# Question #d138d

Feb 4, 2015

You're dealing with a solution that contains hydrogen iodide ($\text{HI}$), a very strong acid, and hypoiodous acid ($\text{HOI}$), a very weak acid. In order to determine the pH of the solution, you must determine the concentration of ${H}^{+}$ in solution.

Hydrogen iodide dissociates completely in aqueous solution

$H {I}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + {I}_{\left(a q\right)}^{-}$

This means that 1 mole of $\text{HI}$ will produce 1 mole of protons in solution; therefore, the concentration of ${H}^{+}$ coming from $\text{HI}$ is

$\left[{H}^{+}\right] = \left[H I\right] = \text{0.15 M}$

Now for the weak acid. The acid dissociation constant for $\text{HOI}$ is ${K}_{a} = 2.0 \cdot {10}^{- 11}$. It is safe to assume that $\text{HOI}$ will not contribute to the overall concentration of ${H}^{+}$ in solution because of the very small value of its acid dissociation constant.

In order to determine whether or not $\text{HOI}$ contributes to the solution's ${H}^{+}$ concentration, you must use the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart)

...$H O {I}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + O {I}_{\left(a q\right)}^{-}$
I...0.15....................0..........0
C...(-x)....................(+x)......(+x)
E...(0.15-x)..............x..........x

${K}_{a} = \frac{x \cdot x}{0.15 - x} = {x}^{2} / \left(0.15 - x\right) = 2.0 \cdot {10}^{- 11}$

Solving for x will produce two values, a positive one and a negative one. SInce $x$ expresses concentration, which cannot be negative, the positive value is the solution. So, $\text{HOI}$ will produce

$\left[{H}^{+}\right] = \text{0.00000173 M}$

The assumption is confirmed, which means that the overall concentration of ${H}^{+}$ in solution will be

$\left[{H}^{+}\right] = \text{0.15 M}$

Therefore, the pH will be

$p {H}_{\text{sol}} = - \log \left(\left[{H}^{+}\right]\right) = - \log \left(0.15\right) = 0.834$