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2 March, 17:35

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

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  1. 2 March, 18:03
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    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
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