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What is [H+] in a solution with a pH = 1.82?

1 Answer

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\\ \tt\Rrightarrow pH=-log[H^+]


\\ \tt\Rrightarrow 1.82=-log[H^+]


\\ \tt\Rrightarrow log[H^+]=-1.82


\\ \tt\Rrightarrow [H^+]=10^(-1.82)


\\ \tt\Rrightarrow [H^+]=0.015M

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User Aaron Schumacher
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