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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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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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