The `pH` of a solution `7.00`. To this solution, sufficient base is added to increase the `pH` to `12.0`. The increase in `overset(Theta)OH` ion concentration is

Options

A. `5 times`
B. `100 times`
C. `10^(5)times`
D. `4 times`

Explanation

To solve the problem, we need to determine the increase in hydroxide ion concentration (OH⁻) when the pH of a solution is increased from 7.00 to 12.00.

### Step-by-Step Solution:

1. **Determine the initial concentration of H⁺ ions:**
- The pH of the initial solution is 7.00.
- We use the formula for pH:
\[ \text{pH} = -\log[H^+] \]
- Rearranging gives:
\[ [H^+] = 10^{-\text{pH}} = 10^{-7} \text{ M} \]

2. **Calculate the initial concentration of OH⁻ ions:**
- We know that the product of the concentrations of H⁺ and OH⁻ ions at 25°C is:
\[ [H^+][OH^-] = 10^{-14} \]
- Using the concentration of H⁺ we found:
\[ [OH^-] = \frac{10^{-14}}{[H^+]} = \frac{10^{-14}}{10^{-7}} = 10^{-7} \text{ M} \]

3. **Determine the new concentration of H⁺ ions after increasing the pH to 12.00:**
- The new pH is 12.00.
- Again using the pH formula:
\[ [H^+] = 10^{-12} \text{ M} \]

4. **Calculate the new concentration of OH⁻ ions:**
- Using the relationship between H⁺ and OH⁻:
\[ [OH^-] = \frac{10^{-14}}{[H^+]} = \frac{10^{-14}}{10^{-12}} = 10^{-2} \text{ M} \]

5. **Calculate the increase in OH⁻ ion concentration:**
- The increase in OH⁻ concentration is given by:
\[ \Delta[OH^-] = [OH^-]_{\text{final}} - [OH^-]_{\text{initial}} = 10^{-2} - 10^{-7} \]
- Since \(10^{-2}\) is much larger than \(10^{-7}\), we can approximate:
\[ \Delta[OH^-] \approx 10^{-2} \text{ M} \]

6. **Express the increase in terms of powers of ten:**
- To find the increase in concentration in terms of powers of ten:
\[ \Delta[OH^-] = 10^{-2} - 10^{-7} \approx 10^{-2} \text{ M} \]
- To express this as a ratio of the initial concentration:
\[ \text{Increase} = \frac{10^{-2}}{10^{-7}} = 10^{5} \]

### Final Answer:
The increase in OH⁻ ion concentration is \(10^{5} \text{ M}\).