How do you calculate the hydronium ion concentration given both the ph (8.5) and the temperature (37C). Thanks!?

Hydrogen ion concentration.....

pH = -log [H+] .... and ..... [H+] = 10^-pH
[H+] = 10^-8.5 = 3x10^-9M ...................... The hydrogen ion concentration can be expressed to only one significant digit. When computing a pH, only the digits to the right of the decimal reflect the precision in the original number. Therefore, a pH of 8.5 has only one significant digit.

As the temperature rises so does the value of Kw of a neutral solution. The value of Kw, like any equilibrium constant, varies with temperature. This means that the pH of a neutral decreases as T increases.
T .........pH ......... Kw
0......... 7.47....... 0.114 x 10-14
10....... 7.27....... 0.293 x 10-14
20....... 7.08....... 0.681 x 10-14
25....... 7.00....... 1.008 x 10-14
30....... 6.92....... 1.471 x 10-14
40....... 6.77....... 2.916 x 10-14
50....... 6.63....... 5.476 x 10-14
100..... 6.14....... 51.3 x 10-14

At 37C the value of Kw is about 2.5x10^-14, and a neutral solution has a hydrogen ion concentration of 1.6x10^-7M and a pH of 6.80.

But since you can have a hydrogen ion concentration of only 1 significant digit, then the slight variation in temperature will make essentially no difference.

The given temperature is dedundance.

At any temperature, pH = -log[H₃O⁺]
Hence, [H₃O⁺] = 10^(-pH) M = 10^(-8.5) M = 3.2 × 10⁻⁹ M (to 2 sig. fig.)