Answer:
(a) \[[{{H}^{+}}]=Anitlog\,[-pH]\]
\[=\text{Antilog}\,[-6.83]\]
\[=1.479\times {{10}^{-7}}M\]
(b) \[[{{H}^{+}}]\]= Antilog [-1.2]
= 0.063 M
(c) \[[{{H}^{+}}]=Antilog[-7.38]\]
\[=4.168\times {{10}^{-8}}M\]
(d) \[[{{H}^{+}}]=Antilog[-6.4]\]
\[=3.98\times
{{10}^{-7}}M\]
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