How many moles of `NaOH` must be removed from 1 litre of aqueous solution to change its `pH` from 12 to 11

Options

A. `0.009`
B. `0.01`
C. `0.02`
D. `0.1`

Explanation

To solve the problem of how many moles of NaOH must be removed from 1 liter of aqueous solution to change its pH from 12 to 11, we can follow these steps:

### Step 1: Understand the relationship between pH and pOH
The pH and pOH of a solution are related by the equation:
\[ \text{pH} + \text{pOH} = 14 \]
When the pH is 12, we can calculate the pOH:
\[ \text{pOH} = 14 - 12 = 2 \]

### Step 2: Calculate the concentration of hydroxide ions (OH⁻) at pH 12
Since NaOH is a strong base, the concentration of hydroxide ions (OH⁻) is equal to the concentration of NaOH. The relationship between pOH and hydroxide ion concentration is given by:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Thus, we can find [OH⁻] when pOH = 2:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} \, \text{mol/L} \]

### Step 3: Calculate the concentration of hydroxide ions (OH⁻) at pH 11
Next, we calculate the pOH when the pH is changed to 11:
\[ \text{pOH} = 14 - 11 = 3 \]
Now, we find [OH⁻] when pOH = 3:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-3} \, \text{mol/L} \]

### Step 4: Determine the change in concentration of OH⁻
Now we can determine how much the concentration of OH⁻ must change:
- Initial concentration at pH 12: \( 10^{-2} \, \text{mol/L} \)
- Final concentration at pH 11: \( 10^{-3} \, \text{mol/L} \)

The change in concentration of OH⁻ is:
\[ \Delta [\text{OH}^-] = 10^{-2} - 10^{-3} = 10^{-2} - 0.1 \times 10^{-2} = 0.9 \times 10^{-2} \, \text{mol/L} \]

### Step 5: Calculate the number of moles to be removed
Since we are dealing with a 1-liter solution, the number of moles of NaOH to be removed is equal to the change in concentration:
\[ \text{Moles of NaOH to be removed} = 0.009 \, \text{mol} \]

### Final Answer
To change the pH from 12 to 11, **0.009 moles of NaOH must be removed** from the solution.

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