To analyze the pH and the role of sodium hydroxide in the solution, we can start by determining the concentration of hydroxide ions (\(OH^-\)) from the given pH.
A pH of 12 indicates a basic solution. We can convert pH to pOH using the following relationship: \[ \text{pH} + \text{pOH} = 14 \] So, we can find the pOH: \[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]
Now, we use the pOH to find the concentration of hydroxide ions: \[ \text{pOH} = -\log[OH^-] \] This implies: \[ 2 = -\log[OH^-] \] Thus, taking the antilog, we have: \[ [OH^-] = 10^{-2} , \text{M} = 0.01 , \text{M} \]
Since sodium hydroxide (NaOH) is a strong base, it completely dissociates in water to produce hydroxide ions. Therefore, we can conclude that sodium hydroxide is indeed a strong base.
Now, looking at the statements provided:
- The negative log of the concentration of \(OH^-\) ions is 12. Sodium hydroxide is a strong base. (Incorrect: the pOH is 2, not 12)
- The negative log of the concentration of \(OH^-\) ions is 12. Sodium hydroxide is a strong acid. (Incorrect: same reason, plus NaOH is not an acid)
- The negative log of the concentration of \(OH^-\) ions is 2. Sodium hydroxide is a strong acid. (Incorrect: NaOH is a base)
- The negative log of the concentration of \(OH^-\) ions is 2. Sodium hydroxide is a strong base. (Correct)
Therefore, the correct statement is: The negative log of the concentration of \(OH^-\) ions is 2. Sodium hydroxide is a strong base.