Let H₁ be the concentration of a solution with a pH of 1, and H₂ be the concentration of a solution with a pH of 2, then:
![\begin{gathered} 1=-\log(H_1), \\ 2=-\log(H_2). \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/z9oojznt7gfk5o2sxkkldt0r4eho88d5by.png)
Then:
![\begin{gathered} H_1=10^(-1)=0.1, \\ H_2=10^(-2)=0.01. \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hpccjetbj24q61cuc68is6nwn99zzq3e5x.png)
Therefore:
![H_1=0.1H_2.](https://img.qammunity.org/2023/formulas/mathematics/college/zdvdmm57mgkh9t0vnjfcb9r1lz2bttphqk.png)
Therefore the H concentration in a solution with a pH of 2 is 0.1 times of a solution with a pH of 1.
Answer: Second option.