Answer:
C.

Step-by-step explanation:
pH =
![\displaystyle -log[H_3O^+]](https://img.qammunity.org/2020/formulas/biology/high-school/w0q8koskblz5jdibcuej7o6pyunnampjic.png)
![\displaystyle [H_3O^+] = 10^(-pH)](https://img.qammunity.org/2020/formulas/biology/high-school/f6jqsx59m417mkj7ueg0ird38q0ik2isg0.png)
If pH = 2, then
![\displaystyle [H_3O^+] = \displaystyle 10^(-2)](https://img.qammunity.org/2020/formulas/biology/high-school/g0m7vsbkc2qxotclz5n7muzrl5wutussjn.png)
If pH = 4, then
![\displaystyle [H_3O^+] = 10^(-4)](https://img.qammunity.org/2020/formulas/biology/high-school/y5mqa6hz1i8eszvxk91sumtky0xysn8amz.png)
In consideration of
the concentration of
is 100 times as massive at pH = 2 than at pH = 4, so the acid is 100 times as puissant at pH = 2 than at pH = 4.
* The pH scale is logarithmic to base 10, so a change in one instrument in pH equals a change in the concentration by a factor of 10
![[H_3O^+].](https://img.qammunity.org/2020/formulas/biology/high-school/hox80rl65awti8mqt6tqtniyln669p3r8u.png)
* Hydronium →
![\displaystyle [H_3O^+]](https://img.qammunity.org/2020/formulas/biology/high-school/9scyibe4ptvu1atgobnqrocqi9tv0aa6vs.png)
I am joyous to assist you anytime.