The pH of a 0.002 50 mol/L solution of benzoic acid is 3.65. Calculate the Ka for benzoic acid.
pH = 3.65 = -log(H^+).
(H^+) = 2.24E-4
Call benzoic acid (C6H5COOH) HBz.
...........HBz ==> H^+ + Bz^-
initial..2.24E-4...0.....0
change.....-x.......x.....x
equil....2.24E-4-x...x....x
Ka = (H^+)(Bz^-)/(HBz)
Substitute from the ICE chart above into the Ka expression and solve for Ka.
To calculate the Ka (acid dissociation constant) for benzoic acid, we can use the formula for pH in terms of concentration and Ka:
pH = -log[H+]
pH = -log(Ka) + log([HA])
Given:
[H+] = 10^(-pH) = 10^(-3.65) = 2.89 x 10^(-4) M
[HA] = 0.00250 M
Now, we can rearrange the equation to solve for Ka:
[H+] = Ka * [HA]
Ka = [H+] / [HA]
Ka = (2.89 x 10^(-4) M) / (0.00250 M)
Ka = 0.1156
Therefore, the Ka for benzoic acid is 0.1156.
To calculate the Ka (acid dissociation constant) for benzoic acid, we need to first understand the chemical equilibrium that occurs when benzoic acid (C6H5COOH) dissolves in water.
The chemical equation for the dissociation of benzoic acid in water is:
C6H5COOH (aq) ⇌ C6H5COO- (aq) + H+ (aq)
The Ka expression for benzoic acid is:
Ka = [C6H5COO-] [H+] / [C6H5COOH]
Given that the pH of the solution is 3.65, we can determine the concentration of H+ ions ([H+]) using the equation:
[H+] = 10^(-pH)
[H+] = 10^(-3.65)
[H+] = 2.84 x 10^(-4) mol/L
Therefore, the concentration of benzoic acid ([C6H5COOH]) is 0.00250 mol/L (as mentioned in the question).
Now, let's denote [C6H5COO-] as x (since it is a concentration that we need to determine), and rewrite the Ka expression substituting the known values:
Ka = x * (2.84 x 10^(-4)) / 0.00250
Rearranging the equation to solve for x:
x = (Ka * 0.00250) / (2.84 x 10^(-4))
x = 8.8 * Ka
Now, we can substitute the value of pH (3.65) into the equation for x using the concentration of H+ ions ([H+]):
x = (Ka * 0.00250) / (2.84 x 10^(-4)) = 2.84 x 10^(-4)
Simplifying the equation, we get:
Ka = (2.84 x 10^(-4)) / 0.00250
Ka ≈ 1.14 x 10^(-1)
Therefore, the Ka value for benzoic acid is approximately 1.14 x 10^(-1).