Question

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) Responses 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 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 2. Sodium hydroxide is a strong acid. The negative log of the concentration of OH – ions is 2. Sodium hydroxide is a strong acid. The negative log of the concentration of OH– ions is 12. Sodium hydroxide is a strong base.

Answer 1

To analyze the situation, let's first convert the given pH to the concentration of hydroxide ions \([OH^-]\).

1. The pH of the solution is 12.

2. The relationship between pH and pOH is given by: \[ pH + pOH = 14 \] Hence, we can find the pOH: \[ pOH = 14 - pH = 14 - 12 = 2 \]

3. Now, we can find the concentration of hydroxide ions \([OH^-]\) using the equation: \[ pOH = -\log[OH^-] \] So, if \(pOH = 2\): \[ 2 = -\log[OH^-] \]

   To find \([OH^-]\): \[ [OH^-] = 10^{-2} = 0.01 , \text{M} \]

Now, let's analyze the options provided in the question based on this information:

• The negative log of the concentration of \(OH^–\) ions being 2 indicates that the concentration of \(OH^–\) is \(0.01 , \text{M}\).
• Sodium hydroxide (NaOH) is a strong base, not a strong acid.

Now, let’s find the correct statement:

The only correct and complete statement from the options would be: "The negative log of the concentration of \(OH–\) ions is 2. Sodium hydroxide is a strong base."

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