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