Hypochlorous Acid (HClO) in 0.145 M Solution: Calculating [H₃O⁺], pH, [ClO⁻], and [HClO]

To find the concentrations of [H₃O⁺], pH, [ClO⁻], and [HClO] in a 0.145 M solution of hypochlorous acid (HClO), we first need to use the pKa value provided, which is 7.54.

1. **Calculating [H₃O⁺]**:
The pKa value is related to the Ka (acid dissociation constant) using the formula:

pKa = -log(Ka)

Therefore,
Ka = 10^(-pKa) = 10^(-7.54) ≈ 2.88 × 10^-8.

2. **Setting up the equilibrium expression**:
For the dissociation of HClO:
HClO ⇌ H⁺ + ClO^–
Initial concentration of HClO = 0.145 M
If we let x be the change in concentration at equilibrium, we have:

   Initial:  [HClO] = 0.145, [H+] = 0, [ClO-] = 0
   Change:   [HClO] = -x, [H+] = +x, [ClO-] = +x
   Equilibrium: [HClO] = 0.145 - x, [H+] = x, [ClO-] = x
   

3. **Applying the equilibrium expression**:
According to the Ka expression:
Ka = [H⁺][ClO^–]/[HClO]
Substituting the equilibrium values gives:
2.88 × 10^-8 = (x)(x)/(0.145 – x)
Since Ka is small, we can assume that x is much smaller than 0.145, allowing us to write:
2.88 × 10^-8 ≈ x²/0.145

4. **Solving for x**:
Rearranging gives:
x² = 2.88 × 10^-8 × 0.145
x² = 4.176 × 10^-9
x = √(4.176 × 10^-9) ≈ 6.45 × 10^-5 M

5. **Finding pH**:
pH is calculated using the concentration of H⁺:
pH = -log([H⁺]) = -log(6.45 × 10^-5) ≈ 4.19

6. **Finding [ClO^–] and [HClO]**:
Since at equilibrium [ClO^–] = x ≈ 6.45 × 10^-5 M and [HClO] = 0.145 – x ≈ 0.145 M (since x is very small compared to 0.145).

In summary, for a 0.145 M solution of hypochlorous acid (HClO):
– [H₃O⁺] ≈ 6.45 × 10^-5 M
– pH ≈ 4.19
– [ClO⁻] ≈ 6.45 × 10^-5 M
– [HClO] ≈ 0.145 M