A) 3.7
B) 5.7
C) 10.3
D) 8.3
Correct Answer: C
Solution :
[c] \[8\times {{10}^{-13}}=[F{{e}^{3+}}]{{[O-H]}^{3}}\] |
\[[O{{H}^{-}}]=2\times {{10}^{-4}}\] |
\[pOH=4-\log 2=3.7\] |
\[pH=10.3\] |
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