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An acid solution of $\mathrm{pH}=6$ is diluted 1000 times, the pH of the final solution becomes
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Verified Answer
The correct answer is:
$6.99$
$\because \mathrm{pH}=6$
$\therefore$
$\left[\mathrm{H}^{+}\right]=10^{-6} \mathrm{M}$
After dilution $\left[\mathrm{H}^{+}\right]=\frac{10^{-6}}{1000}=10^{-9} \mathrm{M}$
$\therefore\left[\mathrm{H}^{+}\right]$ from $\mathrm{H}_{2} \mathrm{O}$ cannot be neglected.
Total $\left[\mathrm{H}^{+}\right]=10^{-9}+10^{-7}$
$\begin{aligned} &=10^{-7}\left(10^{-2}+1\right) \\ &=10^{-7}(1.01) \\ \mathrm{pH} &=-\log \left(1.01 \times 10^{-7}\right) \\ &=7-0.0043 \\ &=6.9957 \end{aligned}$
$\therefore$
$\left[\mathrm{H}^{+}\right]=10^{-6} \mathrm{M}$
After dilution $\left[\mathrm{H}^{+}\right]=\frac{10^{-6}}{1000}=10^{-9} \mathrm{M}$
$\therefore\left[\mathrm{H}^{+}\right]$ from $\mathrm{H}_{2} \mathrm{O}$ cannot be neglected.
Total $\left[\mathrm{H}^{+}\right]=10^{-9}+10^{-7}$
$\begin{aligned} &=10^{-7}\left(10^{-2}+1\right) \\ &=10^{-7}(1.01) \\ \mathrm{pH} &=-\log \left(1.01 \times 10^{-7}\right) \\ &=7-0.0043 \\ &=6.9957 \end{aligned}$
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