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What is the pH of a 0.13 M solution of hypoiodous acid?

What is the pH of a 0.13 M solution of hypoiodous acid?

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Answer #1

H+ + IO-

(0.13 - X) X X

Ka = 2.3 x 10-11

Ka = (H+) x (IO-) / (HIO)

Ka = X2 / (0.13 - X)

0.13Ka - XKa = X2

2.99x10-12 - 2.3x10-11X = X2

X2 + 2.3x10-11X - 2.99x10-12=0

X = 1.73x10-6

(H+) = 1.73x10-6

pH = -Lg(H+)

pH = -Lg(1.73x10-6)

pH = 5.76

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