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#
