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.

1 answer

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:

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