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A solution with $\mathrm{pH}=2$ is more acidic than one with a $\mathrm{pH}=6$, by a factor of
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Verified Answer
The correct answer is:
10000
$$
\begin{aligned}
& \text { } \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
& \Rightarrow \mathrm{pH}=2 \Rightarrow\left[\mathrm{H}^{+}\right] \text {for } \mathrm{pH}=2 \text { is } 10^{-2} \\
& \mathrm{pH}=6 \Rightarrow\left[\mathrm{H}^{+}\right] \text {for } \mathrm{pH}=6 \text { is } 10^{-6} \\
& \Rightarrow \quad \frac{10^{-2}}{10^{-6}}=10^4=10,000 \\
&
\end{aligned}
$$
\begin{aligned}
& \text { } \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\
& \Rightarrow \mathrm{pH}=2 \Rightarrow\left[\mathrm{H}^{+}\right] \text {for } \mathrm{pH}=2 \text { is } 10^{-2} \\
& \mathrm{pH}=6 \Rightarrow\left[\mathrm{H}^{+}\right] \text {for } \mathrm{pH}=6 \text { is } 10^{-6} \\
& \Rightarrow \quad \frac{10^{-2}}{10^{-6}}=10^4=10,000 \\
&
\end{aligned}
$$
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