You're dealing with a solution that contains hydrogen iodide (#"HI"#), a very strong acid, and hypoiodous acid (#"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

#HI_((aq)) rightleftharpoons H_((aq))^(+) + I_((aq))^(-)#

This means that 1 mole of #"HI"# will produce 1 mole of protons in solution; therefore, the concentration of #H^(+)# coming from #"HI"# is

#[H^(+)] = [HI] = "0.15 M"#

Now for the weak acid. The acid dissociation constant for #"HOI"# is #K_a = 2.0 * 10^(-11)#. It is safe to assume that #"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 #"HOI"# contributes to the solution's #H^(+)# concentration, you must use the **ICE table method** (more here: http://en.wikipedia.org/wiki/RICE_chart)

...#HOI_((aq)) rightleftharpoons H_((aq))^(+) + OI_((aq))^(-)#

**I**...0.15....................0..........0

**C**...(-x)....................(+x)......(+x)

**E**...(0.15-x)..............x..........x

#K_a = (x * x)/(0.15 - x) = x^2/(0.15-x) = 2.0 * 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, #"HOI"# will produce

#[H^(+)] = "0.00000173 M"#

The assumption is confirmed, which means that the overall concentration of #H^(+)# in solution will be

#[H^(+)] = "0.15 M"#

Therefore, the pH will be

#pH_("sol") = -log([H^(+)]) = -log(0.15) = 0.834#