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The number of $\mathrm{H}^{+}$ions present in $1 \mathrm{~mL}$ of a solution whose $\mathrm{pH}$ is 13
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2172 Upvotes
Verified Answer
The correct answer is:
$6.022 \times 10^{7}$
$\mathrm{pH}=13$
$$
\begin{aligned}
{[\mathrm{H}]^{+} } &=10^{-\mathrm{pH}}=10^{-13} \mathrm{~mol} \mathrm{} \mathrm{I}^{-1} \\
\Rightarrow 1000 \mathrm{~mL} \text { solution contains } \\
&=10^{-13} \times 6.022 \times 10^{23} \mathrm{H}^{+} \text {ions } \\
&=6.022 \times 10^{10} \mathrm{H}^{+} \text {ions }
\end{aligned}
$$
$1 \mathrm{~mL}$ solution contains
$$
=\frac{6.022 \times 10^{10}}{1000}=6.022 \times 10^{7} \mathrm{H}^{+} \text {ions. }
$$
$$
\begin{aligned}
{[\mathrm{H}]^{+} } &=10^{-\mathrm{pH}}=10^{-13} \mathrm{~mol} \mathrm{} \mathrm{I}^{-1} \\
\Rightarrow 1000 \mathrm{~mL} \text { solution contains } \\
&=10^{-13} \times 6.022 \times 10^{23} \mathrm{H}^{+} \text {ions } \\
&=6.022 \times 10^{10} \mathrm{H}^{+} \text {ions }
\end{aligned}
$$
$1 \mathrm{~mL}$ solution contains
$$
=\frac{6.022 \times 10^{10}}{1000}=6.022 \times 10^{7} \mathrm{H}^{+} \text {ions. }
$$
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