Benzoic Acid (HC6H5OO, 122.0 g/mol) has a Ka = 6.3 x 10-5. What is the pH of a solution that has 3.050 g of benzoic acid in enough water to make a 0.250 L solution?
Let's call benzoic acid HB.
mols HB = grams/molar mass
M HB = mols/L. I calculate approximately 0.1M
............HB ==> H^+ + B^-
initial....0.1.....0.....0
change.......x....x.......x
equil.....0.1-x....x.......x
Ka = (H^+)(B^-)/(HB)
Subsitute into the Ka expression and solve for H^+, then convert to pH.
To find the pH of the solution, we need to determine the concentration of the benzoic acid (C6H5COOH) in the solution and use the equilibrium expression for its dissociation.
Step 1: Calculate the amount of benzoic acid in moles.
Given the molar mass of benzoic acid (C6H5COOH) is 122.0 g/mol, and the mass of benzoic acid in the solution is 3.050 g, we can use the following formula to calculate the number of moles:
moles = mass / molar mass
moles = 3.050 g / 122.0 g/mol = 0.02500 mol
Step 2: Calculate the concentration of benzoic acid in the solution.
Since the volume of the solution is given as 0.250 L, we can use the formula:
concentration (in mol/L) = moles / volume (in L)
concentration = 0.02500 mol / 0.250 L = 0.100 mol/L
Step 3: Write the balanced equation for the dissociation of benzoic acid in water and the expression for its equilibrium constant.
The balanced equation is:
C6H5COOH(aq) ⇌ C6H5COO-(aq) + H+(aq)
The equilibrium expression for Ka is:
Ka = [C6H5COO-][H+]/[C6H5COOH]
Step 4: Calculate the concentration of C6H5COO- ion.
Since benzoic acid is a weak acid, we can assume that at equilibrium, the concentration of C6H5COO- formed is equal to the concentration of H+ ions. Therefore, [C6H5COO-] = x, where x is the concentration of H+ ions in mol/L.
Step 5: Set up the equilibrium expression and solve for x.
Ka = x * x / (0.100 - x)
6.3 x 10^-5 = x^2 / (0.100 - x)
Since the value of x is expected to be small compared to 0.100, we can assume that (0.100 - x) is approximately equal to 0.100:
6.3 x 10^-5 = x^2 / 0.100
To solve this equation, we can rearrange it to:
x^2 = (6.3 x 10^-5) * 0.100
x^2 = 6.3 x 10^-6
x ≈ 2.51 x 10^-3
Step 6: Calculate the pH.
pH is calculated as the negative logarithm (base 10) of the concentration of H+ ions in mol/L:
pH = -log10(x)
pH = -log10(2.51 x 10^-3)
pH ≈ 2.60
Therefore, the pH of the solution is approximately 2.60.