Calculate the [OH-] for a solution in which [H+] is 1000 times greater than [OH-]?
and will the final result be acidic, basic, or neutral?
log 1000 is 3
so pOH-3+POH=14
pOH=17/2=8.5
PH=5.5
To calculate the [OH-] for a solution in which [H+] is 1000 times greater than [OH-], we need to use the relationship between [H+] and [OH-] ions in water, which is given by the equation:
Kw = [H+][OH-]
Kw is the ion product of water and is equal to 1.0 x 10^-14 at 25°C.
In this case, we are told that [H+] is 1000 times greater than [OH-]. Let's represent the concentration of [OH-] as x, so the concentration of [H+] would be 1000x.
Substituting these values into the equation for Kw, we have:
(1000x)(x) = 1.0 x 10^-14
Simplifying the equation:
1000x^2 = 1.0 x 10^-14
Now, we can solve for x:
x^2 = (1.0 x 10^-14) / 1000
x^2 = 1.0 x 10^-17
Taking the square root of both sides:
x = √(1.0 x 10^-17)
Calculating the square root of 1.0 x 10^-17 gives us x = 1.0 x 10^-9.
So, the concentration of [OH-] is 1.0 x 10^-9 M.
Now, to determine the acidity or alkalinity of the solution, we compare the concentration of [H+] and [OH-].
Since [H+] is 1000 times greater than [OH-], it means [H+] is 1.0 x 10^-6 M.
Based on the concentration, the solution is acidic because the concentration of [H+] is greater than the concentration of [OH-].