Answer :
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]
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]
