What's the pH of a 0.1 M HIO solution knowing that its pKa = 10.4?

pKa = 10.4, and thus Ka = 10⁻¹⁰·⁴

                   HIO + H₂O(ℓ) ⇌ IO⁻(aq) + H₃O⁺(aq)  Ka = 10⁻¹⁰·⁴
Initial:       0.1 M                     0 M          0 M
Change:     -y M                    +y M       +y M
Eqm:     (0.1 - y) M                 y M          y M
               ≈ 0.1 M

At equilibrium :
Ka = [IO⁻] [H₃O⁺] / [HIO]
10⁻¹⁰·⁴ = y² / 0.1
y = √(0.1 ×10⁻¹⁰·⁴) = 2.00 × 10⁻⁶
pH = -log[H₃O⁺] = -log(2.00 × 10⁻⁶) = 5.7

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OR:

pH = (1/2)(pKa - log[HIO]ₒ) = (1/2)(10.4 - log(0.1)) = 5.7

with weak acids we can use this simple equation 

H+ = sqrt ( Ka * c) 
H+= sqrt 10 ^-10.4* 0.1) = 2.00E-06

pH = 5.7