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 a 1-liter aqueous solution to change its pH from 12 to 11, we can follow these steps:

### Step 1: Calculate the initial pOH
Given that the initial pH is 12, we can calculate the pOH using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
So,
\[ \text{pOH} = 14 - 12 = 2 \]

### Step 2: Calculate the initial concentration of OH⁻ ions
Using the pOH, we can find the concentration of hydroxide ions (OH⁻) in the solution:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Thus,
\[ 2 = -\log[\text{OH}^-] \]
This implies:
\[ [\text{OH}^-] = 10^{-2} \, \text{mol/L} \]

### Step 3: Calculate the desired pOH and OH⁻ concentration
Next, we need to find the pOH when the pH is changed to 11:
\[ \text{pOH} = 14 - 11 = 3 \]
Now, we can calculate the new concentration of OH⁻ ions:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Thus,
\[ 3 = -\log[\text{OH}^-] \]
This implies:
\[ [\text{OH}^-] = 10^{-3} \, \text{mol/L} \]

### Step 4: Calculate the change in OH⁻ concentration
To find out how many moles of OH⁻ need to be removed, we subtract the final concentration from the initial concentration:
\[ \Delta [\text{OH}^-] = [\text{OH}^-]_{\text{initial}} - [\text{OH}^-]_{\text{final}} \]
Substituting the values:
\[ \Delta [\text{OH}^-] = 10^{-2} - 10^{-3} \]
\[ = 10^{-2} (1 - 10^{-1}) \]
\[ = 10^{-2} \times \frac{9}{10} \]
\[ = 0.009 \, \text{mol/L} \]

### Step 5: Calculate the total moles to be removed
Since the volume of the solution is 1 liter, the number of moles of OH⁻ to be removed is:
\[ \text{Moles of OH}^- = 0.009 \, \text{mol/L} \times 1 \, \text{L} = 0.009 \, \text{mol} \]

### Conclusion
Thus, to change the pH from 12 to 11, **0.009 moles of NaOH must be removed from the solution**.

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