Calculate the pH and pOH of a solution of sodium hypoiodite NaOI, with a formal concentration of 0.05784 mol L-1. Note that the compound is

Question

Calculate the pH and pOH of a solution of sodium hypoiodite NaOI, with a formal concentration of 0.05784 mol L-1. Note that the compound is soluble, dissolves into sodium and hypoiodite ions, and these react with water. The p K a of hypoiodous acid at 298 K is 10.64. Calculate the pH and the pOH of the resulting solution. Use all the approximations you can, be sure to verify them.

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Maris 3 years 2021-08-31T05:44:37+00:00 1 Answers 20 views 0

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  1. Dulcie
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    2021-08-31T05:46:13+00:00
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    Answer:

    pH = 11.70

    pOH = 2.298

    Explanation:

    As we know

    NaOI + H2O  NaOH + HOI

    pH = 7 + ½ (pKa + log (HOI))

    Substituting the given values, we get –

    pH = 7 + ½ (10.64 + log (0. 0.05784))

    pH = 7 + ½ (10.64 -1.2377)

    pH = 11.70

    pOH = 14 – pH  

    pOH = 14 -11.70

    pOH = 2.298

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