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The pH of a 0.001 M solution of HCl is
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Verified Answer
The correct answer is:
3
\(\begin{aligned}
{\left[\mathrm{H}^{+}\right] } & =0.001 \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right] \\
& =-\log \left[10^{-3}\right] \\
& =3
\end{aligned}\)
{\left[\mathrm{H}^{+}\right] } & =0.001 \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right] \\
& =-\log \left[10^{-3}\right] \\
& =3
\end{aligned}\)
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