If HOCl has a Ka=3.2*10^-12, what is the pH of 0.1M KOCl?
........OCl^- + HOH --> HOCl + OH^-
initial..0.1............0......0
change...-x..............x.....x
equil...0.1-x...........x.......x
Kb = (Kw/Ka) = (HOCl)(OH^-)/(OCl^-)
Substitute equilibrium numbers from the ICE chart into the Kb expression and solve for OH^-, then convert to pH.
To find the pH of a solution of KOCl, we need to consider the dissociation of the compound and the acidity of the resulting species.
First, let's write the dissociation reaction for KOCl:
KOCl ⇌ K+ + OCl-
Since OCl- is derived from HOCl (hypochlorous acid), it can act as a weak acid and produce H+ ions through the following reaction:
OCl- + H2O ⇌ HOCl + OH-
Given that the Ka for HOCl is 3.2 × 10^-12, we can assume that the concentration of OH- in the solution is negligible compared to the concentration of HOCl.
Now, let x be the concentration of HOCl that is formed when OCl- reacts with water. This means that the concentration of OH- formed is also x.
So, at equilibrium, the concentrations of HOCl, OCl-, and H+ are all x, and the concentration of OH- is also x.
Since the solution is 0.1 M in KOCl, we can assume that the concentration of OCl- is 0.1 M (as it is the only source of OCl-).
Applying the equilibrium expression for the reaction of HOCl with water, we have:
Ka = [HOCl][OH-] / [OCl-]
Substituting the respective concentrations and given Ka value:
3.2 × 10^-12 = x^2 / 0.1
Rearranging the equation:
x^2 = 0.1 × 3.2 × 10^-12
Taking the square root of both sides:
x ≈ 1.79 × 10^-6
Since the concentration of HOCl (and also H+) is x, the concentration of H+ is approximately 1.79 × 10^-6 M.
The pH is defined as the negative logarithm (base 10) of the concentration of H+ ions.
Therefore, the pH of the solution is approximately -log10(1.79 × 10^-6) = 5.75.