H+] = 10^(-pH)
[H+] = 10^(-8.65)
Now, we can substitute the concentration of the sodium chlorobenzoate solution and the concentration of H+ ions into the Ka expression to calculate the concentration of C7H4ClO2-:
Ka = [H+][C7H4ClO2-] / [HC7H4ClO2]
Ka = (10^(-8.65))(0.20 M) / (0.20 M)
Next, we need to solve for the concentration of [C7H4ClO2-]:
[C7H4ClO2-] = (Ka)([HC7H4ClO2]) / [H+]
[C7H4ClO2-] = (Ka)(0.19 M) / (10^(-8.65))
Finally, we can calculate the pH of the 0.19 M chlorobenzoic acid solution using the concentration of [C7H4ClO2-]:
pH = -log[H+]
Note: pH is calculated as the negative logarithm of the [H+] concentration.
Please provide the value of Ka for chlorobenzoic acid.
To calculate the pH of the chlorobenzoic acid solution, we need to consider the acid dissociation constant of chlorobenzoic acid (Ka) and the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where:
- pH is the measure of acidity or basicity of a solution.
- pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka).
- [A-] is the concentration of the conjugate base.
- [HA] is the concentration of the acid.
In this case, we are provided with a sodium salt (NaC7H4ClO2) solution containing the conjugate base (C7H4ClO2-) with a pH of 8.65 and a concentration of 0.20 M.
The pH of the solution can be related to the pKa and concentration of the conjugate base using the Henderson-Hasselbalch equation.
Rearranging the equation, we have:
pH - pKa = log ([A-]/[HA])
Since the sodium salt solution contains only the conjugate base, the concentration of the conjugate base [A-] equals 0.20 M.
Now, we can substitute the known values into the equation:
8.65 - pKa = log (0.20/[HA])
To find the pKa value, we need to utilize the reverse logarithmic function and solve for pKa:
pKa = 8.65 - log (0.20/[HA])
Next, we need to calculate the concentration of the chlorobenzoic acid ([HA]) solution, which is 0.19 M. Substituting this value into the equation, we can calculate the pH of the 0.19 M chlorobenzoic acid solution:
pH = pKa + log (0.20/0.19)