To find the concentration of \([H_3O^+]\) in an aqueous solution, we can use the relationship between the concentrations of \([H_3O^+]\) and \([OH^-]\) given by the ion product of water (\(K_w\)) at 25°C:
\[ K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14} \]
Given that the concentration of hydroxide ions \([OH^-]\) is \(9.9 \times 10^{-13}\) M, we can rearrange the equation to solve for \([H_3O^+]\):
\[ [H_3O^+] = \frac{K_w}{[OH^-]} \]
Substituting the known values:
\[ [H_3O^+] = \frac{1.0 \times 10^{-14}}{9.9 \times 10^{-13}} \]
Now, performing the division:
\[ [H_3O^+] = \frac{1.0}{9.9} \times 10^{-14 + 13} = \frac{1.0}{9.9} \times 10^{-1} \]
Calculating \(1.0 / 9.9\):
\[ 1.0 / 9.9 \approx 0.10101 \]
So,
\[ [H_3O^+] \approx 0.10101 \times 10^{-1} \approx 1.0101 \times 10^{-2} \]
Now rounding according to significant figures:
The value \(9.9 \times 10^{-13}\) has 2 significant figures, so we round \(1.0101 \times 10^{-2}\) to 2 significant figures as well:
\[ [H_3O^+] \approx 1.0 \times 10^{-2} \]
Thus, the concentration of \([H_3O^+]\) is:
\[ \boxed{1.0 \times 10^{-2} \text{ M}} \]
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