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what is the pH of 0.120 M Fe(NO_3)_3 (Ka of Fe^3+=3.00x10^-3)? Express your answer to two...

what is the pH of 0.120 M Fe(NO_3)_3 (Ka of Fe^3+=3.00x10^-3)? Express your answer to two decimal places.

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Expert Solution

[Fe(H2O)6]+3   ---------------> [Fe(H2O)5OH]2+   + H+

0.120                                              0                         0

0.120 - x                                          x                         x

Ka = x^2 / 0.120 - x

3.00 x 10^-3 = x^2 / 0.120 - x

x = 0.0175

[H+] = 0.0175 M

pH = -log [H+] = -log (0.0175)

pH = 1.76

queen_honey_blossom answered 2 years ago
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