The pH of z solution at `25^(0)C` is `2`. If its pH is to be changed to 4, conc. of `H^(+)` of the original has to be

Options

A. Doubled
B. Halved
C. Increased hundred times
D. Decreased hundred times

Answer

To solve the problem step by step, we will follow the process of calculating the concentration of hydrogen ions at two different pH levels and then determine the change in concentration.

### Step-by-Step Solution:

1. **Understanding pH and Hydrogen Ion Concentration**:
The pH of a solution is related to the concentration of hydrogen ions \([H^+]\) by the formula:
\[ \text{pH} = -\log[H^+] \]

2. **Calculating Initial Hydrogen Ion Concentration**:
Given that the initial pH is 2, we can find the initial concentration of hydrogen ions:
\[ \text{pH} = 2 \implies [H^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M} \]
So, the initial concentration \([H^+]_1 = 10^{-2} \, \text{M}\).

3. **Calculating Final Hydrogen Ion Concentration**:
Now, we need to find the concentration of hydrogen ions when the pH is changed to 4:
\[ \text{pH} = 4 \implies [H^+] = 10^{-\text{pH}} = 10^{-4} \, \text{M} \]
Thus, the final concentration \([H^+]_2 = 10^{-4} \, \text{M}\).

4. **Finding the Change in Concentration**:
To find out how the concentration has changed, we can compare the final concentration to the initial concentration:
\[ \text{Change in concentration} = \frac{[H^+]_2}{[H^+]_1} = \frac{10^{-4}}{10^{-2}} = 10^{-4 + 2} = 10^{-2} \]
This means the concentration of hydrogen ions has decreased by a factor of 100 (since \(10^{-2} = \frac{1}{100}\)).

5. **Conclusion**:
Therefore, the concentration of hydrogen ions of the original solution has decreased 100 times when the pH is changed from 2 to 4.

### Final Answer:
The concentration of \(H^+\) of the original solution has to decrease by a factor of 100.

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