A small amount of sodium hydroxide is added to water. When the chemical reaction reaches equilibrium, the pH is 12. Which statement and conclusion about sodium hydroxide is correct?(1 point)

Question

Responses

The negative log of the concentration of OH– ions is 2. Sodium hydroxide is a strong base.
The negative log of the concentration of OH – ions is 2. Sodium hydroxide is a strong base.

The negative log of the concentration of OH– ions is 12. Sodium hydroxide is a strong acid.
The negative log of the concentration of OH – ions is 12. Sodium hydroxide is a strong acid.

The negative log of the concentration of OH– ions is 12. Sodium hydroxide is a strong base.
The negative log of the concentration of OH – ions is 12. Sodium hydroxide is a strong base.

The negative log of the concentration of OH– ions is 2. Sodium hydroxide is a strong acid.

Answer 1

To determine the accurate statements regarding sodium hydroxide (NaOH) and the pH of the solution:

1. The pH of the solution is given as 12.
2. The relationship between pH and pOH can be defined using the equation: \[ pH + pOH = 14 \] From this we can calculate pOH: \[ pOH = 14 - pH = 14 - 12 = 2 \]
3. The concentration of hydroxide ions (OH⁻) can be calculated using the pOH: \[ OH^- = 10^{-pOH} = 10^{-2} = 0.01 , \text{M} \]
4. Since sodium hydroxide is a strong base, it fully dissociates in water. Therefore, the concentration of NaOH is also 0.01 M.

Now let's analyze the given options:

• The negative log of the concentration of OH– ions is 2. Sodium hydroxide is a strong base. (This is correct)
• The negative log of the concentration of OH– ions is 12. Sodium hydroxide is a strong acid. (This is incorrect)
• The negative log of the concentration of OH– ions is 12. Sodium hydroxide is a strong base. (This is incorrect)
• The negative log of the concentration of OH– ions is 2. Sodium hydroxide is a strong acid. (This is incorrect)

The correct statement is: The negative log of the concentration of OH– ions is 2. Sodium hydroxide is a strong base.