The pH of a solution is 5.0 To this solution sufficient acid is added to decreases the pH to 2.0. The no. of times the concentration of `H^(+)` increased is

Answer

To solve the problem, we need to determine how many times the concentration of \( H^+ \) ions increases when the pH of a solution changes from 5.0 to 2.0.

### Step-by-Step Solution:

1. **Understanding pH and \( H^+ \) Concentration**:
The pH of a solution is defined as:
\[ \text{pH} = -\log[H^+] \]
From this, we can derive the concentration of \( H^+ \) ions:
\[ [H^+] = 10^{-\text{pH}} \]

2. **Calculate Initial \( H^+ \) Concentration**:
For the initial pH of 5.0:
\[ [H^+]_1 = 10^{-5} \text{ M} \]

3. **Calculate Final \( H^+ \) Concentration**:
For the final pH of 2.0:
\[ [H^+]_2 = 10^{-2} \text{ M} \]

4. **Determine the Increase in \( H^+ \) Concentration**:
To find out how many times the concentration of \( H^+ \) has increased, we can use the ratio of the final concentration to the initial concentration:
\[ \text{Increase Factor} = \frac{[H^+]_2}{[H^+]_1} \]

5. **Substituting the Values**:
\[ \text{Increase Factor} = \frac{10^{-2}}{10^{-5}} = 10^{-2 + 5} = 10^{3} \]

6. **Final Calculation**:
\[ 10^{3} = 1000 \]

Thus, the concentration of \( H^+ \) ions increased by a factor of **1000**.

### Summary:
The number of times the concentration of \( H^+ \) increased when the pH changed from 5.0 to 2.0 is **1000**.