What is [H+], [OH-], pH, and pOH for a 0.048mol/L solution of the weak benzoic acid, C6H5COOH?
To find the concentrations of [H+], [OH-], pH, and pOH for a solution of benzoic acid, we need to use the dissociation constant of the acid, also known as Ka.
First, let's write the dissociation equation for benzoic acid:
C6H5COOH ⇌ C6H5COO- + H+
The Ka expression for benzoic acid is:
Ka = [C6H5COO-][H+] / [C6H5COOH]
Since benzoic acid is a weak acid, we can assume that the concentration of [C6H5COO-] is approximately equal to the concentration of [H+]. So we have:
Ka = [H+]^2 / [C6H5COOH]
Rearranging the equation, we get:
[H+]^2 = Ka * [C6H5COOH]
[H+] = sqrt(Ka * [C6H5COOH])
Now, we can calculate the values:
Step 1: Calculate [H+]:
[H+] = sqrt(Ka * [C6H5COOH])
= sqrt(6.3 * 10^-5 * 0.048)
≈ 0.0035 mol/L
Step 2: Calculate [OH-] using the equation:
[OH-] = Kw / [H+]
= 1.0 * 10^-14 / 0.0035
≈ 2.9 * 10^-12 mol/L
Step 3: Calculate pH using the equation:
pH = -log[H+]
= -log(0.0035)
≈ 2.46
Step 4: Calculate pOH using the equation:
pOH = -log[OH-]
= -log(2.9 * 10^-12)
≈ 11.54
Therefore, for the given solution of benzoic acid (0.048 mol/L), the values are:
[H+] ≈ 0.0035 mol/L
[OH-] ≈ 2.9 * 10^-12 mol/L
pH ≈ 2.46
pOH ≈ 11.54
C6H5COOH = HB
........HB ==> H^+ + B^-
I.......0.048...0....0
C........-x.....x....x
E.....0.048-x...x.....x
Ka = ? = (H^+)(B^-)/(HB)
You should find Ka in your text or notes. Substitute and solve for x = (H^+) and convert that to pH, pOH, and (OH^-).