Questions: An aqueous solution at 25°C has a H3O+ concentration of 7.6 x 10^-8 M. Calculate the OH- concentration. Be sure your answer has the correct number of significant digits.

An aqueous solution at 25°C has a H3O+ concentration of 7.6 x 10^-8 M. Calculate the OH- concentration. Be sure your answer has the correct number of significant digits.

Transcript text: An aqueous solution at $25^{\circ} \mathrm{C}$ has a $\mathrm{H}_{3} \mathrm{O}^{+}$ concentration of $7.6 \times 10^{-8} \mathrm{M}$. Calculate the $\mathrm{OH}^{-}$ concentration. Be sure your answer has the correct number of significant digits.

Solution

Solution Steps

Step 1: Understanding the Relationship Between $\mathrm{H}_3\mathrm{O}^+$ and $\mathrm{OH}^-$

In an aqueous solution at $25^{\circ} \mathrm{C}$, the product of the concentrations of $\mathrm{H}_3\mathrm{O}^+$ and $\mathrm{OH}^-$ ions is a constant, known as the ion product of water ($K_w$). At $25^{\circ} \mathrm{C}$, $K_w$ is $1.0 \times 10^{-14}$.

\[ K_w = [\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14} \]

Step 2: Solving for $[\mathrm{OH}^-]$

Given the $[\mathrm{H}_3\mathrm{O}^+]$ concentration is $7.6 \times 10^{-8} \, \mathrm{M}$, we can solve for $[\mathrm{OH}^-]$ using the equation:

\[ [\mathrm{OH}^-] = \frac{K_w}{[\mathrm{H}_3\mathrm{O}^+]} \]

Substitute the given values:

\[ [\mathrm{OH}^-] = \frac{1.0 \times 10^{-14}}{7.6 \times 10^{-8}} \]

Step 3: Calculating the $[\mathrm{OH}^-]$ Concentration

Perform the division to find $[\mathrm{OH}^-]$:

\[ [\mathrm{OH}^-] = \frac{1.0 \times 10^{-14}}{7.6 \times 10^{-8}} = 1.3158 \times 10^{-7} \, \mathrm{M} \]

Step 4: Rounding to the Correct Number of Significant Digits

The given $[\mathrm{H}_3\mathrm{O}^+]$ concentration has two significant digits, so the $[\mathrm{OH}^-]$ concentration should also be reported with two significant digits:

\[ [\mathrm{OH}^-] = 1.3 \times 10^{-7} \, \mathrm{M} \]