At 40 c,the value of Kw is 2.92
10-14.calculate the [H+] and [OH-] of pure water at 40 c.
This question has been answered before.
"the value of Kw is 2.92
10-14" ?? units are missing
Given that [H+][OH-]=Kw and that
[H+] = [OH-] for pure water
then
[H+][OH-]=[H+][H+]=2.92 x 10^-14
[H+]^2= 2.92 x 10^-14
[H+] = sqrt(2.92 x 10^-14)
but this is missing the units.
To calculate the [H+] and [OH-] concentrations of pure water at 40°C, we can use the equation for the autoionization of water:
Kw = [H+][OH-]
Given that the value of Kw at 40°C is 2.92 * 10^(-14), we can divide this value equally between [H+] and [OH-] because pure water is neutral, meaning [H+] is equal to [OH-].
Let's divide Kw by 2 to get the concentration of either [H+] or [OH-].
Kw/2 = [H+] = [OH-]
Now, we can substitute the value of Kw/2 into the equation:
[H+] = [OH-] = 2.92 * 10^(-14) / 2
[H+] = [OH-] = 1.46 * 10^(-14)
Therefore, the [H+] and [OH-] concentrations of pure water at 40°C are both approximately 1.46 * 10^(-14) M.