Questions: Calculate the value of [OH-] from the given [H3O+] in each solution and label the solution as acidic, basic, or neutral. Note: The ion product of water = 1.0 x 10^-14. [ [H3O+]=1.0 x 10^-3 M [OH-]= square M ] This solution is square (Choose one).

Calculate the value of [OH-] from the given [H3O+] in each solution and label the solution as acidic, basic, or neutral.
Note: The ion product of water = 1.0 x 10^-14.

[
[H3O+]=1.0 x 10^-3 M
[OH-]= square M
]

This solution is square (Choose one).

Transcript text: Calculate the value of $\left[\mathrm{OH}^{-}\right]$from the given $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$in each solution and label the solution as acidic, basic, or neutral. Note: The ion product of water $=1.0 \times 10^{-14}$. \[ \begin{array}{l} {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}} \\ {\left[\mathrm{OH}^{-}\right]=\square \mathrm{M}} \end{array} \] This solution is $\square$ (Choose one) .

Solution

Solution Steps

Step 1: Use the Ion Product of Water

The ion product of water is given by the equation: \[ [\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14} \]

Step 2: Substitute the Given $[\mathrm{H}_3\mathrm{O}^+]$

Substitute the given $[\mathrm{H}_3\mathrm{O}^+]$ concentration into the equation: \[ (1.0 \times 10^{-3})[\mathrm{OH}^-] = 1.0 \times 10^{-14} \]

Step 3: Solve for $[\mathrm{OH}^-]$

Rearrange the equation to solve for $[\mathrm{OH}^-]$: \[ [\mathrm{OH}^-] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-3}} \]

Step 4: Calculate $[\mathrm{OH}^-]$

Perform the division to find $[\mathrm{OH}^-]$: \[ [\mathrm{OH}^-] = 1.0 \times 10^{-11} \, \mathrm{M} \]

Step 5: Determine the Nature of the Solution

Compare the concentrations of $[\mathrm{H}_3\mathrm{O}^+]$ and $[\mathrm{OH}^-]$:

Since $[\mathrm{H}_3\mathrm{O}^+] = 1.0 \times 10^{-3} \, \mathrm{M}$ is greater than $[\mathrm{OH}^-] = 1.0 \times 10^{-11} \, \mathrm{M}$, the solution is acidic.