Determine the pH for a 0.100 mol/L solution of benzoic acid
Ka for HSO4 is 1.0 x 10^-2
Ka for HClO is 4.0 x 10^-8
To determine the pH of a 0.100 mol/L solution of benzoic acid, we need to determine the concentration of H+ ions in the solution.
Benzoic acid (C6H5COOH) is a weak acid that partially dissociates in water, meaning it does not completely ionize. The dissociation equation for benzoic acid is as follows:
C6H5COOH (aq) ⇌ C6H5COO- (aq) + H+ (aq)
The equilibrium constant for this dissociation is represented by Ka. To find the concentration of H+ ions, we must first calculate the concentration of C6H5COOH that has dissociated.
Let x be the concentration of C6H5COOH that has dissociated. Then, the concentration of C6H5COOH remaining in the solution is 0.100 - x, and the concentration of C6H5COO- and H+ ions is also x.
The dissociation constant (Ka) can be expressed as follows:
Ka = [C6H5COO-][H+]/[C6H5COOH]
We can substitute the concentrations into this equation:
1.0 x 10^-2 = (x)(x) / (0.100 - x)
As discussed before, the concentration of C6H5COOH remaining in the solution is 0.100 - x. Thus, we can rearrange the equation to solve for x:
1.0 x 10^-2 (0.100 - x) = x^2
0.001 - 0.01x + x^2 = x^2
0.001 - 0.01x = 0
0.01x = 0.001
x = 0.001 / 0.01
x = 0.1
Therefore, the concentration of H+ ions in the solution is 0.1 mol/L.
To determine the pH, we can use the formula:
pH = -log[H+]
pH = -log[0.1]
Using a calculator, we find:
pH ≈ 1.000
Thus, the pH for a 0.100 mol/L solution of benzoic acid is approximately 1.000.