What is the ph of a 0.286 m solution of hypochlorous acid (ka 2.9 x 10-8)?

4.0406

Explanation:

Given that:

Concentration = 0.286 M

Consider the ICE take for the dissociation of acetic acid as:

                                      HClO    ⇄     H⁺ +        ClO⁻

At t=0                            0.286                 -              -

At t =equilibrium        (0.286-x)                x           x            

The expression for dissociation constant of acetic acid is:

x is very small, so (0.286 - x) ≅ 0.286

Solving for x, we get:

x = 9.1071×10⁻⁵  M

pH = -log[H⁺] = -log(9.1071×10⁻⁵) = 4.0406

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