How to solve titration problem?

A sample of 40.0 mL of a .100 molar HIO solution is titrated with a .150 molar NaOH solution. Ka
for hypoiodous acid = 2.3 x ^ (-11)
What is the *pH of the solution*?


I have the following work:


[![work](https://i.stack.imgur.com/pN4Vv.png)](https://i.stack.imgur.com/pN4Vv.png)


As you can see I was given half credit. Can anyone explain why?




This is a case of salt hydrolysis of a salt of weak acid and strong base
$$\ce{HIO + NaOH -> NaOI + H2O}$$
$$\ce{NaOI -> IO^- + Na^+}$$
$$\ce{IO^- + H2O <=> HIO + OH^-}$$


You can calculate the final concentration of $NaOI$ to be $0.06M$ (denote this by $c$).  

Let $h$ be the hydrolysis constant.  

Thus in the equilibrium state, $[IO^-] = (c-ch)M$, $[HIO] = (ch)M$ and $[OH^-] = (ch)M$.  

Now $$ k\_h = \frac{[OH^-][HIO]}{[IO^-]} = \frac{(ch)(ch)}{(c-ch)} = \frac{ch^2}{(1-h)}$$
Here $k\_h$ is the hydrolysis constant. As the hydrolysis constant is small hence we can approximate $(1-h)\approx 1$.  

Therefore $$h = \sqrt{\frac{k\_h}{c}}$$
So $$[OH^-] = ch = \sqrt{ck\_h}$$
As $k\_h = \frac{k\_w}{k\_a}$
Thus $$[OH^-] = \sqrt{c\frac{k\_w}{k\_a}}$$
As $[H^+] = \frac{k\_w}{[OH^-]}$
So finally we can write $$[H^+] = \sqrt{\frac{k\_{w}k\_{a}}{c}}$$
Thus $$pH = -\frac{1}{2}log(\frac{k\_{w}k\_a}{c}) = \frac{1}{2}(-log(k\_w)-log(k\_a)+log(c)) = \frac{1}{2}(pk\_w + pk\_a + log(c))$$
Putting the values of $k\_a$, $k\_w$ and $c$, we have
$$ pH = \frac{1}{2}(14 + 10.64 - 1.22) = 11.71$$


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