Sodium iodate is the salt of a strong base (NaOH) and a weak acid (HIO3). So this salt hydrolyzes as in the last problem I posted. The only difference is that this is a STRONG weak acid with Ka a relatively large value of about 0.18 or so. You need to confirm that. That means that you will not be able to neglect the y term; i.e.,
IO3^- + HOH ==> HIO3 + OH^-
Kb = Kw/Ka = (y)(y)/(0.067-y).
You must solve the quadratic because the concn of HIO3 and its Ka are so close together.
How do you calculate the pH, pOH, [H3O] and [OH] for a 0.067 M solution of sodium iodate
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