Hypochlorous acid, HOCl, is a weak acid with Ka = 3.0 × 10-8. Calculate the
pH of a 0.0125 M solution of HOCl.
....................HOCl ==> H^+ + OCl^-
I..................0.0125M.....0...........0
C................ - x...............x...........x
E................0.0125-x.......x...........x
Ka = 3.0 × 10-8 = (H^+)(OCl^-)/(HOCl)
Substitute the E line into the Ka expression and solve for x = (H^+)
Then pH = -log(H^+)
Post your work if you get stuck.
To calculate the pH of a solution of a weak acid like HOCl, we need to use the expression for the acid dissociation constant (Ka) and the equilibrium expression for the ionization of the acid.
The ionization reaction for HOCl is as follows:
HOCl ⇌ H+ + OCl-
According to the law of chemical equilibrium, the equilibrium constant (Keq) for this reaction is equal to the product of the concentrations of the products divided by the product of the concentrations of the reactants.
Keq = [H+][OCl-] / [HOCl]
Since we know Ka at equilibrium is equal to [H+][OCl-] / [HOCl], we can write:
Ka = [H+][OCl-] / [HOCl]
Given that Ka = 3.0 × 10^-8 and [HOCl] = 0.0125 M, we can solve for [H+] and [OCl-].
3.0 × 10^-8 = [H+][OCl-] / 0.0125
Now, suppose x is the concentration of [H+] or [OCl-] formed, we can substitute it in the equation as follows:
3.0 × 10^-8 = x * x / 0.0125
Rearranging the equation, we get:
x^2 = 0.0125 * 3.0 × 10^-8
Taking the square root of both sides, we find:
x = √ (0.0125 * 3.0 × 10^-8)
Calculating this value, we find x ≈ 1.37 × 10^-4 M.
The pH of a solution is equal to the negative logarithm (base 10) of the hydrogen ion concentration:
pH = -log[H+]
Therefore, the pH of the solution is:
pH = -log(1.37 × 10^-4)
Calculating this value, we find pH ≈ 3.86.
Therefore, the pH of a 0.0125 M solution of HOCl is approximately 3.86.
To calculate the pH of a solution of HOCl, we need to use the dissociation constant (Ka) to determine the concentration of H+ ions in the solution.
The dissociation of HOCl can be represented by the equation: HOCl ⇌ H+ + OCl-
The Ka value of 3.0 × 10-8 tells us that the equilibrium constant for the dissociation reaction is 3.0 × 10-8. This constant relates the concentrations of the products to the concentration of the reactant.
Let's assign the concentration of H+ ions as x, and the concentration of HOCl and OCl- ions as (0.0125 - x) each, assuming that x is a very small value compared to 0.0125.
Applying the equilibrium expression for the dissociation reaction:
Ka = [H+][OCl-]/[HOCl]
Substituting the known values:
3.0 × 10-8 = x*(0.0125 - x)/(0.0125)
Since x is small compared to 0.0125, we can neglect the -x term in the numerator:
3.0 × 10-8 = x*(0.0125)/(0.0125)
3.0 × 10-8 = x
Now, we can solve for x:
x = 3.0 × 10-8
Since x represents the concentration of H+ ions, we can use the definition of pH to calculate the pH of the solution:
pH = -log[H+]
pH = -log(3.0 × 10-8)
Using a logarithmic calculator or software, we find:
pH = 7.52
Therefore, the pH of a 0.0125 M solution of HOCl is approximately 7.52.