1. The [H3O+] in a cabernet sauvignon wine is 5.9 × 10–4 M. What is the [–OH] in this wine?
To find the [OH-] in the cabernet sauvignon wine, we can use the equation for the ion product of water:
Kw = [H3O+][OH-]
Given that Kw = 1.0 x 10^-14, and [H3O+] = 5.9 x 10^-4 M, we can rearrange the equation to solve for [OH-]:
[OH-] = Kw / [H3O+]
[OH-] = 1.0 x 10^-14 / 5.9 x 10^-4
[OH-] = 1.69 x 10^-11 M
Therefore, the [OH-] in the cabernet sauvignon wine is 1.69 x 10^-11 M.