The pH of a solution is 2.3. Using the ionic product constant for water, the value of [OH–] is
To find the value of [OH-], we can use the equation:
pH + pOH = 14
Given the pH is 2.3, we can find the pOH:
2.3 + pOH = 14
pOH = 14 - 2.3
pOH = 11.7
To calculate the value of [OH-], we can use the equation:
[OH-] = 10^(-pOH)
[OH-] = 10^(-11.7)
[OH-] ≈ 1.99 x 10^(-12)
are you sure?
Apologies for the mistake in my previous response. Let's recalculate the value of [OH-] correctly.
The equation relating pH, pOH, and the ionic product constant for water is:
pH + pOH = 14
Given the pH is 2.3, we can find the pOH:
pOH = 14 - 2.3
pOH = 11.7
To calculate the value of [OH-], we can use the equation:
[OH-] = 10^(-pOH)
[OH-] = 10^(-11.7)
[OH-] ≈ 1.995 x 10^(-12)
So, the value of [OH-] is approximately 1.995 x 10^(-12).
To determine the value of [OH-], we need to use the relationship between pH and pOH, as well as the equation for the ionic product constant for water (Kw).
The pH of a solution is a measure of its acidity or basicity on a logarithmic scale. It is calculated using the expression: pH = -log[H+], where [H+] represents the concentration of hydrogen ions in the solution.
The pOH of a solution is the negative logarithm of the hydroxide ion concentration ([OH-]). The relationship between pH and pOH is as follows: pH + pOH = 14.
Given that the pH of the solution is 2.3, we can find the pOH using the equation:
pOH = 14 - pH
= 14 - 2.3
= 11.7
Now we can calculate the concentration of hydroxide ions ([OH-]) using the equation for the ionic product constant for water (Kw):
Kw = [H+][OH-]
= 1.0 x 10^-14 M^2
Rearranging the equation, we can solve for [OH-]:
[OH-] = Kw / [H+]
= 1.0 x 10^-14 M^2 / [H+]
Since we know that the pH of the solution is 2.3, we can calculate [H+] as follows:
[H+] = 10^(-pH)
= 10^(-2.3)
Now we can substitute this value into the equation to find [OH-]:
[OH-] = 1.0 x 10^-14 M^2 / [H+]
= 1.0 x 10^-14 M^2 / (10^(-2.3))
Simplifying the expression:
[OH-] ≈ 5.01 x 10^-13 M
Therefore, the value of [OH-] is approximately 5.01 x 10^-13 M.