The pH of a solution is 5. to this solution acid was added so that its pH value bcomes 2.0. The increase in `H^(+)` concentration is :

Options

A. 100 times
B. 5 times
C. 2.5 times
D. 1000 times

Answer

To solve the problem, we need to determine the increase in the concentration of hydrogen ions \((H^+)\) when the pH of a solution changes from 5 to 2. Here’s how we can approach this step by step:

### Step 1: Understand the relationship between pH and \((H^+)\) concentration
The pH of a solution is related to the concentration of hydrogen ions \((H^+)\) by the formula:
\[ \text{pH} = -\log[H^+] \]

### Step 2: Calculate the initial \((H^+)\) concentration
Given that the initial pH is 5, we can find the initial concentration of \((H^+)\):
\[ \text{pH} = 5 \implies [H^+] = 10^{-\text{pH}} = 10^{-5} \, \text{M} \]

### Step 3: Calculate the final \((H^+)\) concentration
After adding acid, the pH changes to 2. We can now calculate the final concentration of \((H^+)\):
\[ \text{pH} = 2 \implies [H^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M} \]

### Step 4: Determine the increase in \((H^+)\) concentration
To find the increase in \((H^+)\) concentration, we subtract the initial concentration from the final concentration:
\[ \text{Increase in } [H^+] = [H^+]_{\text{final}} - [H^+]_{\text{initial}} = 10^{-2} - 10^{-5} \]
However, since \(10^{-2}\) is much larger than \(10^{-5}\), we can consider the increase in terms of a ratio:
\[ \text{Increase factor} = \frac{[H^+]_{\text{final}}}{[H^+]_{\text{initial}}} = \frac{10^{-2}}{10^{-5}} = 10^{3} = 1000 \]

### Conclusion
The increase in \((H^+)\) concentration is 1000 times.