What is the ph of 0.0950 m fe(no3)3 (ka of fe3+ = 3.00x10-3)? express your answer to two decimal places?

Fe³⁺ = 0.0950 M
Fe³⁺ + H₂O → Fe(OH)²⁺ + H⁺
0.0950                 0             0
0.0950-x              x             x
Ka = x² / (0.0950-x)
3.00 x 10⁻³ = x² / (0.0950-x)
x² + (3.00 x 10⁻³) x - (2.85 x 10⁻⁴) = 0
By solving the value of x = 0.0154
[H⁺] = 0.0154 M
pH = - log [H⁺]
pH = - log (0.0154)
pH = 1.81