A solution with a pH of 7 has a [OH1-] of
pH + pOH = pKw = 14. You know pKw and pH, solve for pOH.
Then pOH = -log(OH^-)
Solve for OH^-
7M. Is that correct?
no. pOH is 7 (but not molar).
pH + pOH = 14
pH = 7; therefore, pOH = 7
Now do pOH = -log(OH^-) and solve for OH^-
1 x 10-7 M?? Is that correct?
To determine the [OH-] concentration of a solution with a pH of 7, we can use the equation:
pOH = 14 - pH
Since the pH of the solution is 7, we can substitute this value into the equation:
pOH = 14 - 7
pOH = 7
Next, we can convert the pOH value to the [OH-] concentration:
[OH-] = 10^(-pOH)
Substituting the value of pOH into the equation:
[OH-] = 10^(-7)
Using a calculator, we can find that 10^(-7) is equal to 0.0000001.
Therefore, a solution with a pH of 7 has an [OH-] concentration of 0.0000001 (or 1 x 10^(-7)) moles per liter.