If the OH- ion concentration in an aqueous solution at 25.0 oC is measured as 6.0 x 10‑4 M, then the pH of the solution is
Two steps.
Solve for pOH = -Log (OH^-) = -log(6.0 x 10‑4 M) = ?
Then pH + pOH = pKw = 14. You know pOH and pKw so solve for pH.
Post your work if you get stuck.
To determine the pH of a solution, we need to know the concentration of the hydrogen ion concentration (H+). However, in this question, the OH- ion concentration (hydroxide ion) is given.
The equation for the autoionization of water is:
2H2O ⇌ H3O+ + OH-
In a neutral solution, the concentration of H+ ions and OH- ions are equal, which means that [H+] = [OH-]. However, in this case, [OH-] is given as 6.0 x 10^-4 M. So, to find the concentration of H+, we can use the fact that [H+] × [OH-] = 1 x 10^-14 at 25°C.
Let's calculate the concentration of H+:
[H+] x 6.0 x 10^-4 M = 1 x 10^-14 M^2
[H+] = (1 x 10^-14 M^2) / (6.0 x 10^-4 M)
[H+] ≈ 1.67 x 10^-11 M
Now that we know the concentration of H+ ions, we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(1.67 x 10^-11) ≈ 10.78
Therefore, the pH of the solution is approximately 10.78.