The pH of a 0.100M solution of HIO is 5.82. Calculate the Ka hypoidioc acid.?

HIO --> H+ + IO- Ka = {products} / {reactants} = {H+} {IO-} / {HIO}

Ka = (x) (x) / 1x10^-1 -x ( Ignore as it is too) small. From the pH the H+ = 1.52 x10^-6 ( 1.52x10^-6)^2 / 1x10^-1

Ka = x2 = 2.25 x10^-12 / 1x10^-1

Ka= x2 = 2.25x10^-11

Ka= = 22.5x10^-12

Ka= x - 4.74 x10^-6