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A solution has a [tex]$\left[ H_3O^+ \right]$[/tex] of [tex]$1 \times 10^{-3} M$[/tex]. What is the [tex]$\left[ OH^- \right]$[/tex] of the solution?

A. [tex]$11 M$[/tex]
B. [tex]$14 M$[/tex]
C. [tex]$1 \times 10^{-14} M$[/tex]
D. [tex]$1 \times 10^{-11} M$[/tex]


Sagot :

To find the concentration of \([OH^-]\) in the solution given the hydronium ion concentration \([H_3O^+]\), we'll use the water dissociation constant \(K_w\). Let's go through the process step-by-step:

1. Identify the given data:
- The concentration of \([H_3O^+]\) is \(1 \times 10^{-3} \, M\).

2. Recall the relationship between \([H_3O^+]\) and \([OH^-]\):
- The water dissociation constant (\(K_w\)) at 25°C is given by:
[tex]\[ K_w = [H_3O^+][OH^-] = 1 \times 10^{-14} \][/tex]

3. Set up the equation:
- We need to find \([OH^-]\), so we can rearrange the formula to solve for \([OH^-]\):
[tex]\[ [OH^-] = \frac{K_w}{[H_3O^+]} \][/tex]

4. Substitute the known values into the equation:
- Plug in the values:
[tex]\[ [OH^-] = \frac{1 \times 10^{-14}}{1 \times 10^{-3}} \][/tex]

5. Perform the division:
- When we divide these exponential terms:
[tex]\[ [OH^-] = 1 \times 10^{-14} \div 1 \times 10^{-3} = 1 \times 10^{-11} \][/tex]

6. Conclusion:
- The concentration of \([OH^-]\) in the solution is \(1 \times 10^{-11} \, M\).

Therefore, the correct answer is:

[tex]\[ \boxed{1 \times 10^{-11} \, M} \][/tex]