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
Now,
Fe³?
+ H?O
------> Fe(OH)²? +
H?
IC:
0.0950
0
0
C:
-
x
+
x
+ x
EC: 0.0950 – x
x
x
Ka = x² / (0.0950-x)
3.00 x 10-3 = x² / (0.0950-x)
(3.00 x 10-3) (0.0950-x) = x²
x² + (3.00 x 10?³) x - (2.85 x 10??) = 0
By solving the quadratic equation, we have
x = 0.0154 and x = - 0.0184
Discarding the negative value of x, we have
x = 0.0154
So, [H?] = x = 0.0154 M
pH = - log [H?]
pH = - log (0.0154)
pH = 1.81
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