Question

Calculate the [H3O+] of an aqueous solution that is 7.8 × 10-13 M in OH. Be sure your answer has the correct number of significant figures.

Note: Reference the Fundamental constants table for additional information.

Answer 1

To find the concentration of hydronium ions \([H_3O^+]\) in an aqueous solution given the concentration of hydroxide ions \([OH^-]\), we can use the relationship provided by the ion product of water (\(K_w\)) at 25°C, which is:

\[
K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}
\]

Given that the concentration of hydroxide ions \([OH^-]\) is \(7.8 \times 10^{-13} \, M\), we can rearrange the equation to solve for \([H_3O^+]\):

\[
[H_3O^+] = \frac{K_w}{[OH^-]}
\]

Now substituting the values:

\[
[H_3O^+] = \frac{1.0 \times 10^{-14}}{7.8 \times 10^{-13}}
\]

Calculating this gives:

\[
[H_3O^+] = \frac{1.0}{7.8} \times 10^{-14 + 13} = \frac{1.0}{7.8} \times 10^{-1}
\]

Now performing the division:

\[
\frac{1.0}{7.8} \approx 0.1282
\]

Thus:

\[
[H_3O^+] \approx 0.1282 \times 10^{-1} = 1.282 \times 10^{-2}
\]

Now, we need to express this number in proper significant figures. The concentration of \([OH^-] = 7.8 \times 10^{-13}\) has 2 significant figures. Therefore, our answer should also be expressed with 2 significant figures:

\[
[H_3O^+] \approx 1.3 \times 10^{-2} \, M
\]

Thus, the final answer is:

\[
\boxed{1.3 \times 10^{-2} \, M}
\]