What is the [OH¯] for a solution at 25 $\degree$C that has pH = 4.29?

Question

Quizplus · Accepted Answer

From the pH of 4.29, we can use the equation:

pH = -log[H+]

4.29 = -log[H+]

[H+] = 1.9 x 10^-5 M

Since water autoionizes as H2O ⇌ H+ + OH-, we can use the ion product constant of water (Kw = [H+][OH-]) to find [OH-]:

Kw = [H+][OH-]

1.0 x 10^-14 = (1.9 x 10^-5)[OH-]

[OH-] = 5.26 x 10^-10 M, which is equivalent to 1.9 x 10^-4 M or 1.9 x 10^-10 mol/L or M.

Therefore, the best answer is C.

What Is the [OH¯] for a Solution at 25 °\degree° C That Has PH = 4