calculate the PH of [OH-]

= 4.5×10−3 M

Please show how you work this problem.

pOH = -log(OH^-)

pOH = -log(4.5E-3)
Then pH + pOH = pKw = 14 so
pH = 14-pOH
You should end up with 11.65

thank you

To calculate the pH of a solution, we need to first find the concentration of H+ ions in the solution. However, in this case, we are given the concentration of OH- ions ([OH-] = 4.5×10−3 M) instead.

To relate the concentrations of H+ and OH- ions, we can use the equation for the ion product of water:

Kw = [H+][OH-]

At 25°C, the value of Kw is 1.0×10^-14. Since we know the concentration of OH- ions ([OH-] = 4.5×10−3 M), we can rearrange the equation to solve for [H+]:

[H+] = Kw / [OH-]
[H+] = (1.0×10^-14) / (4.5×10−3 M)
[H+] ≈ 2.22×10^-12 M

Now that we have the concentration of H+ ions, we can calculate the pH using the formula:

pH = -log[H+]
pH = -log(2.22×10^-12)
pH ≈ 11.65

Therefore, the pH of a solution with a concentration of [OH-] = 4.5×10−3 M is approximately 11.65.