Answer
The $[H_3O^+]$ are, respectively: 0.0032, 0.0031, 0.0030.
As we can see by this results, when we change the number in the second decimal place of the pH value, the concentration only changes the number in the second significant figure. This will happen due to the properties of the log 10 function. Therefore, it is important to use the significant figures of pH in this way, so we do not round the value too much, and do not lose information in the process.
Work Step by Step
We find:
$$[H_3O^+] = 10^{-pH} = 10^{-2.50} = 3.2 \times 10^{-3} \space M$$
$$[H_3O^+] = 10^{-pH} = 10^{-2.51} = 3.1 \times 10^{-3} \space M$$
$$[H_3O^+] = 10^{-pH} = 10^{-2.52} = 3.0 \times 10^{-3} \space M$$
