if the [H] of a solution at 25degrees is 1.7*10^-4 mol/L, calculate the [OH] in mol/L, the PH, and the pOH?
pH = -log(H^+)
pOH = -log(OH^-)
pH + pOH = pKw = 14 and
(H^+)(OH^-) = Kw.
One or more of the four above will calculate all you ever need to know abouat H^+), (OH^-), pH and pOH.
THANK YOU :)
To calculate the [OH] in mol/L, we can use the equation:
[H] * [OH] = 1.0 * 10^-14 (for water at 25 degrees Celsius)
Given that the [H] of the solution is 1.7 * 10^-4 mol/L, we can substitute the value into the equation:
(1.7 * 10^-4) * [OH] = 1.0 * 10^-14
Now, we can rearrange the equation to solve for [OH]:
[OH] = (1.0 * 10^-14) / (1.7 * 10^-4)
Simplifying this, we get:
[OH] = 5.882 * 10^-11 mol/L
Thus, the [OH] in mol/L is approximately 5.882 * 10^-11 mol/L.
To calculate the pH of the solution, we can use the equation:
pH = -log[H]
Given that the [H] of the solution is 1.7 * 10^-4 mol/L, we can substitute the value into the equation:
pH = -log(1.7 * 10^-4)
Using a scientific calculator, the pH is approximately 3.77.
To calculate the pOH of the solution, we can use the equation:
pOH = -log[OH]
Given that the [OH] of the solution is 5.882 * 10^-11 mol/L, we can substitute the value into the equation:
pOH = -log(5.882 * 10^-11)
Using a scientific calculator, the pOH is approximately 10.23.
Therefore, the [OH] in mol/L is 5.882 * 10^-11 mol/L, the pH is 3.77, and the pOH is 10.23.