Calculate the pH of a 0.0065 M solution of NaOH. Report answer to two decimal places.

Answer 1

We are given a solution of NaOH (a strong base) that is 0.0065 M. Because NaOH dissociates into Na⁺ and OH^-, the [OH^-] will also be 0.0065 M

If we recall that pH + pOH = 14, then if we find pOH, we can easily find pH. The pOH will be the negative log of the [OH-], so that = -log 0.0065 = 2.187

pH = 14 - pOH = 14 - 2.187 = 11.8

Another way to do this same problem is to first find the [H+] and then take the negative log to get pH. If we recall Kw = 1x10^-14 = [H+][OH-] we can solve for [H+]....

[H+][OH-] = 1x10^-14

[H+] = 1x10^-14 / 0.0065 = 1.538x10^-12

pH = -log 1.538x10^-12

pH = 11.8