pH of 0.048 M HClO is 4.35.
<u>Explanation:</u>
HClO is a weak acid and it is dissociated as,
HClO ⇄ H⁺ + ClO⁻
We can write the equilibrium expression as,
Ka = ![$\frac{[H^{+}] [ClO^{-}] }{[HClO]}](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%20%5BClO%5E%7B-%7D%5D%20%20%7D%7B%5BHClO%5D%7D)
Ka = 4.0 × 10⁻⁸ M
4.0 × 10⁻⁸ M = 
Now we can find x by rewriting the equation as,
x² = 4.0 × 10⁻⁸ × 0.048
= 1.92 × 10⁻⁹
Taking sqrt on both sides, we will get,
x = [H⁺] = 4.38 × 10⁻⁵
pH = -log₁₀[H⁺]
= - log₁₀[ 4.38 × 10⁻⁵]
= 4.35
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