hey guys I posted this awhile ago im not sure if anyone has read it or whatever but here it is again find the PH of a .00580 M solution of the strong base KOH.
is it -log(.00580)= 2.24
pOH = -log[OH-]
pH + pOH = 14.00
Therefore,
pOH = -log(.00580)
pH + 2.24 = 14
pH = 11.76
Remember, a pH above 7 is a base, and below is an acid.
Hope this helps.
right. Your 2.24 is ok. But you must remember that is pOH.
Then pH + pOH = pKw = 14.
To find the pH of a solution of KOH, you can follow these steps:
Step 1: Calculate pOH
pOH is determined using the formula pOH = -log[OH-]. In this case, the concentration of OH- in the solution is 0.00580 M. So, pOH = -log(0.00580) = 2.24.
Step 2: Calculate pH
The pH and pOH of a solution add up to 14. So, pH + pOH = 14. Since you already found the pOH to be 2.24 in Step 1, you can solve for pH:
pH = 14 - 2.24 = 11.76.
Therefore, the pH of the 0.00580 M solution of KOH is 11.76.
Remember that a pH above 7 indicates a base, while a pH below 7 indicates an acid. In this case, since the pH is 11.76, the solution is strongly basic.