A solution is found to contain 1 x 10-4 OH− ions, determine the pH of this solution.
I did this problem and got 4 but it is not going in as correct.
-log(1*10^-4)=4
It is; however, your calculation gives you pOH. If you want pH, use
pH + pOH = pKw = 14 and solve for pH.
Or you can use (H^+)(OH^-) = Kw = 1E-14
pH = -log(H^+)
pOH = - log(OH^-)
I did it and got 10 and it is saying that its incorrect
pOH = 4
pH = 10
Or
(H^+) = 1E-14/1E-4 = 1E-10
pH = 10
The data base is wrong or you posted the problem incorrectly. The pH is 10 for the above problem.
To determine the pH of a solution, you need to find the negative logarithm (base 10) of the concentration of hydrogen ions, denoted as [H+]. In this case, you were given the concentration of hydroxide ions, denoted as [OH-]. Remember that the concentrations of [H+] and [OH-] in water are related by the Kw constant, which is 1 x 10^-14 at 25°C.
To find [H+], you can use the fact that [H+][OH-] = Kw. Since [OH-] = 1 x 10^-4, you can substitute it into the equation and solve for [H+]:
[H+][OH-] = Kw
[H+](1 x 10^-4) = 1 x 10^-14
[H+] = (1 x 10^-14)/(1 x 10^-4)
[H+] = 1 x 10^-10
Now that you have the concentration of [H+], you can find the pH by taking the negative logarithm (base 10) of [H+]:
pH = -log([H+])
pH = -log(1 x 10^-10)
pH = -(-10)
pH = 10
Therefore, the pH of the solution is 10.