Chemistry: acids and bases help

Calculate the pH of the solution formed when 80.5cm^3 of hydrogen iodide at108.0 kPa and 25degrees Celsius is dissolved in 5.600L.

Please help! thanks :)

First, we need to calculate the number of moles of HI present in 0.0805 L of the gas at 108.0 kPa and 25°C. We can use the universal ideal gas equation: PV = nRT or n = [PV] / [RT]
n = ?
P = [(108.0 kPa)/1][(1 atm)/(101.3 kPa) = 1.066140 atm. or 1.066 atm rounded to four significant figures
V = 0.0805 L
[Note: cm^3 are the same as mL, and 1 L = 1000 mL]
R = 0.0821 L∙atm/mol∙K
T = 25°C + 273.15 = 298.15 K or 298 rounded to no places to the right of the decimal point
n = [(P)(V)] / [(R)(T)]
n = [(1.066140 atm)(0.0805 L)] / [(0.0821 L∙atm/mol∙K)(298.15 K)]
n = 0.0035061 mol or 0.00361 mol rounded to three significant figures
Next, we need to calculate the molarity of HI solution.
molarity HI = (moles of HI) / (Liters of solution)
molarity HI = (0.0035061 mol) / (5.600 L)
molarity HI = 0.000626 M or 6.26 × 10^-4 M
Molar concentration of HI is also the molar concentration of H^+(aq), because HI is a strong (completely ionized) acid.
pH = – log [H^+]
pH = – log [6.26 × 10^-4]
pH = – [– 3.203425667]
pH = 3.203 rounded to three places to the right of the decimal point
Answer: The pH of the solution formed when 80.5cm^3 of hydrogen iodide at108.0 kPa and 25degrees Celsius is dissolved in 5.600L is about 3.203.